Re: [R] transform a list of arrays to tibble
In fact, I realized this is not recommended to give names to rows. A
better approach is to add a column with all names as row. The
following does the job:
asset.stats <- as_tibble_col(unlist(my.ret.lst), column_name =
'Annualized_return')
asset.stats <- rownames_to_column(asset.stats, var = 'Assets')
asset.stats$Assets <- names(my.ret.lst)
asset.stats <- structure(list(Assets = c("BTCUSDT", "ETHUSDT",
"TRXUSDT"), Annualized_return = c(BTCUSDT = 15.36,
ETHUSDT = 4.06, TRXUSDT = 10.9)), row.names = c(NA, -3L), class = c("tbl_df",
"tbl", "data.frame"))
On Tue, Oct 17, 2023 at 3:33 PM wrote:
>
> Arnaud,
>
>
> Short answer may be that the tibble data structure will not be supporting row
> names and you may want to simply save those names in an additional column or
> externally.
>
> My first thought was to simply save the names you need and then put them back
> on the tibble. In your code, something like this:
>
> save.names <- names(my.ret.lst)
> result.tib <- as_tibble_col(unlist(my.ret.lst), column_name = 'return')
> rownames(result.tib) <- save.names
>
> Unfortunately, I got an error message:
>
> > save.names
> [1] "BTCUSDT" "ETHUSDT" "TRXUSDT"
> > rownames(result.tib) <- save.names
> Warning message:
> Setting row names on a tibble is deprecated.
> Error in exists(cacheKey, where = .rs.WorkingDataEnv, inherits = FALSE) :
> invalid first argument
> Error in assign(cacheKey, frame, .rs.CachedDataEnv) :
> attempt to use zero-length variable name
>
> If a tibble deprecates row names, it may not be the ideal storage for you. A
> plain data.frame works:
>
> > result.df <- as.data.frame(result.tib)
> > rownames(result.df) <- save.names
> > result.df
> return
> BTCUSDT 15.36
> ETHUSDT 4.06
> TRXUSDT 10.90
>
> Trying to convert it to a tibble, as anticipated, is not working for me:
>
> > as.tibble(result.df)
> # A tibble: 3 × 1
> return
>
> 1 15.4
> 2 4.06
> 3 10.9
> Warning message:
> `as.tibble()` was deprecated in tibble 2.0.0.
> ℹ Please use `as_tibble()` instead.
> ℹ The signature and semantics have changed, see `?as_tibble`.
> This warning is displayed once every 8 hours.
> Call `lifecycle::last_lifecycle_warnings()` to see where this
> warning was generated.
>
> You can, instead, create a matrix and assign the row and column names you
> save or create:
>
> result.mat <- matrix(my.ret.lst)
> colnames(result.mat) <- c("return")
> rownames(result.mat) <- save.names
>
> > result.mat
> return
> BTCUSDT 15.36
> ETHUSDT 4.06
> TRXUSDT 10.9
>
> But saving a matrix to reuse has other considerations.
>
> So, if I may make a suggestion, if you really want a tibble that allows you
> to know what each row is for, consider one of many methods for saving the
> previous row names as a new column. I used that to take the data.frame
> version I made above and got:
>
> > temp <- as_tibble(result.df, rownames="rows")
> > temp
> # A tibble: 3 × 2
> rowsreturn
>
> 1 BTCUSDT 15.4
> 2 ETHUSDT 4.06
> 3 TRXUSDT 10.9
>
> Note the above uses as_tibble with an underscore, but many other ways to make
> a column exist.
>
>
> -Original Message-
> From: R-help On Behalf Of arnaud gaboury
> Sent: Tuesday, October 17, 2023 4:30 AM
> To: r-help
> Subject: [R] transform a list of arrays to tibble
>
> I work with a list of crypto assets daily closing prices in a xts
> class. Here is a limited example:
>
> asset.xts.lst <- list(BTCUSDT = structure(c(26759.63, 26862, 26852.48,
> 27154.15,
> 27973.45), dim = c(5L, 1L), index = structure(c(1697068800, 1697155200,
> 1697241600, 1697328000, 1697414400), tzone = "UTC", tclass = "Date"),
> class = c("xts",
> "zoo")), ETHUSDT = structure(c(1539.61, 1552.16, 1554.94, 1557.77,
> 1579.73), dim = c(5L, 1L), index = structure(c(1697068800, 1697155200,
> 1697241600, 1697328000, 1697414400), tzone = "UTC", tclass = "Date"),
> class = c("xts",
> "zoo")), TRXUSDT = structure(c(0.08481, 0.08549, 0.08501, 0.08667,
> 0.08821), dim = c(5L, 1L), index = structure(c(1697068800, 1697155200,
> 1697241600, 1697328000, 1697414400), tzone = "UTC", tclass = "Date"),
> class = c("xts",
> "zoo")))
>
> I will compute some function from PerformanceAnalytics package and
> write all results in a tibble. Let's apply a first function,
> Return.annualized() (at first I computed returns from daily prices). I
> hav
[R] transform a list of arrays to tibble
I work with a list of crypto assets daily closing prices in a xts
class. Here is a limited example:
asset.xts.lst <- list(BTCUSDT = structure(c(26759.63, 26862, 26852.48, 27154.15,
27973.45), dim = c(5L, 1L), index = structure(c(1697068800, 1697155200,
1697241600, 1697328000, 1697414400), tzone = "UTC", tclass = "Date"),
class = c("xts",
"zoo")), ETHUSDT = structure(c(1539.61, 1552.16, 1554.94, 1557.77,
1579.73), dim = c(5L, 1L), index = structure(c(1697068800, 1697155200,
1697241600, 1697328000, 1697414400), tzone = "UTC", tclass = "Date"),
class = c("xts",
"zoo")), TRXUSDT = structure(c(0.08481, 0.08549, 0.08501, 0.08667,
0.08821), dim = c(5L, 1L), index = structure(c(1697068800, 1697155200,
1697241600, 1697328000, 1697414400), tzone = "UTC", tclass = "Date"),
class = c("xts",
"zoo")))
I will compute some function from PerformanceAnalytics package and
write all results in a tibble. Let's apply a first function,
Return.annualized() (at first I computed returns from daily prices). I
have now a list of arrays named my.ret.lst:
my.ret.lst <- list(BTCUSDT = structure(15.36, dim = c(1L, 1L), dimnames = list(
"Annualized Return", NULL)), ETHUSDT = structure(4.06, dim = c(1L,
1L), dimnames = list("Annualized Return", NULL)), TRXUSDT =
structure(10.9, dim = c(1L,
1L), dimnames = list("Annualized Return", NULL)))
Now I can't find how to build a tibble in a specific format (asset
names as row names and observations as column names) .
I can of course run:
> mytb <- as_tibble(unlist(my.ret.lst)
but I loose row and column names.
> as_tibble_col(unlist(my.ret.lst), column_name = 'return')
will give me the wanted column name but row names (in my case asset
names) are missing.
Thank you for help
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Re: [R] replace character by numeric value
On Thu, Sep 28, 2023 at 8:18 AM Ivan Calandra wrote:
>
> Dear Arnaud,
>
> I don't quite unterstand why you have imbricated ifelse() statements. Do
> you have more that BUY (1) and SELL (-1)? If not, why not simply:
> mynewdf2 <- mydf2 |> dplyr::mutate(side = ifelse(side == 'BUY', 1, -1))
Yes it works indeed.
I found another solution :
df <- df |> mutate(side = as.numeric(ifelse(side == 'BUY', 1,
ifelse(side == 'SELL', -1, side
but yours is much simpler. and I can use dplyr::if_else() or
fifelse(). which looks safer and faster.
>
> That would solve the problem. I'm not quite sure exactly what happens,
> but this is probably related to the intermediary result after the first
> ifelse(), where characters and numeric are mixed. But conversion to
> numeric works properly, so I'm not sure what you meant:
> as.numeric(mynewdf2$side)
>
> More generally, why are you trying to convert to 1 and -1?
I am working on a crypto asset portfolio and want to compute profit &
loss. To achieve it, I need to convert (BUY * quantity) to quantity
and
(SELL * quantity) to -quantity.
Thank you for your answer.
Why not use
> factors? Are you trying to test contrasts maybe? I would be surprised if
> the function for the statistical test you are trying to use does not
> deal with that already on its own.
>
> HTH,
> Ivan
>
>
> On 27/09/2023 13:01, arnaud gaboury wrote:
> > I have two data.frames:
> >
> > mydf1 <- structure(list(symbol = "ETHUSDT", cummulative_quote_qty =
> > 1999.9122, side = "BUY", time = structure(1695656875.805, tzone = "", class
> > = c("POSIXct", "POSIXt"))), row.names = c(NA, -1L), class = c("data.table",
> > "data.frame"))
> >
> > mydf2 <- structure(list(symbol = c("ETHUSDT", "ETHUSDT", "ETHUSDT"),
> > cummulative_quote_qty = c(1999.119408,
> > 0, 2999.890985), side = c("SELL", "BUY", "BUY"), time =
> > structure(c(1695712848.487,
> > 1695744226.993, 1695744509.082), class = c("POSIXct", "POSIXt"
> > ), tzone = "")), row.names = c(NA, -3L), class = c("data.table",
> > "data.frame"))
> >
> > I use this line to replace 'BUY' by numeric 1 and 'SELL' by numeric -1 in
> > mydf1 and mydf2:
> > mynewdf <- mydf |> dplyr::mutate(side = ifelse(side == 'BUY', 1,
> > ifelse(side == 'SELL', -1, side)))
> >
> > This does the job but I am left with an issue: 1 and -1 are characters for
> > mynewdf2 when it is numeric for mynewdf1. The result I am expecting is
> > getting numeric values.
> > I can't solve this issue (using as.numeric(1) doesn't work) and don't
> > understand why I am left with num for mynewdf1 and characters for mynewdf2.
> >
> >> mynewdf1 <- mydf1 |> dplyr::mutate(side = ifelse(side == 'BUY', 1,
> > ifelse(side == 'SELL', -1, side)))
> >> str(mynewdf1)
> > Classes ‘data.table’ and 'data.frame': 1 obs. of 4 variables:
> > $ symbol : chr "ETHUSDT"
> > $ cummulative_quote_qty: num 2000
> > $ side : num 1 <<<--
> > $ time : POSIXct, format: "2023-09-25 17:47:55"
> > - attr(*, ".internal.selfref")=
> >
> >> mynewdf2 <- mydf2 |> dplyr::mutate(side = ifelse(side == 'BUY', 1,
> > ifelse(side == 'SELL', -1, side)))
> >> str(mynewdf2)
> > Classes ‘data.table’ and 'data.frame': 3 obs. of 4 variables:
> > $ symbol : chr "ETHUSDT" "ETHUSDT" "ETHUSDT"
> > $ cummulative_quote_qty: num 1999 0 3000
> > $ side : chr "-1" "1" "1" <<<--
> > $ time : POSIXct, format: "2023-09-26 09:20:48"
> > "2023-09-26 18:03:46" "2023-09-26 18:08:29"
> > - attr(*, ".internal.selfref")=
> >
> > Thank you for help
> >
> > [[alternative HTML version deleted]]
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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> and provide commented, minimal, self-contained, reproducible code.
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[R] replace character by numeric value
I have two data.frames:
mydf1 <- structure(list(symbol = "ETHUSDT", cummulative_quote_qty =
1999.9122, side = "BUY", time = structure(1695656875.805, tzone = "", class
= c("POSIXct", "POSIXt"))), row.names = c(NA, -1L), class = c("data.table",
"data.frame"))
mydf2 <- structure(list(symbol = c("ETHUSDT", "ETHUSDT", "ETHUSDT"),
cummulative_quote_qty = c(1999.119408,
0, 2999.890985), side = c("SELL", "BUY", "BUY"), time =
structure(c(1695712848.487,
1695744226.993, 1695744509.082), class = c("POSIXct", "POSIXt"
), tzone = "")), row.names = c(NA, -3L), class = c("data.table",
"data.frame"))
I use this line to replace 'BUY' by numeric 1 and 'SELL' by numeric -1 in
mydf1 and mydf2:
mynewdf <- mydf |> dplyr::mutate(side = ifelse(side == 'BUY', 1,
ifelse(side == 'SELL', -1, side)))
This does the job but I am left with an issue: 1 and -1 are characters for
mynewdf2 when it is numeric for mynewdf1. The result I am expecting is
getting numeric values.
I can't solve this issue (using as.numeric(1) doesn't work) and don't
understand why I am left with num for mynewdf1 and characters for mynewdf2.
> mynewdf1 <- mydf1 |> dplyr::mutate(side = ifelse(side == 'BUY', 1,
ifelse(side == 'SELL', -1, side)))
> str(mynewdf1)
Classes ‘data.table’ and 'data.frame': 1 obs. of 4 variables:
$ symbol : chr "ETHUSDT"
$ cummulative_quote_qty: num 2000
$ side : num 1 <<<--
$ time : POSIXct, format: "2023-09-25 17:47:55"
- attr(*, ".internal.selfref")=
> mynewdf2 <- mydf2 |> dplyr::mutate(side = ifelse(side == 'BUY', 1,
ifelse(side == 'SELL', -1, side)))
> str(mynewdf2)
Classes ‘data.table’ and 'data.frame': 3 obs. of 4 variables:
$ symbol : chr "ETHUSDT" "ETHUSDT" "ETHUSDT"
$ cummulative_quote_qty: num 1999 0 3000
$ side : chr "-1" "1" "1" <<<--
$ time : POSIXct, format: "2023-09-26 09:20:48"
"2023-09-26 18:03:46" "2023-09-26 18:08:29"
- attr(*, ".internal.selfref")=
Thank you for help
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[R] apply a function to a list of lists
I have a list of 9 lists called my.list. Each one of these 9 lists is
itself a list of 6 data.frames. Most of these data.frames have 0 rows and 0
columns.
I want to return all data.frames from the list with row numbers different
from 0.
I first created the following function:
non_empty_df <- function(l) {
lapply(l, function(df) df[sapply(df, function(df) nrow(df) !=0)])
}
If I test this way: non_empty_df(my.list[1]) it does the job. It will
return the data.frame from the first list of my_list with rows.
Now I can't find a way to iterate the above function to all the 9 nested
lists.
Thank you for help
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[R] getting historical data from cryptocurrency exchange
I need to download basic OHLCV (Open, High, Low, Close, Volume) in a hourly format from various cryptocurrency exchanges. There is the crypto package[0] but it has been removed from CRAN. Then there is the coinmarketcapr[1] package on CRAN, but it is limited to a paid service, coinmarketcap, which sell data. Browsing the web, I mainly found Python scripts to fetch the data, which I would like to avoid. Can anyone point me to any github or website with material about crypto data fetching with R? Thank you [0]https://cran.r-project.org/web/packages/crypto/index.html [1]https://github.com/amrrs/coinmarketcapr __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] change col types of a df/tbl_df
Here is a sample of my data frame, obtained with read_csv2 from readr package.
myDf <- structure(list(X15 = c("30.09.2015", "05.10.2015", "30.09.2015",
"29.09.2015", "10.10.2015"), X16 = c("02.10.2015", "06.10.2015",
"01.10.2015", "01.10.2015", "13.10.2015"), X17 = c("Grains",
"Grains", "Grains", "Grains", "Grains"), X18 = c("Soyabeans",
"Soyabeans", "Soyabeans", "Soyabeans", "Soyabeans"), X19 = c("20,000",
"20,000", "20,000", "29,930", "26,000")), .Names = c("X15", "X16",
"X17", "X18", "X19"), class = c("tbl_df", "data.frame"), row.names = c(NA,
-5L))
gabx@hortensia [R] str(myDf)
Classes ‘tbl_df’ and 'data.frame': 5 obs. of 5 variables:
$ X15: chr "30.09.2015" "05.10.2015" "30.09.2015" "29.09.2015" ...
$ X16: chr "02.10.2015" "06.10.2015" "01.10.2015" "01.10.2015" ...
$ X17: chr "Grains" "Grains" "Grains" "Grains" ...
$ X18: chr "Soyabeans" "Soyabeans" "Soyabeans" "Soyabeans" ...
$ X19: chr "20,000" "20,000" "20,000" "29,930" ...
I want to change date to date class and numbers (X19) to numeric, and
keep the class of my object.
This code works:
myDf$X19 <- as.numeric(gsub(",", "", myDf$X19))
myDf$X15 <- as.Date(myDf$X15, format = "%d.%m.%Y"))
myDf$X16 <- as.Date(myDf$X16, format = "%d.%m.%Y"))
Now, as I have more than 5 columns, this can be fastidious and slowing
code (?), even if I can group by type. Columns are only types of char,
num and Date, so it could be OK.
I tried with lapply for the Date columns. It works BUT will place NA
in any columns with numbers as characters.
The reuslt will be this for X19: num NA NA NA NA NA NA NA NA NA NA ..
How can I target my goal with something else than lapply or writing a
line for each type ?
Thank you for hints.
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Re: [R] change col types of a df/tbl_df
On Thu, Dec 10, 2015 at 12:54 PM, Duncan Murdoch <murdoch.dun...@gmail.com>
wrote:
> On 10/12/2015 6:12 AM, arnaud gaboury wrote:
>
>> Here is a sample of my data frame, obtained with read_csv2 from readr
>> package.
>>
>> myDf <- structure(list(X15 = c("30.09.2015", "05.10.2015", "30.09.2015",
>>
>> "29.09.2015", "10.10.2015"), X16 = c("02.10.2015", "06.10.2015",
>> "01.10.2015", "01.10.2015", "13.10.2015"), X17 = c("Grains",
>> "Grains", "Grains", "Grains", "Grains"), X18 = c("Soyabeans",
>> "Soyabeans", "Soyabeans", "Soyabeans", "Soyabeans"), X19 = c("20,000",
>> "20,000", "20,000", "29,930", "26,000")), .Names = c("X15", "X16",
>> "X17", "X18", "X19"), class = c("tbl_df", "data.frame"), row.names = c(NA,
>> -5L))
>>
>> gabx@hortensia [R] str(myDf)
>> Classes ‘tbl_df’ and 'data.frame': 5 obs. of 5 variables:
>> $ X15: chr "30.09.2015" "05.10.2015" "30.09.2015" "29.09.2015" ...
>> $ X16: chr "02.10.2015" "06.10.2015" "01.10.2015" "01.10.2015" ...
>> $ X17: chr "Grains" "Grains" "Grains" "Grains" ...
>> $ X18: chr "Soyabeans" "Soyabeans" "Soyabeans" "Soyabeans" ...
>> $ X19: chr "20,000" "20,000" "20,000" "29,930" ...
>>
>> I want to change date to date class and numbers (X19) to numeric, and
>> keep the class of my object.
>>
>> This code works:
>>
>> myDf$X19 <- as.numeric(gsub(",", "", myDf$X19))
>> myDf$X15 <- as.Date(myDf$X15, format = "%d.%m.%Y"))
>> myDf$X16 <- as.Date(myDf$X16, format = "%d.%m.%Y"))
>>
>> Now, as I have more than 5 columns, this can be fastidious and slowing
>> code (?), even if I can group by type. Columns are only types of char,
>> num and Date, so it could be OK.
>>
>> I tried with lapply for the Date columns. It works BUT will place NA
>> in any columns with numbers as characters.
>> The reuslt will be this for X19: num NA NA NA NA NA NA NA NA NA NA ..
>>
>> How can I target my goal with something else than lapply or writing a
>> line for each type ?
>>
>
> I don't see how a function could reliably detect the types,
In fact, I only have 25 columns, so it is not difficult to list them in the
3 types: char, num and Date. No need of a function thus.
> but it might be good enough to use a regular expression, possibly just on
> the first line of the result. Once you've identified columns, e.g.
>
> numcols <- 19
> datecols <- c(15:16)
>
> etc, you can use lapply:
>
> myDf[,numcols] <- lapply(myDf[, numcools, drop=FALSE], function(x)
> as.numeric(gsub(",", "", x)))
>
> You can simplify myDf[,numcols] to myDf[numcols] if you want, but I think
> it makes it less clear.
Thank you.
>
>
> Duncan Murdoch
>
>
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Re: [R] change col types of a df/tbl_df
On Thu, Dec 10, 2015 at 1:47 PM, Giorgio Garziano <
giorgio.garzi...@ericsson.com> wrote:
> my_convert <- function(col) {
> v <- grep("[0-9]{2}.[0-9]{2}.[0-9]{4}", col);
> w <- grep("[0-9]+,[0-9]+", col)
> col2 <- col
> if (length(v) == length(col)){
> col2 <- as.Date(col, format="%d.%m.%y")
> } else if (length(w) == length(col)) {
> col2 <- as.numeric(gsub(",", "", col))
> }
> col2
> }
>
> myDf <- as.data.frame(lapply(myDf, my_convert), stringsAsFactors = FALSE)
>
> Sounds good. I will give a try,
Thank you
> --
> GG
>
>
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>
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[R] R environment variables
I was doing some cleaning ony my linux machine and, among others, I
try to clean my R environment variables accordingly [0] and [1].
I am not really sure how to declare in a clean manner these startup variables.
Here is my setup:
1- my home folder
-$ ls ~/.config/R
env/ helper/ Renviron Rprofile.r
2- system R install
/usr/lib/R
3- user library
/developement/language/r/library
4- system configuration files: (NOTE: /usr/lib/R/etc/ are symlnks to
the below files)
$ ls /etc/R/
javaconf ldpaths Makeconf Renviron repositories
Most important R environment variables for my system:
R_HOME, R_LIBS, R_LIBS_SITE,R_LIBS_USER,R_ENVIRON,R_ENVIRON_USER,R_PROFILE_USER
As far I can understand, in my setup, these above variables would be:
R_LIBS_SITE=/usr/lib/R/library
R_LIBS_USER=/developement/language/r/library
R_ENVIRON=/usr/lib/R/etc/Renviron
R_ENVIRON_USER=~/.config/R/Renviron
R_PROFILE_USER=~/.config/R/Rprofile.r
I have a doubt about two variables:
R_HOME=/usr/lib/R/ right ? Is there any need to export it in my environment?
R_LIBS=${R_LIBS_USER}:${R_LIBS_SITE} right ?
Do I need to export all these mentioned variables in my user
environment? Until now, I jsut export R_ENVIRON_USER,R_PROFILE_USER
via my /etc/profile file.
$ cat ~/.config/r/Renviron
R_HOME=/usr/lib/R
R_HOME_USER=/developement/language/r
R_LIBS_USER=${R_HOME_USER}/library
R_LIBS=${R_LIBS_USER}:${R_HOME}/library
R_HISTFILE=/developement/language/r/R.Rhistory
R_HELPER=/home/gabx/.config/r/helper
R_HISTSIZE=5000
Are the above variables correctly set ?
On a R session, Sys.getenv() returns correctly everything, except
R_LIBS_SITE which is empty. Why? Do I need to export it somewhere ?
R_LIBS/developement/language/r/library:/usr/lib/R/library Is
this the correct way for R to see R_LIBS? Then, when I upgrade
packages, do I need to upgrade separatly site library and user
library, or just one ligne upgardaing jusr R_LIBS?
Thank you for any advice on my current variable declaration setup.
[0]http://stat.ethz.ch/R-manual/R-devel/library/base/html/EnvVar.html
[1]http://stat.ethz.ch/R-manual/R-devel/library/base/html/Startup.html
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[R] Build R with optimized BLAS library
I want to build R with an optimized BLAS library. My OS: Fedora 22 Hardware: CPU op-mode(s): 32-bit, 64-bit Byte Order: Little Endian CPU(s): 8 Thread(s) per core: 2 Vendor ID: GenuineIntel Model name: Intel(R) Core(TM) i7-2600K CPU @ 3.40GHz I am a little confused when it comes to choose a method and would like to hear your experiences. If I am right, I have 3 possibilities: - OpenBLAS: opensource and free, but I came across some posts describing seg faults issues and bugs. These posts are 2 years old and I wonder if it is still the case. - ATLAS: can't see any reason to not use it - Intel MKL: this is part of Intel Parallel Studio and is a paid software. Now, there is the MKL package distributed by Revolutionanalytics, but I am not certain how this can be distributed for free. Is there any kind of difference? In case of use of this package, do I need to install RRO or can I just build R from GNU against these libraries? Thank you for advices. -- google.com/+arnaudgabourygabx __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Build R with optimized BLAS library
On Sat, Aug 22, 2015, 5:12 PM Jeff Newmiller jdnew...@dcn.davis.ca.us wrote: Questions about compiling generally belong on R-devel. Ok. Sorrx fpr the noise --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On August 22, 2015 7:51:39 AM PDT, arnaud gaboury arnaud.gabo...@gmail.com wrote: I want to build R with an optimized BLAS library. My OS: Fedora 22 Hardware: CPU op-mode(s): 32-bit, 64-bit Byte Order: Little Endian CPU(s): 8 Thread(s) per core: 2 Vendor ID: GenuineIntel Model name: Intel(R) Core(TM) i7-2600K CPU @ 3.40GHz I am a little confused when it comes to choose a method and would like to hear your experiences. If I am right, I have 3 possibilities: - OpenBLAS: opensource and free, but I came across some posts describing seg faults issues and bugs. These posts are 2 years old and I wonder if it is still the case. - ATLAS: can't see any reason to not use it - Intel MKL: this is part of Intel Parallel Studio and is a paid software. Now, there is the MKL package distributed by Revolutionanalytics, but I am not certain how this can be distributed for free. Is there any kind of difference? In case of use of this package, do I need to install RRO or can I just build R from GNU against these libraries? Thank you for advices. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running R in Server
On Sun, Aug 16, 2015, 2:29 PM Swagato Chatterjee swagato1...@gmail.com wrote: Hello, I have written a R script which runs a regression of a dataset and saves the result in a csv file. Now this dataset has to be edited periodically which is done in a server. I need to run the R script in a server so that the results can also be shared in a server and used in a web application. Have a look at deployR on Revolutioanalytic website. There is a free open-source solution for windows. I have coded in R and have used R in windows. I have never used Ubuntu/Linux. Is there a step by step guide on how to run a R code in server? Thanks and Regards, Swagato [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] misbehavior with extract_numeric() from tidyr
R 3.2.0 on Linux library(tidyr) playerStats - c(LVL 10, 5,671,448 AP l6,000,000 AP, Unique Portals Visited 1,038, XM Collected 15,327,123 XM, Hacks 14,268, Resonators Deployed 11,126, Links Created 1,744, Control Fields Created 294, Mind Units Captured 2,995,484 MUs, Longest Link Ever Created 75 km, Largest Control Field 189,731 MUs, XM Recharged 3,006,364 XM, Portals Captured 1,204, Unique Portals Captured 486, Resonators Destroyed 12,481, Portals Neutralized 1,240, Enemy Links Destroyed 3,169, Enemy Control Fields Destroyed 1,394, Distance Walked 230 km, Max Time Portal Held 240 days, Max Time Link Maintained 15 days, Max Link Length x Days 276 km-days, Max Time Field Held 4days, Largest Field MUs x Days 83,226 MU-days) --- extract_numeric(playerStats) [1] 10 5671448600 1038 15327123 14268 11126 17442942995484 [10] 75 1897313006364 1204 486 12481 1240 3169 1394 [19]230240 15 NA 4 NA playerStats[c(22,24)] [1] Max Link Length x Days 276 km-days Largest Field MUs x Days 83,226 MU-days I do not understand why these two vectors return NA when the function extract_numeric() works well for others, Any wrong settings in my env? Thank you for hints. -- google.com/+arnaudgabourygabx __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] misbehavior with extract_numeric() from tidyr
On Mon, Apr 20, 2015 at 12:09 PM, Jim Lemon drjimle...@gmail.com wrote:
Hi arnaud,
At a guess, it is the two hyphens that are present in those strings. I
think that the function you are using interprets them as subtraction
operators and since the string following the hyphen would produce NA,
the result would be NA.
I was thinking of 'x' as being the culprit (interpreted as multiply)
but you are right indeed
noHyphens - str_replace(playerStats[c(22,24)],'-','')
extract_numeric(noHyphens)
[1] 276 83226
in fact:
-
extract_numeric
function (x)
{
as.numeric(gsub([^0-9.-]+, , as.character(x)))
}
environment: namespace:tidyr
-
Is there any particular reason for the hyphen in gsub() ? Why not
remove it thus ?
TY much Jim
Jim
On Mon, Apr 20, 2015 at 7:46 PM, arnaud gaboury
arnaud.gabo...@gmail.com wrote:
On Mon, Apr 20, 2015 at 9:10 AM, arnaud gaboury
arnaud.gabo...@gmail.com wrote:
R 3.2.0 on Linux
library(tidyr)
playerStats - c(LVL 10, 5,671,448 AP l6,000,000 AP, Unique
Portals Visited 1,038,
XM Collected 15,327,123 XM, Hacks 14,268, Resonators Deployed 11,126,
Links Created 1,744, Control Fields Created 294, Mind Units
Captured 2,995,484 MUs,
Longest Link Ever Created 75 km, Largest Control Field 189,731 MUs,
XM Recharged 3,006,364 XM, Portals Captured 1,204, Unique Portals
Captured 486,
Resonators Destroyed 12,481, Portals Neutralized 1,240, Enemy
Links Destroyed 3,169,
Enemy Control Fields Destroyed 1,394, Distance Walked 230 km,
Max Time Portal Held 240 days, Max Time Link Maintained 15 days,
Max Link Length x Days 276 km-days, Max Time Field Held 4days,
Largest Field MUs x Days 83,226 MU-days)
---
extract_numeric(playerStats)
[1] 10 5671448600 1038 15327123
14268 11126 17442942995484
[10] 75 1897313006364 1204
486 12481 1240 3169 1394
[19]230240 15 NA
4 NA
playerStats[c(22,24)]
[1] Max Link Length x Days 276 km-days Largest Field MUs x
Days 83,226 MU-days
I do not understand why these two vectors return NA when the function
extract_numeric() works well for others,
Any wrong settings in my env?
-
as.numeric(gsub([^0-9], ,playerStats))
[1] 10 5671448600 1038 15327123
14268 11126 17442942995484
[10] 75 1897313006364 1204
486 12481 1240 3169 1394
[19]230240 15276
4 83226
The above command does the job, but I still can not figure out why
extract_numeric() returns two NA
Thank you for hints.
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Re: [R] misbehavior with extract_numeric() from tidyr
On Mon, Apr 20, 2015 at 9:10 AM, arnaud gaboury arnaud.gabo...@gmail.com wrote: R 3.2.0 on Linux library(tidyr) playerStats - c(LVL 10, 5,671,448 AP l6,000,000 AP, Unique Portals Visited 1,038, XM Collected 15,327,123 XM, Hacks 14,268, Resonators Deployed 11,126, Links Created 1,744, Control Fields Created 294, Mind Units Captured 2,995,484 MUs, Longest Link Ever Created 75 km, Largest Control Field 189,731 MUs, XM Recharged 3,006,364 XM, Portals Captured 1,204, Unique Portals Captured 486, Resonators Destroyed 12,481, Portals Neutralized 1,240, Enemy Links Destroyed 3,169, Enemy Control Fields Destroyed 1,394, Distance Walked 230 km, Max Time Portal Held 240 days, Max Time Link Maintained 15 days, Max Link Length x Days 276 km-days, Max Time Field Held 4days, Largest Field MUs x Days 83,226 MU-days) --- extract_numeric(playerStats) [1] 10 5671448600 1038 15327123 14268 11126 17442942995484 [10] 75 1897313006364 1204 486 12481 1240 3169 1394 [19]230240 15 NA 4 NA playerStats[c(22,24)] [1] Max Link Length x Days 276 km-days Largest Field MUs x Days 83,226 MU-days I do not understand why these two vectors return NA when the function extract_numeric() works well for others, Any wrong settings in my env? - as.numeric(gsub([^0-9], ,playerStats)) [1] 10 5671448600 1038 15327123 14268 11126 17442942995484 [10] 75 1897313006364 1204 486 12481 1240 3169 1394 [19]230240 15276 4 83226 The above command does the job, but I still can not figure out why extract_numeric() returns two NA Thank you for hints. -- google.com/+arnaudgabourygabx -- google.com/+arnaudgabourygabx __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] misbehavior with extract_numeric() from tidyr
On Mon, Apr 20, 2015 at 6:09 PM, William Dunlap wdun...@tibco.com wrote: The hyphen without a following digit confuses tidyr::extract_numeric(). E.g., extract_numeric(23 ft-lbs) Warning message: In extract_numeric(23 ft-lbs) : NAs introduced by coercion [1] NA extract_numeric(23 ft*lbs) [1] 23 See[0] for the reason on the minus in the regex. It is not a bug but a wish. I am honestly very surprised the maintainer decided to go with such a so simple solution for negative numbers. [0]https://github.com/hadley/tidyr/issues/20 Contact the BugReports address for the package packageDescription(tidyr)$BugReports [1] https://github.com/hadley/tidyr/issues; or package's maintainer maintainer(tidyr) [1] Hadley Wickham had...@rstudio.com to report problems in a user-contributed package. Bill Dunlap TIBCO Software wdunlap tibco.com On Mon, Apr 20, 2015 at 12:10 AM, arnaud gaboury arnaud.gabo...@gmail.com wrote: R 3.2.0 on Linux library(tidyr) playerStats - c(LVL 10, 5,671,448 AP l6,000,000 AP, Unique Portals Visited 1,038, XM Collected 15,327,123 XM, Hacks 14,268, Resonators Deployed 11,126, Links Created 1,744, Control Fields Created 294, Mind Units Captured 2,995,484 MUs, Longest Link Ever Created 75 km, Largest Control Field 189,731 MUs, XM Recharged 3,006,364 XM, Portals Captured 1,204, Unique Portals Captured 486, Resonators Destroyed 12,481, Portals Neutralized 1,240, Enemy Links Destroyed 3,169, Enemy Control Fields Destroyed 1,394, Distance Walked 230 km, Max Time Portal Held 240 days, Max Time Link Maintained 15 days, Max Link Length x Days 276 km-days, Max Time Field Held 4days, Largest Field MUs x Days 83,226 MU-days) --- extract_numeric(playerStats) [1] 10 5671448600 1038 15327123 14268 11126 17442942995484 [10] 75 1897313006364 1204 486 12481 1240 3169 1394 [19]230240 15 NA 4 NA playerStats[c(22,24)] [1] Max Link Length x Days 276 km-days Largest Field MUs x Days 83,226 MU-days I do not understand why these two vectors return NA when the function extract_numeric() works well for others, Any wrong settings in my env? Thank you for hints. -- google.com/+arnaudgabourygabx __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- google.com/+arnaudgabourygabx https://plus.google.com/_/notifications/emlink?emr=05814804238976922326emid=CKiv-v6PvboCFcfoQgod6msAAApath=%2F116159236040461325607%2Fop%2Fudt=1383086841306ub=50 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error when loading shared library - stringini()
On Thu, Apr 16, 2015 at 3:25 PM, arnaud gaboury arnaud.gabo...@gmail.com
wrote:
On a Linux 64 bits, R.3.1.2, with tidyr() loaded.
gabx@hortensia [R] separate(rawStats, 'toto')
Error in dyn.load(file, DLLpath = DLLpath, ...) :
unable to load shared object
'/developement/language/r/library/stringi/libs/stringi.so':
libicui18n.so.54: cannot open shared object file: No such file or
directory
When trying to upgrade stringini(), it is not available for 3.1.2
install_github('Rexamine/stringi')
did the trick
My box run :icu 55.1-1 :lib32-icu 55.1-1
If I am right, I need to downgrade to 54 to be able to run separate()
from tidyr package? Or is there any other way?
Thank you for hint.
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[R] Error when loading shared library - stringini()
On a Linux 64 bits, R.3.1.2, with tidyr() loaded. gabx@hortensia [R] separate(rawStats, 'toto') Error in dyn.load(file, DLLpath = DLLpath, ...) : unable to load shared object '/developement/language/r/library/stringi/libs/stringi.so': libicui18n.so.54: cannot open shared object file: No such file or directory When trying to upgrade stringini(), it is not available for 3.1.2 My box run :icu 55.1-1 :lib32-icu 55.1-1 If I am right, I need to downgrade to 54 to be able to run separate() from tidyr package? Or is there any other way? Thank you for hint. -- google.com/+arnaudgabourygabx __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running R Remotely on LINUX
On Tue, Apr 14, 2015 at 10:09 PM, John Sorkin
jsor...@grecc.umaryland.edu wrote:
I suggest that you investigate installing RStudio server on the Linux
Box. If you do this, you can logon to RStudio (on the Linux server), and
it will look exactly like RStudio running on a windows box. You may need
some help configuring the Linux box to allow access to port 8787, which
is the default port that RStudio Server uses. You may also have to set
port forwarding on your cable modem or firewall.
John
In case you just need a R server, with no IDE (like Rstudio provides
it), please have a look at this project[0]
[0]http://www.obiba.org/?q=node/63
John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and
Geriatric Medicine
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Jeff Newmiller jdnew...@dcn.davis.ca.us 04/14/15 2:54 PM
You should investigate using the parallel package. You have to have R
installed on the Linux machine along with any contributed packages you
use, but you can delegate tasks to it from within an Rgui or RStudio
session running on your Windows box. There are even tutorials online
that step you through setting up a cloud computer to serve this
purpose.
It is possible to install R under your own account on a Linux server,
but there are more hiccups to overcome in doing so. There is a small
charge for using cloud servers, but if you use it right then the cost
can be quite cheap.
Note that while you can set up Windows servers in the cloud, they are
not as well suited to this remote use as Linux servers are, so it is
worth the effort to learn enough to do that.
---
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DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
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---
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On April 14, 2015 9:57:28 AM PDT, Michael Haenlein
haenl...@escpeurope.eu wrote:
Dear all,
I am used to running R locally on my Windows-based PC. Since some of my
computations are taking a lot of time I am now trying to move to a
remote R
session on a LINUX server but I am having trouble to getting things
work.
I am able to access the LINUX server using PuTTY and SSH. Once I have
access I can log in with my username and password (which is asked
through
keyboard-interactive authentication). I can then open an R session.
Since I am not used to working with LINUX, I have several questions:
(1) Ideally I am looking for a Windows-based software that would allow
me
to work on R as I am used to with the difference that the computations
are
run remotely on the LINUX server. Does a software like this exist?
Please
note that I do not think that I can install any software on the LINUX
server. But I can install stuff on my Windows-based PC.
(2) I am running an extensive simulation that takes about one week to
run.
Right now it seems that when I log out of R on LINUX and close PuTTY,
the R
session closes as well. Is there a way to let R run in the background
for
the week and just check into the progress 1-2 times a day?
(3) Can I open several instances of R in parallel? On my PC I sometimes
have 2-3 windows open in parallel that work on different calculations
to
save time. Not sure to which extent this is possible on LINUX.
I assume that this questions are very naïve. But since I’m only used to
working with Windows I’m quite stuck at the moment. Any help would be
very
appreciated!
Thanks in advance,
Michael
Michael Haenlein
Professor of Marketing
ESCP Europe
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running R Remotely on LINUX
On Tue, Apr 14, 2015 at 7:09 PM, Sarah Goslee sarah.gos...@gmail.com wrote: Hi Michael, On Tue, Apr 14, 2015 at 12:57 PM, Michael Haenlein haenl...@escpeurope.eu wrote: Dear all, I am used to running R locally on my Windows-based PC. Since some of my computations are taking a lot of time I am now trying to move to a remote R session on a LINUX server but I am having trouble to getting things work. Please have a look at this link[0]. It may help I am able to access the LINUX server using PuTTY and SSH. Once I have access I can log in with my username and password (which is asked through keyboard-interactive authentication). I can then open an R session. Since I am not used to working with LINUX, I have several questions: (1) Ideally I am looking for a Windows-based software that would allow me to work on R as I am used to with the difference that the computations are run remotely on the LINUX server. Does a software like this exist? Please note that I do not think that I can install any software on the LINUX server. But I can install stuff on my Windows-based PC. I'm not even sure what this question means, sorry. (2) I am running an extensive simulation that takes about one week to run. Right now it seems that when I log out of R on LINUX and close PuTTY, the R session closes as well. Is there a way to let R run in the background for the week and just check into the progress 1-2 times a day? Start screen on the linux system, then start R. Once the R job is running, you can disconnect from the screen with Ctrl-A Ctrl-D and the R job will continue in the background. Google linux screen for more information. (3) Can I open several instances of R in parallel? On my PC I sometimes have 2-3 windows open in parallel that work on different calculations to save time. Not sure to which extent this is possible on LINUX. Sure. Using screen you can have multiple sessions going. Sarah -- Sarah Goslee http://www.functionaldiversity.org [0]https://www.opencpu.org/ __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- google.com/+arnaudgabourygabx __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] replace values in one df by values from key pairs in another df
I have been searching for a while now, but can't put all pieces of the
puzzle together.
Goal : I want to replace all these kinds of patterns @U032FHV3S by
this @agreenmamba. In a more generic way, it is replacing 'id' by
user 'name'.
I have two df:
The first, 'history', is some message history where I want to do the
replacement. It is a chat history log,
The second one, idName, is a list of id:name. All are unique. This df
rows are exctracted from a bigger one called 'users' this way :
users[users$id %in% ext].
Please find below my two data.frames:
-
history - structure(list(name = c(xaeth, agreenmamba, poisonivy,
agreenmamba, cruzecontrol, poisonivy, poisonivy, vairis,
vairis, poisonivy), text = c(and people told that agent their
geocities experience would never amount to anything (the convo
yesterday) ,
@U032FHV3S http://youtu.be/oGIFublvDes, for me there is no such
question,
@U03AEKWTL|agreenmamba uploaded a file:
https://enlightened.slack.com/files/agreenmamba/F03KGRF3W/screenshot_2015-02-09-14-31-15.png|regarding:
should I stay or should I go?,
Lol., ha ha sorry, some testing with my irc client, anybody
knows how does \Passcode circuitry too hot. Wait for cool down to
enter another passcode.\ works?,
what is the cooldown time or how does it work...;, @U03AEKYL4:
what app are you talking about ?
)), .Names = c(name, text), class = c(data.table, data.frame
), row.names = c(NA, -10L)
--
idName - structure(list(id = c(U03AEKWTL, U032FHV3S, U03AEKYL4),
name = c(agreenmamba, poisonivy, vairis)), .Names = c(id,
name), sorted = name, class = c(data.table, data.frame
), row.names = c(NA, -3L))
1- I can find the pattern to be replaced in history this way :
ext - regmatches(history$text,regexpr('(?=@)[^|]{9}(?=|)',history$text,
perl = T)
ext
[1] U032FHV3S U03AEKWTL U03AEKYL4
2- I can select the pairs id:name in my 'users' df:
idName=users[users$id %in% ext]
idname
1: U03AEKWTL agreenmamba
2: U032FHV3S poisonivy
3: U03AEKYL4 vairis
3- i can write a loop to make the changes:
hist.txt - history$text
for (i in 1:3){hist.txt - gsub(ext[i],idName[[2]][i], hist.txt, perl = T)}
But this will not work because of two things:
- 'idName' is not ordered the same way as 'ext'
- if a users speak twice it will leave with with some NA
May you have some hints how to process ?
I was thinking of maybe using sapply(myFun) to my 'history' df where
myFun would be the replacement process.
OR something like with assigning in my env each 'id' to its corresponding 'name'
Thank you for help
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] replace values in one df by values from key pairs in another df
This is the wrong part of my code. idName=users[users$id %in% ext] idname 1: U03AEKWTL agreenmamba 2: U032FHV3S poisonivy 3: U03AEKYL4 vairis Best is to use: idNames - users[pmatch(ext, users$id, duplicates.ok = T)]. This leave me with an ordered and duplicate entries : idNames - users[pmatch(ext, users$id, duplicates.ok = T)] idname 1: U03AEKYL4 vairis 2: U03AEKYL4 vairis 3: U03AEKWTL agreenmamba 4: U032FHV3S poisonivy __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] gsub : replace regex pattern with values from another data.frame
On Thu, Feb 12, 2015 at 3:40 PM, arnaud gaboury
arnaud.gabo...@gmail.com wrote:
I have two df (and dt):
df1
structure(list(name = c(poisonivy, poisonivy, poisonivy,
poisonivy, poisonivy, poisonivy, poisonivy, poisonivy,
cruzecontrol, agreenmamba, agreenmamba, vairis, vairis,
vairis, vairis, vairis, vairis, xaeth), text = c(ok,
need items ?, i didn't submit pass codes for a long now,
ok, @U03AEKYL4: what app are you talking about ?, some testing
with my irc client,
ha ha sorry, for me there is no such question, Lol.,
@U03AEKWTL|agreenmamba uploaded a file:
https://enlightened.slack.com/files/agreenmamba/F03KGRF3W/screenshot_2015-02-09-14-31-15.png|regarding:
should I stay or should I go?,
@U032FHV3S http://youtu.be/oGIFublvDes, ok, see you around,
yeah, I had a procrastination rush so I started to decode a little,
http://ingress.com/intel|ingress.com/intel when you submit passcodes,
intel, what is the cooldown time or how does it work...;,
anybody knows how does \Passcode circuitry too hot. Wait for cool
down to enter another passcode.\ works?,
and people told that agent their geocities experience would never
amount to anything (the convo yesterday)
), ts = c(1423594336.000138, 1423594311.000136, 1423594294.000135,
1423594258.000133, 1423594244.000131, 1423497058.000127,
1423497041.000126, 1423478555.000123, 1423494427.000125,
1423492370.000124, 1423478364.000121, 1423594358.000139,
1423594329.000137, 1423594264.000134, 1423594251.000132,
1423592204.000130, 1423592174.000129, 1423150354.000112
)), .Names = c(name, text, ts), class = c(data.table,
data.frame), row.names = c(NA, -18L))
df2
structure(list(id = c(U03KH8Z52, U02AF1DTJ, U02AF0ZT8,
U03AEKWTL, U02BCJH0G, U033YA1MS, U029QMCRR, U03H139M5,
U02AET1D0, U02A6U41Z, U02B5T4CX, U02B2QU4R, U03F0LQ5X,
U03JNFKLY, U02ASMBMQ, U029QLQC7, U03AEMBQU, U02B4D3Q1,
U02AGDC14, U029A467C, U02A7NFG6, U02AESPPL, U02AQANK7,
U03ADJDFK, U03EYR0KB, U02AW7Q5Q, U02AE8RKD, U02FT84BS,
U02B25M3B, U03EZDQT7, U02AECKFF, U03H2691M, U02DWTJ5V,
U02AFTAHH, U029QQEPM, U03C51Z42, U02CAK2CV, U03AK21DP,
U03FFN8ED, U02B23V03, U029T2143, U02C1LEEX, U03AF2QH2,
U03E0GN0S, U03AG20R9, U02AES8S2, U02AG64S7, U02B5A0R7,
U02AS4SLR, U03C2SG0R, U03AV7CCW, U032XPFDU, U03AUKSSV,
U02C2A61Y, U02AESHJQ, U02BLSKHU, U02E34WM6, U03AK6P26,
U02E6ADRZ, U03FCDQ50, U03EW1CC5, U02BL0DBD, U02FHQZ6D,
U02B47T63, U03H2TTQP, U03AVP71V, U03JLV38V, U02E39HAY,
U02AE5281, U032FHV3S, U03AL2096, U02ARUG6M, U02AECRSP,
U02B42XG4, U03AFQZNS, U02AE7H41, U03G9UNTG, U02GEQ0E6,
U02AGLE5A, U02BQTRC9, U03H0J6GS, U02B3D27F, U02AEKTHV,
U02C52YN3, U02E33MUW, U03AKUT85, U03B53EHG, U02FBN38P,
U03AH3E5W, U02B5PLE0, U02AS4RCK, U03ANE1GZ, U02E8LZQB,
U03EPGJ98, U02E3N220, U03AEKYL4, U02AE7HT1, U02C1RR3G,
U03JH408J, U03KL0FKN, U02B44R92, U03EURWGX), name = c(10k_affair,
1upwuzhere, 4xcss, agreenmamba, ait109, arly69, azkop13,
barcik75, bigolnob, blackrose, blink619, bobaloo23,
bodger, bomb, bootswithdefer, brandizzle, bregalad,
camon, celticrain, ch3mical, checksum, cocothunder,
cruxicon, cruzecontrol, crystalskunk, cscheetah, dabcelin,
deelicious, delthanar, drkaosdk, droidenl-joe, dukeceph,
fillerbunny, flickohmsford, flyingg0d, garaxiel, goby9,
gymbal, hideandseek, hobojr, ijackportals, invalidcharactr,
itso9, j0shs, jarvis, jc0mm5, jencyberchic, jimbobradyson,
joespr0cket, jostrander, jueliet, karlashi, khan99,
kingkonn0r, krispycridder, kritickalmass, lawgiver, maxcorbett,
memory556, meta000x, minkovsky, mistylady, mstephans,
mstrinity, nocarryr, ollietronic, philistine11, pickledpickles,
piercingsbykris, poisonivy, raugmor, remarks999, rheds77,
rhinz, rigiritter, robbie0017, rohdef, ryoziya, s4n1ty,
sacredcow133, samwill, sgtlemonpepper, sivan, spline9,
starwolf, stueliueli, sweetiris, swift2plunder, swissphoenix,
synyck, test, therug, tinja551, trulyjuan, twinster,
vairis, vinylz3ro, watervirus, xaeth, yagamiyukari,
zafo, zexium)), .Names = c(id, name), class = c(data.table,
data.frame), row.names = c(NA, -102L))
I need to replace this regex pattern in df1 :
(?=@)[^|]{9}(?=|) by its corresponding name from df2.
E.g : if @U03KH8Z52 is found in df1, then I want to replace it by
the name which correspond to this id in df2., in this case
10k_affair
I know of replace an expression with gsub:
gsub('(?=@)[^|]{9}(?=|)', 'toto', df1, perl = T)
but I have no idea how to replace it with value from another df.
Thank you for hints
I am gathering some pieces of the puzzles.
regmatches(df1$text,regexpr('(?=@)[^|]{9}(?=|)',df1$text, perl = T))
[1] U032FHV3S U03AEKWTL U03AEKYL4
The above commands extract the needed pattern
df2[grep(U032FHV3S,df2$id),][[2]]
[1] poisonivy
The above command returns the name in the same row than the id. I need
more than one name (in my case, I need 3)
Shall I now write a loop and get a list of my needed name ? Pseudo
code would be something like:
for i %in% regmatches(df1$text,regexpr('(?=@)[^|]{9}(?=|)',df1$text,
perl = T
Re: [R] gsub : replace regex pattern with values from another data.frame
On Thu, Feb 12, 2015 at 7:12 PM, arnaud gaboury
arnaud.gabo...@gmail.com wrote:
On Thu, Feb 12, 2015 at 3:40 PM, arnaud gaboury
arnaud.gabo...@gmail.com wrote:
I have two df (and dt):
df1
structure(list(name = c(poisonivy, poisonivy, poisonivy,
poisonivy, poisonivy, poisonivy, poisonivy, poisonivy,
cruzecontrol, agreenmamba, agreenmamba, vairis, vairis,
vairis, vairis, vairis, vairis, xaeth), text = c(ok,
need items ?, i didn't submit pass codes for a long now,
ok, @U03AEKYL4: what app are you talking about ?, some testing
with my irc client,
ha ha sorry, for me there is no such question, Lol.,
@U03AEKWTL|agreenmamba uploaded a file:
https://enlightened.slack.com/files/agreenmamba/F03KGRF3W/screenshot_2015-02-09-14-31-15.png|regarding:
should I stay or should I go?,
@U032FHV3S http://youtu.be/oGIFublvDes, ok, see you around,
yeah, I had a procrastination rush so I started to decode a little,
http://ingress.com/intel|ingress.com/intel when you submit passcodes,
intel, what is the cooldown time or how does it work...;,
anybody knows how does \Passcode circuitry too hot. Wait for cool
down to enter another passcode.\ works?,
and people told that agent their geocities experience would never
amount to anything (the convo yesterday)
), ts = c(1423594336.000138, 1423594311.000136, 1423594294.000135,
1423594258.000133, 1423594244.000131, 1423497058.000127,
1423497041.000126, 1423478555.000123, 1423494427.000125,
1423492370.000124, 1423478364.000121, 1423594358.000139,
1423594329.000137, 1423594264.000134, 1423594251.000132,
1423592204.000130, 1423592174.000129, 1423150354.000112
)), .Names = c(name, text, ts), class = c(data.table,
data.frame), row.names = c(NA, -18L))
df2
structure(list(id = c(U03KH8Z52, U02AF1DTJ, U02AF0ZT8,
U03AEKWTL, U02BCJH0G, U033YA1MS, U029QMCRR, U03H139M5,
U02AET1D0, U02A6U41Z, U02B5T4CX, U02B2QU4R, U03F0LQ5X,
U03JNFKLY, U02ASMBMQ, U029QLQC7, U03AEMBQU, U02B4D3Q1,
U02AGDC14, U029A467C, U02A7NFG6, U02AESPPL, U02AQANK7,
U03ADJDFK, U03EYR0KB, U02AW7Q5Q, U02AE8RKD, U02FT84BS,
U02B25M3B, U03EZDQT7, U02AECKFF, U03H2691M, U02DWTJ5V,
U02AFTAHH, U029QQEPM, U03C51Z42, U02CAK2CV, U03AK21DP,
U03FFN8ED, U02B23V03, U029T2143, U02C1LEEX, U03AF2QH2,
U03E0GN0S, U03AG20R9, U02AES8S2, U02AG64S7, U02B5A0R7,
U02AS4SLR, U03C2SG0R, U03AV7CCW, U032XPFDU, U03AUKSSV,
U02C2A61Y, U02AESHJQ, U02BLSKHU, U02E34WM6, U03AK6P26,
U02E6ADRZ, U03FCDQ50, U03EW1CC5, U02BL0DBD, U02FHQZ6D,
U02B47T63, U03H2TTQP, U03AVP71V, U03JLV38V, U02E39HAY,
U02AE5281, U032FHV3S, U03AL2096, U02ARUG6M, U02AECRSP,
U02B42XG4, U03AFQZNS, U02AE7H41, U03G9UNTG, U02GEQ0E6,
U02AGLE5A, U02BQTRC9, U03H0J6GS, U02B3D27F, U02AEKTHV,
U02C52YN3, U02E33MUW, U03AKUT85, U03B53EHG, U02FBN38P,
U03AH3E5W, U02B5PLE0, U02AS4RCK, U03ANE1GZ, U02E8LZQB,
U03EPGJ98, U02E3N220, U03AEKYL4, U02AE7HT1, U02C1RR3G,
U03JH408J, U03KL0FKN, U02B44R92, U03EURWGX), name = c(10k_affair,
1upwuzhere, 4xcss, agreenmamba, ait109, arly69, azkop13,
barcik75, bigolnob, blackrose, blink619, bobaloo23,
bodger, bomb, bootswithdefer, brandizzle, bregalad,
camon, celticrain, ch3mical, checksum, cocothunder,
cruxicon, cruzecontrol, crystalskunk, cscheetah, dabcelin,
deelicious, delthanar, drkaosdk, droidenl-joe, dukeceph,
fillerbunny, flickohmsford, flyingg0d, garaxiel, goby9,
gymbal, hideandseek, hobojr, ijackportals, invalidcharactr,
itso9, j0shs, jarvis, jc0mm5, jencyberchic, jimbobradyson,
joespr0cket, jostrander, jueliet, karlashi, khan99,
kingkonn0r, krispycridder, kritickalmass, lawgiver, maxcorbett,
memory556, meta000x, minkovsky, mistylady, mstephans,
mstrinity, nocarryr, ollietronic, philistine11, pickledpickles,
piercingsbykris, poisonivy, raugmor, remarks999, rheds77,
rhinz, rigiritter, robbie0017, rohdef, ryoziya, s4n1ty,
sacredcow133, samwill, sgtlemonpepper, sivan, spline9,
starwolf, stueliueli, sweetiris, swift2plunder, swissphoenix,
synyck, test, therug, tinja551, trulyjuan, twinster,
vairis, vinylz3ro, watervirus, xaeth, yagamiyukari,
zafo, zexium)), .Names = c(id, name), class = c(data.table,
data.frame), row.names = c(NA, -102L))
I need to replace this regex pattern in df1 :
(?=@)[^|]{9}(?=|) by its corresponding name from df2.
E.g : if @U03KH8Z52 is found in df1, then I want to replace it by
the name which correspond to this id in df2., in this case
10k_affair
I know of replace an expression with gsub:
gsub('(?=@)[^|]{9}(?=|)', 'toto', df1, perl = T)
but I have no idea how to replace it with value from another df.
Thank you for hints
I am gathering some pieces of the puzzles.
regmatches(df1$text,regexpr('(?=@)[^|]{9}(?=|)',df1$text, perl = T))
[1] U032FHV3S U03AEKWTL U03AEKYL4
The above commands extract the needed pattern
df2[grep(U032FHV3S,df2$id),][[2]]
[1] poisonivy
The above command returns the name in the same row than the id. I need
more than one name (in my case, I need 3)
Shall I now write a loop and get a list of my needed name ? Pseudo
code would be something like
[R] apply two functions to column
I am little lost between all the possibilities to apply a function to a data.frame or data.table. Here is mine: structure(list(name = c(poisonivy, poisonivy, poisonivy, poisonivy, poisonivy, poisonivy, poisonivy, poisonivy, cruzecontrol, agreenmamba, agreenmamba, vairis, vairis, vairis, vairis, vairis, vairis, xaeth), text = c(ok, need items ?, i didn't submit pass codes for a long now, ok, @U03AEKYL4: what app are you talking about ?, some testing with my irc client, ha ha sorry, for me there is no such question, Lol., @U03AEKWTL|agreenmamba uploaded a file: https://enlightened.slack.com/files/agreenmamba/F03KGRF3W/screenshot_2015-02-09-14-31-15.png|regarding: should I stay or should I go?, @U032FHV3S http://youtu.be/oGIFublvDes, ok, see you around, yeah, I had a procrastination rush so I started to decode a little, http://ingress.com/intel|ingress.com/intel when you submit passcodes, intel, what is the cooldown time or how does it work...;, anybody knows how does \Passcode circuitry too hot. Wait for cool down to enter another passcode.\ works?, and people told that agent their geocities experience would never amount to anything (the convo yesterday) ), ts = c(1423594336.000138, 1423594311.000136, 1423594294.000135, 1423594258.000133, 1423594244.000131, 1423497058.000127, 1423497041.000126, 1423478555.000123, 1423494427.000125, 1423492370.000124, 1423478364.000121, 1423594358.000139, 1423594329.000137, 1423594264.000134, 1423594251.000132, 1423592204.000130, 1423592174.000129, 1423150354.000112 )), .Names = c(name, text, ts), class = c(data.table, data.frame), row.names = c(NA, -18L)) As you can see, it is of class data.frame and data.table. I need to work on last column ts. These are characters and are times in epoch format. I want to have time in Posix. I must thus do two things: - characters to numeric : myData$ts - as.numeric(myData$ts) - epoch to Posix : mapply(as.POSIXct, tz = 'GMT', origin = '1970-01-01', myMerge$ts) I wonder: - if I can apply these two functions at the same time - what will be the fastest way : use dplyr package, or work with data.table ? My real data will be much more important than the one posted. Thank you for advice and hint. -- google.com/+arnaudgabourygabx __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] gsub : replace regex pattern with values from another data.frame
I have two df (and dt):
df1
structure(list(name = c(poisonivy, poisonivy, poisonivy,
poisonivy, poisonivy, poisonivy, poisonivy, poisonivy,
cruzecontrol, agreenmamba, agreenmamba, vairis, vairis,
vairis, vairis, vairis, vairis, xaeth), text = c(ok,
need items ?, i didn't submit pass codes for a long now,
ok, @U03AEKYL4: what app are you talking about ?, some testing
with my irc client,
ha ha sorry, for me there is no such question, Lol.,
@U03AEKWTL|agreenmamba uploaded a file:
https://enlightened.slack.com/files/agreenmamba/F03KGRF3W/screenshot_2015-02-09-14-31-15.png|regarding:
should I stay or should I go?,
@U032FHV3S http://youtu.be/oGIFublvDes, ok, see you around,
yeah, I had a procrastination rush so I started to decode a little,
http://ingress.com/intel|ingress.com/intel when you submit passcodes,
intel, what is the cooldown time or how does it work...;,
anybody knows how does \Passcode circuitry too hot. Wait for cool
down to enter another passcode.\ works?,
and people told that agent their geocities experience would never
amount to anything (the convo yesterday)
), ts = c(1423594336.000138, 1423594311.000136, 1423594294.000135,
1423594258.000133, 1423594244.000131, 1423497058.000127,
1423497041.000126, 1423478555.000123, 1423494427.000125,
1423492370.000124, 1423478364.000121, 1423594358.000139,
1423594329.000137, 1423594264.000134, 1423594251.000132,
1423592204.000130, 1423592174.000129, 1423150354.000112
)), .Names = c(name, text, ts), class = c(data.table,
data.frame), row.names = c(NA, -18L))
df2
structure(list(id = c(U03KH8Z52, U02AF1DTJ, U02AF0ZT8,
U03AEKWTL, U02BCJH0G, U033YA1MS, U029QMCRR, U03H139M5,
U02AET1D0, U02A6U41Z, U02B5T4CX, U02B2QU4R, U03F0LQ5X,
U03JNFKLY, U02ASMBMQ, U029QLQC7, U03AEMBQU, U02B4D3Q1,
U02AGDC14, U029A467C, U02A7NFG6, U02AESPPL, U02AQANK7,
U03ADJDFK, U03EYR0KB, U02AW7Q5Q, U02AE8RKD, U02FT84BS,
U02B25M3B, U03EZDQT7, U02AECKFF, U03H2691M, U02DWTJ5V,
U02AFTAHH, U029QQEPM, U03C51Z42, U02CAK2CV, U03AK21DP,
U03FFN8ED, U02B23V03, U029T2143, U02C1LEEX, U03AF2QH2,
U03E0GN0S, U03AG20R9, U02AES8S2, U02AG64S7, U02B5A0R7,
U02AS4SLR, U03C2SG0R, U03AV7CCW, U032XPFDU, U03AUKSSV,
U02C2A61Y, U02AESHJQ, U02BLSKHU, U02E34WM6, U03AK6P26,
U02E6ADRZ, U03FCDQ50, U03EW1CC5, U02BL0DBD, U02FHQZ6D,
U02B47T63, U03H2TTQP, U03AVP71V, U03JLV38V, U02E39HAY,
U02AE5281, U032FHV3S, U03AL2096, U02ARUG6M, U02AECRSP,
U02B42XG4, U03AFQZNS, U02AE7H41, U03G9UNTG, U02GEQ0E6,
U02AGLE5A, U02BQTRC9, U03H0J6GS, U02B3D27F, U02AEKTHV,
U02C52YN3, U02E33MUW, U03AKUT85, U03B53EHG, U02FBN38P,
U03AH3E5W, U02B5PLE0, U02AS4RCK, U03ANE1GZ, U02E8LZQB,
U03EPGJ98, U02E3N220, U03AEKYL4, U02AE7HT1, U02C1RR3G,
U03JH408J, U03KL0FKN, U02B44R92, U03EURWGX), name = c(10k_affair,
1upwuzhere, 4xcss, agreenmamba, ait109, arly69, azkop13,
barcik75, bigolnob, blackrose, blink619, bobaloo23,
bodger, bomb, bootswithdefer, brandizzle, bregalad,
camon, celticrain, ch3mical, checksum, cocothunder,
cruxicon, cruzecontrol, crystalskunk, cscheetah, dabcelin,
deelicious, delthanar, drkaosdk, droidenl-joe, dukeceph,
fillerbunny, flickohmsford, flyingg0d, garaxiel, goby9,
gymbal, hideandseek, hobojr, ijackportals, invalidcharactr,
itso9, j0shs, jarvis, jc0mm5, jencyberchic, jimbobradyson,
joespr0cket, jostrander, jueliet, karlashi, khan99,
kingkonn0r, krispycridder, kritickalmass, lawgiver, maxcorbett,
memory556, meta000x, minkovsky, mistylady, mstephans,
mstrinity, nocarryr, ollietronic, philistine11, pickledpickles,
piercingsbykris, poisonivy, raugmor, remarks999, rheds77,
rhinz, rigiritter, robbie0017, rohdef, ryoziya, s4n1ty,
sacredcow133, samwill, sgtlemonpepper, sivan, spline9,
starwolf, stueliueli, sweetiris, swift2plunder, swissphoenix,
synyck, test, therug, tinja551, trulyjuan, twinster,
vairis, vinylz3ro, watervirus, xaeth, yagamiyukari,
zafo, zexium)), .Names = c(id, name), class = c(data.table,
data.frame), row.names = c(NA, -102L))
I need to replace this regex pattern in df1 :
(?=@)[^|]{9}(?=|) by its corresponding name from df2.
E.g : if @U03KH8Z52 is found in df1, then I want to replace it by
the name which correspond to this id in df2., in this case
10k_affair
I know of replace an expression with gsub:
gsub('(?=@)[^|]{9}(?=|)', 'toto', df1, perl = T)
but I have no idea how to replace it with value from another df.
Thank you for hints
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[R] upgrading issues with Rcpp
gabx@hortensia [R] sessionInfo() R version 3.1.2 (2014-10-31) Platform: x86_64-unknown-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8 LC_MONETARY=en_US.UTF-8 [6] LC_MESSAGES=en_US.UTF-8LC_PAPER=en_US.UTF-8 LC_NAME=C LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] utils base loaded via a namespace (and not attached): [1] tools_3.1.2 gabx@hortensia [R] search() [1] .GlobalEnvtools:rstudio package:utils package:base -- Now when trying to update Rcpp: gabx@hortensia [R] install.packages(Rcpp) . Error in unloadNamespace(pkg_name) : namespace ‘Rcpp’ is imported by ‘reshape2’, ‘plyr’, ‘dplyr’ so cannot be unloaded .. I can't upgrade my packages because of Rcpp =0.11.3 needed (running actually 0.11.2) What is this namespace imported by 'reshape2’, ‘plyr’, ‘dplyr’ ? How can I get rid of it and upgrade my packages ? Thank you for hints -- google.com/+arnaudgabourygabx __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Best Beginner Books?
On Thu, Oct 2, 2014 at 4:54 AM, Jason Eyerly teamtraders3...@gmail.com wrote: A lot of excellent suggestions! Thank you everyone for the input. I’ve purchased via Amazon: A Beginner's Guide to R by Zuur Data Manipulation with R by Spector “Introductory Statistics with R.” by Peter Dalgaard Are there any suggestions as to which order I should read them in when they arrive? For best comprehension, I’m thinking “Introductory Statistics with R” would be the best to start with, and perhaps the beginners guide after that, or vice versa? I would not start with a whole book, but a good short intro (~ 30-40 pages). Feel free to pick up free publications here: https://github.com/gabx/r-project/tree/master/documentation. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Building R for better performance
I got the benchmark script, which I've attached, from Texas Advanced Computing Center. Here are my results (elapsed times, in secs): Where can we get the benchmark script? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Building R for better performance
The best that I can see for R would be if someone were to post a how to use MKL for compiling R type document. I build R with MKL and ICC on my Archlinux box[1][2]. If I can help in anything, I will do it. [1]https://wiki.archlinux.org/index.php/R [2]https://aur.archlinux.org/packages/r-mkl/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using openBLAS in R under Unix / Linux
On Thu, Aug 28, 2014 at 11:45 AM, Martin Spindler spind...@mea.mpisoc.mpg.de wrote: Dear all, I would like to us openBLAS in R under Linux / Unix. Which steps do I have to undertake? Does someone know a detailed description? (I found some sources on the web, but none was really helpful for me.) Thanks and best, Martin I run it on my Archlinux. Please visit R archwiki[1] Do not hesitate to ask, I will help. [1]https://wiki.archlinux.org/index.php/R -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on installing R packages in a Citrix
On Fri, Aug 22, 2014 at 2:03 PM, S Ellison s.elli...@lgcgroup.com wrote: We are currently trying to migrate 3 users of R to a citrix based environment, but are coming across major issues trying to install the packages to the relevant image. Please be more precised in your issue. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Regex - subsetting parts of a file name.
A directory is full of data.frames cache files. All these files have
the same pattern:
df.some_name.RData
my.cache.list - c(df.subject_test.RData, df.subject_train.RData,
df.y_test.RData,
df.y_train.RData)
I want to keep only the part inside the two points. After lots of
headache using grep() when trying something like this:
grep('.(.*?).','df.subject_test.RData',value=T)
I couldn't find a clean one liner and found this workaround:
my.cache.list - gsub('df.','',my.cache.list)
my.cache.list - gsub('.RData','',my.cache.list)
The two above commands do the trick, but a clean one line with some
regex expression would be a more elegant way.
Does anyone have any suggestion ?
TY for help
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Re: [R] Regex - subsetting parts of a file name.
R as.vector(sapply(my.cache.list, function(x)strsplit(x, \\.)[[1]][2])) [1] subject_test subject_train y_testy_train R gsub(df\\.(.*)\\.RData, \\1, my.cache.list) [1] subject_test subject_train y_testy_train Note that . will match any character, while \\. matches a period. Thank you for your various suggestions. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] compare two data frames of different dimensions and onlykeep unique rows
TY very much for your setdiffDF(). It does the job perfectly.
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Petr Savicky
Sent: lundi 27 février 2012 20:41
To: r-help@r-project.org
Subject: Re: [R] compare two data frames of different dimensions and onlykeep
unique rows
On Mon, Feb 27, 2012 at 07:10:57PM +0100, Arnaud Gaboury wrote:
No, but I tried your way too.
In fact, the only three unique rows are these ones:
Product Price Nbr.Lots
Cocoa 24405
Cocoa 24501
Cocoa 24406
Here is a dirty working trick I found :
df-merge(exportfile,reported,all.y=T)
df1-merge(exportfile,reported)
dff1-do.call(paste,df)
dff-do.call(paste,df)
dff1-do.call(paste,df1)
df[!dff %in% dff1,]
Product Price Nbr.Lots
3 Cocoa 24405
4 Cocoa 24501
My two problems are : I do think it is not so a clean code, then I won't know
by advance which of my two df will have the greates dimension (I can add some
lines to deal with it, but again, seems very heavy).
Hi.
Try the following.
setdiffDF - function(A, B)
{
A[!duplicated(rbind(B, A))[nrow(B) + 1:nrow(A)], ]
}
df1 - setdiffDF(reported, exportfile)
df2 - setdiffDF(exportfile, reported)
rbind(df1, df2)
I obtained
Product Price Nbr.Lots
3Cocoa 24405
4Cocoa 24501
31 Cocoa 24406
Is this correct? I see the row
Cocoa 2440.006
only in exportfile and not in reported.
The trick with paste() is not a bad idea. A variant of it is used also in the
base function duplicated.matrix(), since it contains
apply(x, MARGIN, function(x) paste(x, collapse = \r))
If speed is critical, then possibly the paste() trick written for the whole
columns, for example
paste(df[[1]], df[[2]], df[[3]], sep=\r)
and then setdiff() can be better.
Hope this helps.
Petr Savicky.
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[R] compare two data frames with same columns names but of different dimensions
Dear List, I want to compare and return the rows which are NOT in the two data frames. Classic methods don't work as the df have NOT the same dimensions. Here are one example of my df: reported - structure(list(Product = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 3L, 4L, 5L, 5L), .Label = c(Cocoa, Coffee C, GC, Sugar No 11, ZS), class = factor), Price = c(2331, 2356, 2440, 2450, 204.55, 205.45, 17792, 24.81, 1273.5, 1276.25), Nbr.Lots = c(-61L, -61L, 5L, 1L, 40L, 40L, -1L, -1L, -1L, 1L)), .Names = c(Product, Price, Nbr.Lots), row.names = c(1L, 2L, 3L, 4L, 6L, 7L, 5L, 10L, 8L, 9L), class = data.frame) exportfile - structure(list(Product = c(Cocoa, Cocoa, Cocoa, Coffee C, Coffee C, GC, Sugar No 11, ZS, ZS), Price = c(2331, 2356, 2440, 204.55, 205.45, 17792, 24.81, 1273.5, 1276.25), Nbr.Lots = c(-61, -61, 6, 40, 40, -1, -1, -1, 1)), .Names = c(Product, Price, Nbr.Lots), row.names = c(NA, 9L), class = data.frame) As you can see, they have same column names. My idea was to merge these two df when passing as argument not to take into account duplicate rows, so I will get one df with rows which are not in both df. Is it possible? How to do it? TY for any help. Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Ubuntu jaunty - R can't be install
I don't understand where is your problem. Are you looking for jaunty as a package on R Crans?? Are you looking to install R package on your Ubuntu box? As you know, Jaunty is no more supported by Ubuntu, so packets have been moved in Archives. Arnaud Gaboury A2CT2 Ltd. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Veyssière Marine Sent: lundi 27 février 2012 14:50 To: r-help@r-project.org Subject: [R] Ubuntu jaunty - R can't be install Dear helpers, I would like to install R on a computer with Ubuntu Jaunty. Unfortunately, Jaunty does not appear in any CRAN mirror website I have tried (France, Canada, Italy, Germany). Could you give me any mirror website allowing installation on Jaunty?or could you give me a procedure to passby this problem? Thanks for your help, Best regards, Marine Veyssière __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] compare two data frames of different dimensions and only keep unique rows
Dear list, I am still struggling with something that should be easy: I compare two data frames with a lot of common rows and want to keep only rows that are NOT in both data frames, unique. Here are an example of these data frame. reported - structure(list(Product = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 3L, 4L, 5L, 5L), .Label = c(Cocoa, Coffee C, GC, Sugar No 11, ZS), class = factor), Price = c(2331, 2356, 2440, 2450, 204.55, 205.45, 17792, 24.81, 1273.5, 1276.25), Nbr.Lots = c(-61L, -61L, 5L, 1L, 40L, 40L, -1L, -1L, -1L, 1L)), .Names = c(Product, Price, Nbr.Lots), row.names = c(1L, 2L, 3L, 4L, 6L, 7L, 5L, 10L, 8L, 9L), class = data.frame) exportfile - structure(list(Product = c(Cocoa, Cocoa, Cocoa, Coffee C, Coffee C, GC, Sugar No 11, ZS, ZS), Price = c(2331, 2356, 2440, 204.55, 205.45, 17792, 24.81, 1273.5, 1276.25), Nbr.Lots = c(-61, -61, 6, 40, 40, -1, -1, -1, 1)), .Names = c(Product, Price, Nbr.Lots), row.names = c(NA, 9L), class = data.frame) I can rbind() them, thus resulting in one data frame with duplicated row, but I have no idea how to delete duplicated rows. I have tried plyaing with unique(), duplicated with no success v-rbind(exportfile,reported) v - structure(list(Product = c(Cocoa, Cocoa, Cocoa, Coffee C, Coffee C, GC, Sugar No 11, ZS, ZS, Cocoa, Cocoa, Cocoa, Cocoa, Coffee C, Coffee C, GC, Sugar No 11, ZS, ZS), Price = c(2331, 2356, 2440, 204.55, 205.45, 17792, 24.81, 1273.5, 1276.25, 2331, 2356, 2440, 2450, 204.55, 205.45, 17792, 24.81, 1273.5, 1276.25), Nbr.Lots = c(-61, -61, 6, 40, 40, -1, -1, -1, 1, -61, -61, 5, 1, 40, 40, -1, -1, -1, 1)), .Names = c(Product, Price, Nbr.Lots), row.names = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 21, 31, 41, 61, 71, 51, 10, 81, 91), class = data.frame) TY for your help Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] compare two data frames of different dimensions and only keep unique rows
No, but I tried your way too. In fact, the only three unique rows are these ones: Product Price Nbr.Lots Cocoa 24405 Cocoa 24501 Cocoa 24406 Here is a dirty working trick I found : df-merge(exportfile,reported,all.y=T) df1-merge(exportfile,reported) dff1-do.call(paste,df) dff-do.call(paste,df) dff1-do.call(paste,df1) df[!dff %in% dff1,] Product Price Nbr.Lots 3 Cocoa 24405 4 Cocoa 24501 My two problems are : I do think it is not so a clean code, then I won't know by advance which of my two df will have the greates dimension (I can add some lines to deal with it, but again, seems very heavy). I hoped I could find a better solution. A2CT2 Ltd. -Original Message- From: jim holtman [mailto:jholt...@gmail.com] Sent: lundi 27 février 2012 18:42 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] compare two data frames of different dimensions and only keep unique rows is this what you want: v - rbind(reported, exportfile) v[!duplicated(v), ] ProductPrice Nbr.Lots 1Cocoa 2331.00 -61 2Cocoa 2356.00 -61 3Cocoa 2440.005 4Cocoa 2450.001 6 Coffee C 204.55 40 7 Coffee C 205.45 40 5 GC 17792.00 -1 10 Sugar No 1124.81 -1 8 ZS 1273.50 -1 9 ZS 1276.251 13 Cocoa 2440.006 On Mon, Feb 27, 2012 at 12:36 PM, Arnaud Gaboury arnaud.gabo...@a2ct2.com wrote: Dear list, I am still struggling with something that should be easy: I compare two data frames with a lot of common rows and want to keep only rows that are NOT in both data frames, unique. Here are an example of these data frame. reported - structure(list(Product = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 3L, 4L, 5L, 5L), .Label = c(Cocoa, Coffee C, GC, Sugar No 11, ZS), class = factor), Price = c(2331, 2356, 2440, 2450, 204.55, 205.45, 17792, 24.81, 1273.5, 1276.25), Nbr.Lots = c(-61L, -61L, 5L, 1L, 40L, 40L, -1L, -1L, -1L, 1L)), .Names = c(Product, Price, Nbr.Lots), row.names = c(1L, 2L, 3L, 4L, 6L, 7L, 5L, 10L, 8L, 9L), class = data.frame) exportfile - structure(list(Product = c(Cocoa, Cocoa, Cocoa, Coffee C, Coffee C, GC, Sugar No 11, ZS, ZS), Price = c(2331, 2356, 2440, 204.55, 205.45, 17792, 24.81, 1273.5, 1276.25), Nbr.Lots = c(-61, -61, 6, 40, 40, -1, -1, -1, 1)), .Names = c(Product, Price, Nbr.Lots), row.names = c(NA, 9L), class = data.frame) I can rbind() them, thus resulting in one data frame with duplicated row, but I have no idea how to delete duplicated rows. I have tried plyaing with unique(), duplicated with no success v-rbind(exportfile,reported) v - structure(list(Product = c(Cocoa, Cocoa, Cocoa, Coffee C, Coffee C, GC, Sugar No 11, ZS, ZS, Cocoa, Cocoa, Cocoa, Cocoa, Coffee C, Coffee C, GC, Sugar No 11, ZS, ZS), Price = c(2331, 2356, 2440, 204.55, 205.45, 17792, 24.81, 1273.5, 1276.25, 2331, 2356, 2440, 2450, 204.55, 205.45, 17792, 24.81, 1273.5, 1276.25), Nbr.Lots = c(-61, -61, 6, 40, 40, -1, -1, -1, 1, -61, -61, 5, 1, 40, 40, -1, -1, -1, 1)), .Names = c(Product, Price, Nbr.Lots), row.names = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 21, 31, 41, 61, 71, 51, 10, 81, 91), class = data.frame) TY for your help Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data frame manipulation with condition
Dear list, n00b question, but still can't find any easy answer. Here is a df: df-data.frame(cbind(x=c(AA,BB,CC,AA),y=1:4)) df x y 1 AA 1 2 BB 2 3 CC 3 4 AA 4 I want to modify this df this way : if df$x==AA then df$y=df$y*10 if df$x==BB then df$y=df$y*25 and so on with other conditions. TY for any help. Trading A2CT2 Ltd. Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation with condition
TY Uwe, So I will have to write a line for each condition? Right? In fact I was trying to do something with apply in one line, but couldn't achieve any result. In fact, all my transformation will be multiplying one object by a specific number according to the value of df$x. Arnaud Gaboury A2CT2 Ltd. -Original Message- From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Sent: vendredi 24 février 2012 16:33 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] data frame manipulation with condition On 24.02.2012 16:25, Arnaud Gaboury wrote: Dear list, n00b question, but still can't find any easy answer. Here is a df: Change df-data.frame(cbind(x=c(AA,BB,CC,AA),y=1:4)) to df - data.frame(x = c(AA,BB,CC,AA), y = 1:4) to make your object a sensible data.frame. df x y 1 AA 1 2 BB 2 3 CC 3 4 AA 4 I want to modify this df this way : if df$x==AA then df$y=df$y*10 df$y[df$x==AA] - df$y[df$x==AA] * 25 ... Uwe Ligges if df$x==BB then df$y=df$y*25 and so on with other conditions. TY for any help. Trading A2CT2 Ltd. Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation with condition
df- data.frame(x = c(AA,BB,CC,AA,DD,DD), y = 1:6) mult - c(AA = 10, BB = 25,DD=15) df$y - df$y * mult[df$x] df x y 1 AA 10 2 BB 50 3 CC 45 4 AA 40 5 DD NA 6 DD NA My df is in fact much more longer than the chosen example shown here. It seems your tip didn't do the job. I am expecting this as result : df x y 1 AA 10 if df$x==AA, df$y-1*10 2 BB 50 if df$x==BB, df$y-2*25 3 CC 3 NOTHING 4 AA 40 if df$x==AA, df$y-4*10 5 DD 75 if df$x==DD, df$y-5*15 6 DD 90 if df$x==DD, df$y-6*15 Arnaud Gaboury A2CT2 Ltd. -Original Message- From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Sent: vendredi 24 février 2012 17:07 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] data frame manipulation with condition On 24.02.2012 16:59, Arnaud Gaboury wrote: TY Uwe, So I will have to write a line for each condition? Right? In fact I was trying to do something with apply in one line, but couldn't achieve any result. In fact, all my transformation will be multiplying one object by a specific number according to the value of df$x. In that case: mult - c(AA = 10, BB = 25) Then: df$y - df$y * mult[df$x] Uwe Ligges Arnaud Gaboury A2CT2 Ltd. -Original Message- From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Sent: vendredi 24 février 2012 16:33 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] data frame manipulation with condition On 24.02.2012 16:25, Arnaud Gaboury wrote: Dear list, n00b question, but still can't find any easy answer. Here is a df: Change df-data.frame(cbind(x=c(AA,BB,CC,AA),y=1:4)) to df- data.frame(x = c(AA,BB,CC,AA), y = 1:4) to make your object a sensible data.frame. df x y 1 AA 1 2 BB 2 3 CC 3 4 AA 4 I want to modify this df this way : if df$x==AA then df$y=df$y*10 df$y[df$x==AA]- df$y[df$x==AA] * 25 ... Uwe Ligges if df$x==BB then df$y=df$y*25 and so on with other conditions. TY for any help. Trading A2CT2 Ltd. Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation with condition
OK Uwe, I understand, and I will be more explicit. Here is how could my df be: reported - structure(list(Product = structure(c(1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 5L, 5L, 5L, 6L, 7L, 7L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 12L, 13L, 14L, 14L), .Label = c(CL, Cocoa, Coffee C, GC, HG, HO, NG, PL, RB, SI, Sugar No 11, ZC, ZL, ZW), class = factor), reported.Price = c(105.35, 2380, 2407, 2408, 202.35, 202.8, 202.95, 205.85, 206.05, 206.1, 206.2, 1748, 378.8, 379.25, 379.5, 320.61, 2.538, 2.543, 1669, 1678.5, 304.49, 321.39, 321.6, 321.65, 322.5, 322.55, 322.8, 323.04, 3390, 3397.5, 24.16, 24.2, 24.22, 24.23, 24.54, 25.5, 25.55, 631.75, 638, 53.77, 630.75, 633), reported.Nbr.Lots = c(6L, 3L, -1L, -2L, -40L, -1L, -1L, 10L, 5L, 6L, 19L, 17L, 23L, 12L, 35L, 11L, -54L, -52L, 26L, 26L, 10L, -10L, 1L, 4L, 4L, 1L, 5L, 5L, 17L, 17L, 114L, 71L, 16L, 27L, -3L, 3L, -3L, -89L, -1L, -1L, -51L, -51L)), .Names = c(Product, reported.Price, reported.Nbr.Lots ), row.names = c(7L, 4L, 5L, 6L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 8L, 9L, 10L, 11L, 12L, 20L, 21L, 22L, 23L, 35L, 36L, 37L, 38L, 39L, 40L, 41L, 42L, 31L, 32L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 2L, 3L, 1L, 33L, 34L), class = data.frame) Row will change. I am looking to multiply reported.Price by 100 IF Product=CL, multiply by 10 IF product=GC, multiply by 100 IF product=HG, multiply by 1000 IF Product=NG, multiply by 100 IF product=RB. I hope I am clear enough, and YES I have tried many workarounds myself before posting. Feel free to ignore my post if you think I am lazy and disrespectful to the list. Arnaud Gaboury A2CT2 Ltd. -Original Message- From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Sent: vendredi 24 février 2012 17:41 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] data frame manipulation with condition On 24.02.2012 17:36, Arnaud Gaboury wrote: df- data.frame(x = c(AA,BB,CC,AA,DD,DD), y = 1:6) mult- c(AA = 10, BB = 25,DD=15) df$y- df$y * mult[df$x] df x y 1 AA 10 2 BB 50 3 CC 45 4 AA 40 5 DD NA 6 DD NA My df is in fact much more longer than the chosen example shown here. It seems your tip didn't do the job. I am expecting this as result : This is not the I do the job for you hotline. You are free to think a little bit yourself given you have not managed in two attempts to describe your problem sufficiently well! Uwe Ligges df x y 1 AA 10 if df$x==AA, df$y-1*10 2 BB 50 if df$x==BB, df$y-2*25 3 CC 3 NOTHING 4 AA 40 if df$x==AA, df$y-4*10 5 DD 75 if df$x==DD, df$y-5*15 6 DD 90 if df$x==DD, df$y-6*15 Arnaud Gaboury A2CT2 Ltd. -Original Message- From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Sent: vendredi 24 février 2012 17:07 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] data frame manipulation with condition On 24.02.2012 16:59, Arnaud Gaboury wrote: TY Uwe, So I will have to write a line for each condition? Right? In fact I was trying to do something with apply in one line, but couldn't achieve any result. In fact, all my transformation will be multiplying one object by a specific number according to the value of df$x. In that case: mult- c(AA = 10, BB = 25) Then: df$y- df$y * mult[df$x] Uwe Ligges Arnaud Gaboury A2CT2 Ltd. -Original Message- From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Sent: vendredi 24 février 2012 16:33 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] data frame manipulation with condition On 24.02.2012 16:25, Arnaud Gaboury wrote: Dear list, n00b question, but still can't find any easy answer. Here is a df: Change df-data.frame(cbind(x=c(AA,BB,CC,AA),y=1:4)) to df- data.frame(x = c(AA,BB,CC,AA), y = 1:4) to make your object a sensible data.frame. df x y 1 AA 1 2 BB 2 3 CC 3 4 AA 4 I want to modify this df this way : if df$x==AA then df$y=df$y*10 df$y[df$x==AA]- df$y[df$x==AA] * 25 ... Uwe Ligges if df$x==BB then df$y=df$y*25 and so on with other conditions. TY for any help. Trading A2CT2 Ltd. Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation with condition
TY very much Sarah: your tip is doing the job: reported - structure(list(Product = structure(c(1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 5L, 5L, 5L, 6L, 7L, 7L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 12L, 13L, 14L, 14L), .Label = c(CL, Cocoa, Coffee C, GC, HG, HO, NG, PL, RB, SI, Sugar No 11, ZC, ZL, ZW), class = factor), reported.Price = c(105.35, 2380, 2407, 2408, 202.35, 202.8, 202.95, 205.85, 206.05, 206.1, 206.2, 1748, 378.8, 379.25, 379.5, 320.61, 2.538, 2.543, 1669, 1678.5, 304.49, 321.39, 321.6, 321.65, 322.5, 322.55, 322.8, 323.04, 3390, 3397.5, 24.16, 24.2, 24.22, 24.23, 24.54, 25.5, 25.55, 631.75, 638, 53.77, 630.75, 633), reported.Nbr.Lots = c(6L, 3L, -1L, -2L, -40L, -1L, -1L, 10L, 5L, 6L, 19L, 17L, 23L, 12L, 35L, 11L, -54L, -52L, 26L, 26L, 10L, -10L, 1L, 4L, 4L, 1L, 5L, 5L, 17L, 17L, 114L, 71L, 16L, 27L, -3L, 3L, -3L, -89L, -1L, -1L, -51L, -51L)), .Names = c(Product, reported.Price, reported.Nbr.Lots ), row.names = c(7L, 4L, 5L, 6L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 8L, 9L, 10L, 11L, 12L, 20L, 21L, 22L, 23L, 35L, 36L, 37L, 38L, 39L, 40L, 41L, 42L, 31L, 32L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 2L, 3L, 1L, 33L, 34L), class = data.frame) mult-c(CL=100,GC=10,HG=10,NG=1000,PL=10,RB=100,SI=10,ZL=100,HO=100,KC=1,CC=1,SB=1,ZC=1,ZW=1) reported$reported.Price - reported$reported.Price * mult[reported$Product] reported - structure(list(Product = structure(c(1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 5L, 5L, 5L, 6L, 7L, 7L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 12L, 13L, 14L, 14L), .Label = c(CL, Cocoa, Coffee C, GC, HG, HO, NG, PL, RB, SI, Sugar No 11, ZC, ZL, ZW), class = factor), reported.Price = c(10535, 23800, 24070, 24080, 2023.5, 2028, 2029.5, 2058.5, 2060.5, 2061, 2062, 1748000, 3788, 3792.5, 3795, 32061, 25.38, 25.43, 166900, 167850, 30449, 32139, 32160, 32165, 32250, 32255, 32280, 32304, 3390, 3397.5, 24.16, 24.2, 24.22, 24.23, 24.54, 25.5, 25.55, 631.75, 638, 53.77, 630.75, 633), reported.Nbr.Lots = c(6L, 3L, -1L, -2L, -40L, -1L, -1L, 10L, 5L, 6L, 19L, 17L, 23L, 12L, 35L, 11L, -54L, -52L, 26L, 26L, 10L, -10L, 1L, 4L, 4L, 1L, 5L, 5L, 17L, 17L, 114L, 71L, 16L, 27L, -3L, 3L, -3L, -89L, -1L, -1L, -51L, -51L)), .Names = c(Product, reported.Price, reported.Nbr.Lots ), row.names = c(7L, 4L, 5L, 6L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 8L, 9L, 10L, 11L, 12L, 20L, 21L, 22L, 23L, 35L, 36L, 37L, 38L, 39L, 40L, 41L, 42L, 31L, 32L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 2L, 3L, 1L, 33L, 34L), class = data.frame) Have a good weekend. Arnaud Gaboury A2CT2 Ltd. Trade: +41 22 849 88 63 Fax: +41 22 849 88 66 arnaud.gabo...@a2ct2.com This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. Access to this email by anyone else is unauthorized. If you are not the intended recipient, any disclosure, copying, distribution or any action taken or omitted to be taken in reliance on it, is prohibited and may be unlawful. If you have received this email in error please notify the sender. -Original Message- From: Sarah Goslee [mailto:sarah.gos...@gmail.com] Sent: vendredi 24 février 2012 17:54 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] data frame manipulation with condition You need, as I already suggested, to use a value of 1 for levels you don't want to change. mult - c(AA = 10, BB = 25, CC=1, DD=15) mult[df$x] AA BB CC AA DD DD 10 25 1 10 15 15 df$y * mult[df$x] AA BB CC AA DD DD 10 50 3 40 75 90 On Fri, Feb 24, 2012 at 11:36 AM, Arnaud Gaboury arnaud.gabo...@a2ct2.com wrote: df- data.frame(x = c(AA,BB,CC,AA,DD,DD), y = 1:6) mult - c(AA = 10, BB = 25,DD=15) df$y - df$y * mult[df$x] df x y 1 AA 10 2 BB 50 3 CC 45 4 AA 40 5 DD NA 6 DD NA My df is in fact much more longer than the chosen example shown here. It seems your tip didn't do the job. I am expecting this as result : df x y 1 AA 10 if df$x==AA, df$y-1*10 2 BB 50 if df$x==BB, df$y-2*25 3 CC 3 NOTHING 4 AA 40 if df$x==AA, df$y-4*10 5 DD 75 if df$x==DD, df$y-5*15 6 DD 90 if df$x==DD, df$y-6*15 Arnaud Gaboury A2CT2 Ltd. -Original Message- From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Sent: vendredi 24 février 2012 17:07 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] data frame manipulation with condition On 24.02.2012 16:59, Arnaud Gaboury wrote: TY Uwe, So I will have to write a line for each condition? Right? In fact I was trying to do something with apply in one line, but couldn't achieve any result. In fact, all my transformation will be multiplying one object by a specific number according to the value of df$x. In that case: mult - c(AA = 10, BB = 25) Then: df$y - df$y * mult[df$x] Uwe Ligges Arnaud Gaboury A2CT2 Ltd
Re: [R] data frame manipulation with condition
In fact I need to use William tip: Use mult[as.character(df$x)] instead of mult[df$x]. Let's try again with a shorter df as example: The rule: if AA, then multiply y by 2, if BB multiply y by 5, if CC do nothing, if DD multiply by 2. Let's say day 1 I have df1: df1 - structure(list(x = structure(c(1L, 2L, 2L, 3L), .Label = c(AA, BB, CC), class = factor), y = 1:4), .Names = c(x, y ), row.names = c(NA, -4L), class = data.frame) df1 x y 1 AA 1 2 BB 2 3 BB 3 4 CC 4 mult - c(AA=2,BB=5,CC=1,DD=2) df1$y - df1$y * mult[as.character(df1$x)] df1 x y 1 AA 2 2 BB 10 3 BB 15 4 CC 4 WORKING Now day 2 with df2: df2 - data.frame(x = c(AA,AA,BB,BB,BB,CC,DD,DD), y = 1:8) df2$y - df2$y * mult[as.character(df2$x)] df2 x y 1 AA 2 2 AA 4 3 BB 15 4 BB 20 5 BB 25 6 CC 6 7 DD 14 8 DD 16 WORKING Ty both of you and have a good weekend. Arnaud Gaboury A2CT2 Ltd. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Arnaud Gaboury Sent: vendredi 24 février 2012 18:17 To: Sarah Goslee Cc: r-help@r-project.org Subject: Re: [R] data frame manipulation with condition TY very much Sarah: your tip is doing the job: reported - structure(list(Product = structure(c(1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 5L, 5L, 5L, 6L, 7L, 7L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 12L, 13L, 14L, 14L), .Label = c(CL, Cocoa, Coffee C, GC, HG, HO, NG, PL, RB, SI, Sugar No 11, ZC, ZL, ZW), class = factor), reported.Price = c(105.35, 2380, 2407, 2408, 202.35, 202.8, 202.95, 205.85, 206.05, 206.1, 206.2, 1748, 378.8, 379.25, 379.5, 320.61, 2.538, 2.543, 1669, 1678.5, 304.49, 321.39, 321.6, 321.65, 322.5, 322.55, 322.8, 323.04, 3390, 3397.5, 24.16, 24.2, 24.22, 24.23, 24.54, 25.5, 25.55, 631.75, 638, 53.77, 630.75, 633), reported.Nbr.Lots = c(6L, 3L, -1L, -2L, -40L, -1L, -1L, 10L, 5L, 6L, 19L, 17L, 23L, 12L, 35L, 11L, -54L, -52L, 26L, 26L, 10L, -10L, 1L, 4L, 4L, 1L, 5L, 5L, 17L, 17L, 114L, 71L, 16L, 27L, -3L, 3L, -3L, -89L, -1L, -1L, -51L, -51L)), .Names = c(Product, reported.Price, reported.Nbr.Lots ), row.names = c(7L, 4L, 5L, 6L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 8L, 9L, 10L, 11L, 12L, 20L, 21L, 22L, 23L, 35L, 36L, 37L, 38L, 39L, 40L, 41L, 42L, 31L, 32L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 2L, 3L, 1L, 33L, 34L), class = data.frame) mult-c(CL=100,GC=10,HG=10,NG=1000,PL=10,RB=100,SI=10,ZL=100,HO=100,KC =1,CC=1,SB=1,ZC=1,ZW=1) reported$reported.Price - reported$reported.Price * mult[reported$Product] reported - structure(list(Product = structure(c(1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 5L, 5L, 5L, 6L, 7L, 7L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 12L, 13L, 14L, 14L), .Label = c(CL, Cocoa, Coffee C, GC, HG, HO, NG, PL, RB, SI, Sugar No 11, ZC, ZL, ZW), class = factor), reported.Price = c(10535, 23800, 24070, 24080, 2023.5, 2028, 2029.5, 2058.5, 2060.5, 2061, 2062, 1748000, 3788, 3792.5, 3795, 32061, 25.38, 25.43, 166900, 167850, 30449, 32139, 32160, 32165, 32250, 32255, 32280, 32304, 3390, 3397.5, 24.16, 24.2, 24.22, 24.23, 24.54, 25.5, 25.55, 631.75, 638, 53.77, 630.75, 633), reported.Nbr.Lots = c(6L, 3L, -1L, -2L, -40L, -1L, -1L, 10L, 5L, 6L, 19L, 17L, 23L, 12L, 35L, 11L, -54L, -52L, 26L, 26L, 10L, -10L, 1L, 4L, 4L, 1L, 5L, 5L, 17L, 17L, 114L, 71L, 16L, 27L, -3L, 3L, -3L, -89L, -1L, -1L, -51L, -51L)), .Names = c(Product, reported.Price, reported.Nbr.Lots ), row.names = c(7L, 4L, 5L, 6L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 8L, 9L, 10L, 11L, 12L, 20L, 21L, 22L, 23L, 35L, 36L, 37L, 38L, 39L, 40L, 41L, 42L, 31L, 32L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 2L, 3L, 1L, 33L, 34L), class = data.frame) Have a good weekend. Arnaud Gaboury A2CT2 Ltd. Trade: +41 22 849 88 63 Fax: +41 22 849 88 66 arnaud.gabo...@a2ct2.com This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. Access to this email by anyone else is unauthorized. If you are not the intended recipient, any disclosure, copying, distribution or any action taken or omitted to be taken in reliance on it, is prohibited and may be unlawful. If you have received this email in error please notify the sender. -Original Message- From: Sarah Goslee [mailto:sarah.gos...@gmail.com] Sent: vendredi 24 février 2012 17:54 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] data frame manipulation with condition You need, as I already suggested, to use a value of 1 for levels you don't want to change. mult - c(AA = 10, BB = 25, CC=1, DD=15) mult[df$x] AA BB CC AA DD DD 10 25 1 10 15 15 df$y * mult[df$x] AA BB CC AA DD DD 10 50 3 40 75 90 On Fri, Feb 24, 2012 at 11:36 AM, Arnaud Gaboury arnaud.gabo...@a2ct2.com wrote: df- data.frame(x = c(AA,BB,CC,AA,DD,DD), y = 1:6) mult - c(AA = 10, BB = 25,DD=15) df$y - df$y * mult[df$x] df x y 1 AA 10
Re: [R] install rJava in Ubuntu 11.10
As a Linux user, I know you are not running exactly the same Java as on windows. It exists some copyright issues which prevent you to run Sun Java on your Linux box. As shown, you are running in fact openJDK. If result is the same for you, being able to run Java, it is not the same for most apps. Install folders are different, and I guess R cannot find the right place for your Java environment. Try to google more on this issue (Java on Linux openJDK install folder), or anything about building Android on Linux( same problem as Android needs java). There is something you could try: add the openJDK path in your environment. Hope this help. Arnaud Gaboury A2CT2 Ltd. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of John Kane Sent: vendredi 24 février 2012 19:32 To: r-help@r-project.org Subject: [R] install rJava in Ubuntu 11.10 I am trying to install rJava on a WUBI Ubuntu 11.10 installation of R with no luck. I was originally trying to install the iplots package and encountered this rJava problem. Code used: install.packages(rJava) (CRAN mirrors --Canada(ON) and Canada(QC2) I installed iplots with no problem on Windows 7. I know just about zero about Ubuntu or Linux in general so I have no idea what I am doing. Any suggestions would be gratefully received. I am getting the error/warning messages below. = Cannot compile a simple JNI program. See config.log for details. Make sure you have Java Development Kit installed and correctly registered in R. If in doubt, re-run R CMD javareconf as root. ERROR: configuration failed for package ‘rJava’ * removing ‘/home/john/R/i686-pc-linux-gnu-library/2.13/rJava’ The downloaded packages are in ‘/tmp/RtmphfnJ62/downloaded_packages’ Warning message: In install.packages(rJava) : installation of package 'rJava' had non-zero exit status = EXISTING JAVA INSTALLAION john@ubuntu:~$ java -version java version 1.6.0_23 OpenJDK Runtime Environment (IcedTea6 1.11pre) (6b23~pre11-0ubuntu1.11.10.1) OpenJDK Client VM (build 20.0-b11, mixed mode, sharing) === I have re-run R CMD javareconf as root. Does the R CMC javareconf need something added? (i.e. R CMD javaconf xxx) Do I really need a Java Development Kit installed? sessionInfo() R version 2.13.1 (2011-07-08) Platform: i686-pc-linux-gnu (32-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8 [7] LC_PAPER=en_US.UTF-8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base Publish your photos in seconds for FREE TRY IM TOOLPACK at http://www.imtoolpack.com/default.aspx?rc=if4 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation with conditions
TY Elai for your answer. One solution has been given earlier in this list by Sarah Goslee and William Dunlap. Arnaud Gaboury A2CT2 Ltd. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of ilai Sent: vendredi 24 février 2012 20:14 To: A2CT2 Trading Cc: r-help@r-project.org Subject: Re: [R] data frame manipulation with conditions On Fri, Feb 24, 2012 at 8:11 AM, A2CT2 Trading trad...@a2ct2.com wrote: Dear list, n00b question, but still can't find any easy answer. Here is a df: df-data.frame(cbind(x=c(AA,BB,CC,AA),y=1:4)) # No, your y is a factor str(df) 'data.frame': 4 obs. of 2 variables: $ x: Factor w/ 3 levels AA,BB,CC: 1 2 3 1 $ y: Factor w/ 4 levels 1,2,3,4: 1 2 3 4 # You want to remove the cbind df-data.frame(x=c(AA,BB,CC,AA),y=1:4) str(df) 'data.frame': 4 obs. of 2 variables: $ x: Factor w/ 3 levels AA,BB,CC: 1 2 3 1 $ y: int 1 2 3 4 I want to modify this df this way : if df$x==AA then df$y=df$y*10 if df$x==BB then df$y=df$y*25 and so on with other conditions. df$y- df$y * c(10,25,.5)[df$x] [1] 10.0 50.0 1.5 40.0 # 1*10 2*25 3*.5 4*10 HTH Elai TY for any help. Trading A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data frame colnames through ddply
Dear list, I am trying to pass data frame colnames through ddply without sucess. Here is my command: exportfile-ddply(exportfile,c(Product,Price),summarise,Nbr.Lots=sum(Filled.Qty)) exportfile is a df. I want to apply summarise to Product and Price columns and sum the Filled.Qty column and renamed it Nbr.Lots. This line works. What I would like is changing the col names in the same line, thus avoiding another line with colnames(exportfile)-c(Contract,Price,Nbr.Lots) Is there a possibility to change my col names in the same line? TY Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ifelse
Hello, I need to print a screen message after a test. list-c(10,1,100,10) ifelse(all(list %in% c(1,10,100)==TRUE),cat(YES\n),cat(NO\n)) YES Error in ifelse(all(l %in% c(1, 10, 100) == TRUE), cat(YES\n), cat(NO\n)) : replacement has length zero I have the correct answer, YES, but with an Error. Why? TY for any help Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ifelse
TY much. Works for me. Arnaud Gaboury A2CT2 Ltd. -Original Message- From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com] Sent: jeudi 9 février 2012 17:47 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] ifelse cat() returns a null value so it's problematic inside ifelse. Here you probably need a regular if statement: if(all(list %in% c(1, 10, 100))) cat(YES\n) else cat(NO\n) # The == TRUE is redundant. Michael On Thu, Feb 9, 2012 at 11:37 AM, Arnaud Gaboury arnaud.gabo...@a2ct2.com wrote: Hello, I need to print a screen message after a test. list-c(10,1,100,10) ifelse(all(list %in% c(1,10,100)==TRUE),cat(YES\n),cat(NO\n)) YES Error in ifelse(all(l %in% c(1, 10, 100) == TRUE), cat(YES\n), cat(NO\n)) : replacement has length zero I have the correct answer, YES, but with an Error. Why? TY for any help Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] loading packages in a function
Hello,
I sourced successfully my function().
I need to load libraries, so I wrote this inside my function():
function()
{
#load needed library
library(plyr)
library(car)
/...
}
It is OK, but I have this on my invite command when running the function:
function()
Loading required package: MASS
Loading required package: nnet
YOU DID A GOOD JOB,SEND EMAIL
Last line is the supposed result of my function, so it ok.
How to get rid of the first two lines, only for esthetic purpose?
TY for your time.
Arnaud Gaboury
A2CT2 Ltd.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] loading packages in a function
TY for your answer, but here what i did :
#load needed lybrary
suppressPackageStartupMessages(library(plyr))
suppressPackageStartupMessages(library(car))
library(plyr)
library(car)
But I still get :
Loading required package: MASS
Loading required package: nnet
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: jeudi 9 février 2012 19:26
To: Arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] loading packages in a function
Perhaps the longest function in base R I know of:
suppressPackageStartupMessages
e.g.,
suppressPackageStartupMessages(library(zoo))
Michael
On Thu, Feb 9, 2012 at 12:44 PM, Arnaud Gaboury arnaud.gabo...@a2ct2.com
wrote:
Hello,
I sourced successfully my function().
I need to load libraries, so I wrote this inside my function():
function()
{
#load needed library
library(plyr)
library(car)
/...
}
It is OK, but I have this on my invite command when running the function:
function()
Loading required package: MASS
Loading required package: nnet
YOU DID A GOOD JOB,SEND EMAIL
Last line is the supposed result of my function, so it ok.
How to get rid of the first two lines, only for esthetic purpose?
TY for your time.
Arnaud Gaboury
A2CT2 Ltd.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] loading packages in a function
Ah, I feel stupid!
OK, it works for me.
TY for your help
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: Sarah Goslee [mailto:sarah.gos...@gmail.com]
Sent: jeudi 9 février 2012 20:02
To: Arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] loading packages in a function
That's because you're loading the packages twice, once suppressing the messages
and once not.
You only need the first pair of lines.
On Thu, Feb 9, 2012 at 1:55 PM, Arnaud Gaboury arnaud.gabo...@a2ct2.com wrote:
TY for your answer, but here what i did :
#load needed lybrary
suppressPackageStartupMessages(library(plyr))
suppressPackageStartupMessages(library(car))
library(plyr)
library(car)
But I still get :
Loading required package: MASS
Loading required package: nnet
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: jeudi 9 février 2012 19:26
To: Arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] loading packages in a function
Perhaps the longest function in base R I know of:
suppressPackageStartupMessages
e.g.,
suppressPackageStartupMessages(library(zoo))
Michael
On Thu, Feb 9, 2012 at 12:44 PM, Arnaud Gaboury arnaud.gabo...@a2ct2.com
wrote:
Hello,
I sourced successfully my function().
I need to load libraries, so I wrote this inside my function():
function()
{
#load needed library
library(plyr)
library(car)
/...
}
It is OK, but I have this on my invite command when running the function:
function()
Loading required package: MASS
Loading required package: nnet
YOU DID A GOOD JOB,SEND EMAIL
Last line is the supposed result of my function, so it ok.
How to get rid of the first two lines, only for esthetic purpose?
TY for your time.
Arnaud Gaboury
--
Sarah Goslee
http://www.functionaldiversity.org
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] replace elements of a data frame
names1-recode(df$names,'BO'='BOO';'CL'='CLR';'C'='CC') df1-data.frame(names1,price) df1 names1 price 1BOO10 2 CC25 3CLR20 TY for the tip. Any possibility to write this in one single line? Arnaud Gaboury A2CT2 Ltd. -Original Message- From: Berend Hasselman [mailto:b...@xs4all.nl] Sent: mardi 7 février 2012 20:46 To: Arnaud Gaboury Cc: Jorge I Velez; r-help@r-project.org Subject: Re: [R] replace elements of a data frame On 07-02-2012, at 20:24, Arnaud Gaboury wrote: I did indeed have a look at recode(), and was able to replace, but an error warning : recode(names,BO,BOO,df) Warning message: In recode.default(names, BO, BOO, df) : Name(s) of vars duplicates with an object outside the dataFrame. df names price 1 BOO10 2 C25 3CL20 As you can see, BO has been replaced by BOO, but with a warning! library(car) names-c(BO,C,CL) price-c(10,25,20) df-data.frame(names,price) recode(df$names,'BO'='BOO'; 'CL'='CLO'; 'C'='CR') results in [1] BOO CR CLO Levels: BOO CLO CR Note the single quotes. Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] decimal number format as quarter
Hello, I have to deal with numbers with a decimal part as quarter, coming from two systems with different way to show decimals. I need to tell R these are in fact the same number. On one side my number are formatted this way : 2.2 , 2.4 and 2.6. On the other side, I have 2.25, 2.50 and 2.75. All numbers are in fact 2.1/4, 2.1/2, 2.3/4. How can I tell R 2.2 is 2.25, 2.4 is 2.50 and 2.6 is 2.75 ? TY for any help. Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] decimal number format as quarter
TY Jim, It do the trick. I was trying to play without success with the format() options. No simplest way so? Arnaud Gaboury A2CT2 Ltd. -Original Message- From: jim holtman [mailto:jholt...@gmail.com] Sent: mercredi 8 février 2012 15:36 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] decimal number format as quarter will this do it for you: x - c(2.2, 2.4, 2.6, 3.2, 3.4, 3.6) # get integer part x.i - as.integer(x) # get fractional part x.f - (x * 10) %% 10 # new result result - x.i + ifelse(x.f == 2 + , .25 + , ifelse(x.f == 4 + , .5 + , .75 + ) + ) result [1] 2.25 2.50 2.75 3.25 3.50 3.75 On Wed, Feb 8, 2012 at 9:12 AM, Arnaud Gaboury arnaud.gabo...@a2ct2.com wrote: Hello, I have to deal with numbers with a decimal part as quarter, coming from two systems with different way to show decimals. I need to tell R these are in fact the same number. On one side my number are formatted this way : 2.2 , 2.4 and 2.6. On the other side, I have 2.25, 2.50 and 2.75. All numbers are in fact 2.1/4, 2.1/2, 2.3/4. How can I tell R 2.2 is 2.25, 2.4 is 2.50 and 2.6 is 2.75 ? TY for any help. Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] decimal number format as quarter
David, You are not far indeed, as I trade commodities, and prices are the ones from grain market: Corn, wheat and Soybeans. They are quoted in 1/4, and my trading platform displays them in 2,4,6 and my statements are in 25,50,75. TY Arnaud Gaboury A2CT2 Ltd. -Original Message- From: David Reiner [mailto:david.rei...@xrtrading.com] Sent: mercredi 8 février 2012 15:48 To: Arnaud Gaboury; jim holtman Cc: r-help@r-project.org Subject: RE: [R] decimal number format as quarter Looks like something priced in eighths; we deal with similar notation for bonds and similar instruments. x - c(2.2, 2.4, 2.6, 3.2, 3.4, 3.6) as.integer(x)+10*(x-as.integer(x))/8 [1] 2.25 2.50 2.75 3.25 3.50 3.75 Adjust the 10 and 8 if you have other denominators. -- David -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Arnaud Gaboury Sent: Wednesday, February 08, 2012 8:45 AM To: jim holtman Cc: r-help@r-project.org Subject: Re: [R] decimal number format as quarter TY Jim, It do the trick. I was trying to play without success with the format() options. No simplest way so? Arnaud Gaboury A2CT2 Ltd. -Original Message- From: jim holtman [mailto:jholt...@gmail.com] Sent: mercredi 8 février 2012 15:36 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] decimal number format as quarter will this do it for you: x - c(2.2, 2.4, 2.6, 3.2, 3.4, 3.6) # get integer part x.i - as.integer(x) # get fractional part x.f - (x * 10) %% 10 # new result result - x.i + ifelse(x.f == 2 + , .25 + , ifelse(x.f == 4 + , .5 + , .75 + ) + ) result [1] 2.25 2.50 2.75 3.25 3.50 3.75 On Wed, Feb 8, 2012 at 9:12 AM, Arnaud Gaboury arnaud.gabo...@a2ct2.com wrote: Hello, I have to deal with numbers with a decimal part as quarter, coming from two systems with different way to show decimals. I need to tell R these are in fact the same number. On one side my number are formatted this way : 2.2 , 2.4 and 2.6. On the other side, I have 2.25, 2.50 and 2.75. All numbers are in fact 2.1/4, 2.1/2, 2.3/4. How can I tell R 2.2 is 2.25, 2.4 is 2.50 and 2.6 is 2.75 ? TY for any help. Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. This e-mail and any materials attached hereto, including, without limitation, all content hereof and thereof (collectively, XR Content) are confidential and proprietary to XR Trading, LLC (XR) and/or its affiliates, and are protected by intellectual property laws. Without the prior written consent of XR, the XR Content may not (i) be disclosed to any third party or (ii) be reproduced or otherwise used by anyone other than current employees of XR or its affiliates, on behalf of XR or its affiliates. THE XR CONTENT IS PROVIDED AS IS, WITHOUT REPRESENTATIONS OR WARRANTIES OF ANY KIND. TO THE MAXIMUM EXTENT PERMISSIBLE UNDER APPLICABLE LAW, XR HEREBY DISCLAIMS ANY AND ALL WARRANTIES, EXPRESS AND IMPLIED, RELATING TO THE XR CONTENT, AND NEITHER XR NOR ANY OF ITS AFFILIATES SHALL IN ANY EVENT BE LIABLE FOR ANY DAMAGES OF ANY NATURE WHATSOEVER, INCLUDING, BUT NOT LIMITED TO, DIRECT, INDIRECT, CONSEQUENTIAL, SPECIAL AND PUNITIVE DAMAGES, LOSS OF PROFITS AND TRADING LOSSES, RESULTING FROM ANY PERSON'S USE OR RELIANCE UPON, OR INABILITY TO USE, ANY XR CONTENT, EVEN IF XR IS ADVISED OF THE POSSIBILITY OF SUCH DAMAGES OR IF SUCH DAMAGES WERE FORESEEABLE. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] decimal number format as quarter
Thanks so much Peter. You are the man. Easy way and working for me. Arnaud Gaboury A2CT2 Ltd. -Original Message- From: peter dalgaard [mailto:pda...@gmail.com] Sent: mercredi 8 février 2012 16:15 To: David Reiner Cc: Arnaud Gaboury; jim holtman; r-help@r-project.org Subject: Re: [R] decimal number format as quarter On Feb 8, 2012, at 15:48 , David Reiner wrote: Looks like something priced in eighths; we deal with similar notation for bonds and similar instruments. x - c(2.2, 2.4, 2.6, 3.2, 3.4, 3.6) as.integer(x)+10*(x-as.integer(x))/8 [1] 2.25 2.50 2.75 3.25 3.50 3.75 Adjust the 10 and 8 if you have other denominators. For the case at hand, I expect that you can even get away with ceiling(x*4)/4 [1] 2.25 2.50 2.75 3.25 3.50 3.75 -- Peter Dalgaard, Professor Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] replace elements of a data frame
Hello, I am getting mad at finding a simple way to replace elements of a df. Here is a short df : names-c(BO,C,CL) price-c(10,25,20) df-data.frame(names,price) I want to replace BO by BOB, C by CR, CL by CLO, and the list is more long. I can do that for each element: df[df==BO]-BOB But my df is bigger indeed with other elements. I was thinking using replace(), but can't get any clean result ( NA or all elements replaced with only one), neither with sapply(). TY for any help, and sorry for the n00b question. Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] replace elements of a data frame
I did indeed have a look at recode(), and was able to replace, but an error warning : recode(names,BO,BOO,df) Warning message: In recode.default(names, BO, BOO, df) : Name(s) of vars duplicates with an object outside the dataFrame. df names price 1 BOO10 2 C25 3CL20 As you can see, BO has been replaced by BOO, but with a warning! Arnaud Gaboury A2CT2 Ltd. Trade: +41 22 849 88 63 Fax: +41 22 849 88 66 arnaud.gabo...@a2ct2.com This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. Access to this email by anyone else is unauthorized. If you are not the intended recipient, any disclosure, copying, distribution or any action taken or omitted to be taken in reliance on it, is prohibited and may be unlawful. If you have received this email in error please notify the sender. From: Jorge I Velez [mailto:jorgeivanve...@gmail.com] Sent: mardi 7 février 2012 19:55 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] replace elements of a data frame Hi Arnaud, Take a look at require(car) ?recode HTH, Jorge.- On Tue, Feb 7, 2012 at 1:05 PM, Arnaud Gaboury wrote: Hello, I am getting mad at finding a simple way to replace elements of a df. Here is a short df : names-c(BO,C,CL) price-c(10,25,20) df-data.frame(names,price) I want to replace BO by BOB, C by CR, CL by CLO, and the list is more long. I can do that for each element: df[df==BO]-BOB But my df is bigger indeed with other elements. I was thinking using replace(), but can't get any clean result ( NA or all elements replaced with only one), neither with sapply(). TY for any help, and sorry for the n00b question. Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] replace elements of a data frame
I used in fact recode() from epilcac package, not the one you mentioned! Arnaud Gaboury A2CT2 Ltd. -Original Message- From: Arnaud Gaboury Sent: mardi 7 février 2012 20:25 To: Jorge I Velez Cc: r-help@r-project.org; Arnaud Gaboury Subject: RE: [R] replace elements of a data frame I did indeed have a look at recode(), and was able to replace, but an error warning : recode(names,BO,BOO,df) Warning message: In recode.default(names, BO, BOO, df) : Name(s) of vars duplicates with an object outside the dataFrame. df names price 1 BOO10 2 C25 3CL20 As you can see, BO has been replaced by BOO, but with a warning! Arnaud Gaboury A2CT2 Ltd. From: Jorge I Velez [mailto:jorgeivanve...@gmail.com] Sent: mardi 7 février 2012 19:55 To: Arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] replace elements of a data frame Hi Arnaud, Take a look at require(car) ?recode HTH, Jorge.- On Tue, Feb 7, 2012 at 1:05 PM, Arnaud Gaboury wrote: Hello, I am getting mad at finding a simple way to replace elements of a df. Here is a short df : names-c(BO,C,CL) price-c(10,25,20) df-data.frame(names,price) I want to replace BO by BOB, C by CR, CL by CLO, and the list is more long. I can do that for each element: df[df==BO]-BOB But my df is bigger indeed with other elements. I was thinking using replace(), but can't get any clean result ( NA or all elements replaced with only one), neither with sapply(). TY for any help, and sorry for the n00b question. Arnaud Gaboury A2CT2 Ltd. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] merge two data frames
Dear list, here are my two data frames: av - structure(list(DESCRIPTION = c(COFFEE C Sep/10, COPPER Sep/10, CORN Dec/10, CRUDE OIL miNY Sep/10, GOLD Aug/10, HENRY HUB NATURAL GAS Sep/10, PALLADIUM Sep/10, SILVER Sep/10, SOYBEANS Nov/10, SPCL HIGH GRADE ZINC USD, SUGAR NO.11 Oct/10, WHEAT Sep/10), prix = c(-168.3, -1.62, -773.75, -78.75, -1168.3, -0.0916, -470.75, 1758.5, -975.25, 1964, -19.09, -605.75), pos = c(-1, 0, -2, -1, -1, 0, -1, 1, -1, 1, -1, -1), PL = c(-12.03, -31.68, -43.2, -70.49, -11.88, -95.04, -3.96, -35.64, -30.24, -12.5, -36.09, -4.32)), .Names = c(DESCRIPTION, prix, pos, PL), row.names = c(NA, 12L), class = data.frame) zz - structure(list(DESCRIPTION = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 13L, 11L, 12L), .Label = c(COFFEE C Sep/10, COPPER Sep/10, CORN Dec/10, CRUDE OIL miNY Sep/10, GOLD Aug/10, HENRY HUB NATURAL GAS Sep/10, PALLADIUM Sep/10, PRM HGH GD ALUMINIUM USD, SILVER Sep/10, SOYBEANS Nov/10, SPCL HIGH GRADE ZINC USD, SUGAR NO.11 Oct/10, WHEAT Sep/10), class = factor), pl = c(-8.22, 2.31, -47.25, -0.234, -12.7, -0.236, -2, 11.71000, 14.40001, -34.75, -10.75, 55, -0.668), PL = c(3075.001, -575.0003, 2362.5, 115.0002, 1270.000, 2360, 200, -292.7501, -720.0005, 1737.5, 537.5, -1375, 750.3998), POSITION = c(1, -1, 2, 2, 1, 2, -1, -1, 1, 2, 0, 0, 0), SETTLEMENT = c(167.4, 324.55, 390.75, 76.99, 1160.4, 4.718, 468.75, 2067.71, 1744.1, 978, 0, 0, 0)), .Names = c(DESCRIPTION, pl, PL, POSITION, SETTLEMENT), row.names = c(NA, -13L), class = data.frame) I am looking for one data frame with the column $PL=zz$PL+av$PL. I have been trying using the merge() function and its different arguments with no sucess. Any help is appreciated. Thank You. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] merge two data frames
TY Petr, it works. I will then replace NA by 0. 2010/8/4 Petr PIKAL petr.pi...@precheza.cz: Hi you tried OK result - merge(zz, av, by=DESCRIPTION, all=TRUE) and as you did not specify what to do when one value is NA here is one possible solution rowSums(cbind(result$PL.x, result$PL.y), na.rm=T) Regards Petr r-help-boun...@r-project.org napsal dne 04.08.2010 11:52:00: Dear list, here are my two data frames: av - structure(list(DESCRIPTION = c(COFFEE C Sep/10, COPPER Sep/10, CORN Dec/10, CRUDE OIL miNY Sep/10, GOLD Aug/10, HENRY HUB NATURAL GAS Sep/10, PALLADIUM Sep/10, SILVER Sep/10, SOYBEANS Nov/10, SPCL HIGH GRADE ZINC USD, SUGAR NO.11 Oct/10, WHEAT Sep/10), prix = c(-168.3, -1.62, -773.75, -78.75, -1168.3, -0.0916, -470.75, 1758.5, -975.25, 1964, -19.09, -605.75), pos = c(-1, 0, -2, -1, -1, 0, -1, 1, -1, 1, -1, -1), PL = c(-12.03, -31.68, -43.2, -70.49, -11.88, -95.04, -3.96, -35.64, -30.24, -12.5, -36.09, -4.32)), .Names = c(DESCRIPTION, prix, pos, PL), row.names = c(NA, 12L), class = data.frame) zz - structure(list(DESCRIPTION = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 13L, 11L, 12L), .Label = c(COFFEE C Sep/10, COPPER Sep/10, CORN Dec/10, CRUDE OIL miNY Sep/10, GOLD Aug/10, HENRY HUB NATURAL GAS Sep/10, PALLADIUM Sep/10, PRM HGH GD ALUMINIUM USD, SILVER Sep/10, SOYBEANS Nov/10, SPCL HIGH GRADE ZINC USD, SUGAR NO.11 Oct/10, WHEAT Sep/10), class = factor), pl = c(-8.22, 2.31, -47.25, -0.234, -12.7, -0.236, -2, 11.71000, 14.40001, -34.75, -10.75, 55, -0.668), PL = c(3075.001, -575.0003, 2362.5, 115.0002, 1270.000, 2360, 200, -292.7501, -720.0005, 1737.5, 537.5, -1375, 750.3998), POSITION = c(1, -1, 2, 2, 1, 2, -1, -1, 1, 2, 0, 0, 0), SETTLEMENT = c(167.4, 324.55, 390.75, 76.99, 1160.4, 4.718, 468.75, 2067.71, 1744.1, 978, 0, 0, 0)), .Names = c(DESCRIPTION, pl, PL, POSITION, SETTLEMENT), row.names = c(NA, -13L), class = data.frame) I am looking for one data frame with the column $PL=zz$PL+av$PL. I have been trying using the merge() function and its different arguments with no sucess. Any help is appreciated. Thank You. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- ** Arnaud Gaboury Mobile: +41 79 392 79 56 BBM: 255B488F __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] which function
Dear group, Here is a list of elements : l - c(100415, 100416, 100419, 100420, 100421, 100422, 100423, 100426, 100427, 100428, 100429, 100430, 100503, 100504, 100505, 100506, 100507, 100510, 100511, 100512, 100513, 100514, 100517, 100518, 100519, 100520, 100521, 100524, 100525, 100526, 100527, 100528, 100531) l[3:4] [1] 100419 100420 #result is fine l[(which(100420==l))-1:which(100420==l)] [1] 100419 100416 100415 #can't understand why I do not have same result than the above line Thank You for help __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] which function
The correct line is : l[(which(100420==l)-1):which(100420==l)] [1] 100419 100420 -Original Message- From: arnaud Gaboury [mailto:arnaud.gabo...@gmail.com] Sent: Wednesday, June 02, 2010 9:41 AM To: r-help@r-project.org Cc: 'arnaud Gaboury' Subject: which function Dear group, Here is a list of elements : l - c(100415, 100416, 100419, 100420, 100421, 100422, 100423, 100426, 100427, 100428, 100429, 100430, 100503, 100504, 100505, 100506, 100507, 100510, 100511, 100512, 100513, 100514, 100517, 100518, 100519, 100520, 100521, 100524, 100525, 100526, 100527, 100528, 100531) l[3:4] [1] 100419 100420 #result is fine l[(which(100420==l))-1:which(100420==l)] [1] 100419 100416 100415 #can't understand why I do not have same result than the above line Thank You for help __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation with zero rows
I do really think it is a very good idea.
TY
-Original Message-
From: h.wick...@gmail.com [mailto:h.wick...@gmail.com] On Behalf Of
Hadley Wickham
Sent: Wednesday, June 02, 2010 3:31 PM
To: arnaud Gaboury
Cc: Peter Ehlers; r-help@r-project.org; Prof Brian Ripley
Subject: Re: [R] data frame manipulation with zero rows
Hi Arnaud,
I've added this case to the set of test cases in plyr and it will be
fixed in the next version.
Hadley
On Tue, Jun 1, 2010 at 2:33 PM, arnaud Gaboury
arnaud.gabo...@gmail.com wrote:
Maybe not the cleanest way, but I create a fake data frame with one
row so
ddply() is happy!!
if (nrow(futures)==0) futures-data.frame(...)
-Original Message-
From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Sent: Tuesday, June 01, 2010 12:07 PM
To: arnaud Gaboury
Cc: 'Prof Brian Ripley'; r-help@r-project.org
Subject: Re: [R] data frame manipulation with zero rows
On 2010-06-01 1:53, arnaud Gaboury wrote:
Brian,
If I do understand correctly, I must use in my function something
else than
ddply() if I want to avoid any error each time my df has zero
rows?
Am I correct?
You could define a function to handle the zero-rows case:
f - function(x){
if(nrow(x) 1) out - x[, c(1,3,2)] # or whatever
else
out - ddply(x, c(DESCRIPTION,SETTLEMENT), summarise,
POSITION=sum(QUANTITY))[,c(1,3,2)]
out
}
f(futures)
-Peter Ehlers
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: Tuesday, June 01, 2010 9:47 AM
To: arnaud Gaboury
Subject: Re: [R] data frame manipulation with zero rows
On Tue, 1 Jun 2010, arnaud Gaboury wrote:
Dear group,
Here is the kind of data.frame I obtain every day with my
function
:
futures-
structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10,
CORN Jul/10, CORN Jul/10, CORN Jul/10, LIVE CATTLE
Aug/10,
LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10, SUGAR NO.11
Jul/10,
SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10
), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406,
18407, 18408, 18406, 18407, 18407, 18407, 18407), class =
Date),
QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT
=
c(373.2500,
373.2500, 373.2500, 373.2500, 373.2500, 90.7750,
90.7750, 14.9200, 14.9200, 14.9200, 14.9200,
14.9200
)), .Names = c(DESCRIPTION, CREATED.DATE, QUANTITY,
SETTLEMENT), row.names = c(NA, 12L), class = data.frame)
I need then to apply to the df this following code line :
PosFut=ddply(futures, c(DESCRIPTION,SETTLEMENT), summarise,
POSITION=
sum(QUANTITY))[,c(1,3,2)]
It works perfectly in most of case, BUT I have a new problem: it
can
sometime occurs that my df futures is empty, with zero rows.
futures-
structure(list(DESCRIPTION = character(0), CREATED.DATE =
structure(numeric(0), class = Date),
QUANTITY = numeric(0), SETTLEMENT = character(0)), .Names =
c(DESCRIPTION,
CREATED.DATE, QUANTITY, SETTLEMENT), row.names =
integer(0),
class =
data.frame)
It is not the usual case, but it can happen. With this df, when
I
pass the
above mentione line, I get an error :
PosFut=ddply(futures, c(DESCRIPTION,SETTLEMENT), summarise,
POSITION=
sum(QUANTITY))[,c(1,3,2)]
Error in tapply(1:nrow(data), splitv, list) :
arguments must have same length
How can I avoid this when my df is empty?
Ask the author of the (missing) function ddply() to correct the
error
of using 1:nrow(data) by replacing it by seq_len(nrow(data)).
It's helpful to give example code, but much more helpful if you
test
it: yours cannot work without the function ddply() -- this is
what
'self-contained' means in the footer here.
Any help is appreciated
__
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PLEASE do read the posting guide http://www.R-
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code.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics,
http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
__
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PLEASE do read the posting guide http://www.R-project.org/posting-
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and provide commented, minimal, self-contained, reproducible code.
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz
[R] bind select data frames
Dear group, Here is my environment: ls() [1] DailyPL100419 DailyPL100420 DailyPL100421 ddi l PLglobal Pos100416 Pos100419 Pos100420 Pos100421 position [13] resultsel Trad100416Trad100419 Trad100420Trad100421trade With sel the following element : sel - c(100419, 100420, 100421) DailyPL100419 , DailyPL100420,DailyPL100421 are all data frames with same columns names. I want to rbind them with this condition : for (i in sel[-1]) I have no idea how to write it. TY for any help __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bind select data frames
Yes indeed I just want to rbind the data frames with numbers belonging to
sel, and if possible, avoid a loop.
To be more precise, if sel[-1]-c(100420, 100421), I look for a line which
will give me this result :
dd-rbind(DailyPL100420,DailyPL100421)
-Original Message-
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
Sent: Wednesday, June 02, 2010 4:48 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] bind select data frames
Hello,
Does this do what you are looking for?
output - NULL
for(i in paste(DailyPL, sel, sep=)[-1]){
output - rbind(output, get(i))
}
If you just want to rbind all the data frames, there are ways of doing
it without a loop too, but since you specifically asked with the
condition for(i in sep[-1]) I did it using a loop.
Best regards,
Josh
On Wed, Jun 2, 2010 at 7:24 AM, arnaud Gaboury
arnaud.gabo...@gmail.com wrote:
Dear group,
Here is my environment:
ls()
[1] DailyPL100419 DailyPL100420 DailyPL100421 dd
i
l PLglobal Pos100416 Pos100419
Pos100420
Pos100421 position
[13] resultsel Trad100416Trad100419
Trad100420Trad100421trade
With sel the following element :
sel -
c(100419, 100420, 100421)
DailyPL100419 , DailyPL100420,DailyPL100421 are all data frames
with
same columns names. I want to rbind them with this condition :
for (i in sel[-1])
I have no idea how to write it.
TY for any help
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Senior in Psychology
University of California, Riverside
http://www.joshuawiley.com/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] bind select data frames
I am working with something like this :
for (i in sel[-1]) {
dd-data.frame(do.call(rbind,
mget(paste(DailyPL,i,sep=),envir=.GlobalEnv)),row.names=NULL)
}
But the result is not good. In the case where sel[-1]-c(100420, 100421),
dd is only equal to DailyPL100421
-Original Message-
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
Sent: Wednesday, June 02, 2010 4:48 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] bind select data frames
Hello,
Does this do what you are looking for?
output - NULL
for(i in paste(DailyPL, sel, sep=)[-1]){
output - rbind(output, get(i))
}
If you just want to rbind all the data frames, there are ways of doing
it without a loop too, but since you specifically asked with the
condition for(i in sep[-1]) I did it using a loop.
Best regards,
Josh
On Wed, Jun 2, 2010 at 7:24 AM, arnaud Gaboury
arnaud.gabo...@gmail.com wrote:
Dear group,
Here is my environment:
ls()
[1] DailyPL100419 DailyPL100420 DailyPL100421 dd
i
l PLglobal Pos100416 Pos100419
Pos100420
Pos100421 position
[13] resultsel Trad100416Trad100419
Trad100420Trad100421trade
With sel the following element :
sel -
c(100419, 100420, 100421)
DailyPL100419 , DailyPL100420,DailyPL100421 are all data frames
with
same columns names. I want to rbind them with this condition :
for (i in sel[-1])
I have no idea how to write it.
TY for any help
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Senior in Psychology
University of California, Riverside
http://www.joshuawiley.com/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] bind select data frames
Yes Joshua, it does the trick. Once more, I am surprised by the easiness and
obviousness of R.
Now, I would like to add argument row.names=NULL.
dd - do.call(rbind, mget(paste(DailyPL,sel[-1], sep=),
envir=.GlobalEnv),row.names=NULL)
Error in do.call(rbind, mget(paste(DailyPL, sel[-1], sep = ), envir =
.GlobalEnv), :
unused argument(s) (row.names = NULL)
Why this error?
TY for your help
-Original Message-
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
Sent: Wednesday, June 02, 2010 5:08 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] bind select data frames
This should do the trick:
dd - do.call(rbind, mget(paste(DailyPL,sel[-1], sep=),
envir=.GlobalEnv))
On Wed, Jun 2, 2010 at 8:06 AM, arnaud Gaboury
arnaud.gabo...@gmail.com wrote:
I am working with something like this :
for (i in sel[-1]) {
dd-data.frame(do.call(rbind,
mget(paste(DailyPL,i,sep=),envir=.GlobalEnv)),row.names=NULL)
}
But the result is not good. In the case where sel[-1]-c(100420,
100421), dd is only equal to DailyPL100421
-Original Message-
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
Sent: Wednesday, June 02, 2010 4:48 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] bind select data frames
Hello,
Does this do what you are looking for?
output - NULL
for(i in paste(DailyPL, sel, sep=)[-1]){
output - rbind(output, get(i))
}
If you just want to rbind all the data frames, there are ways of
doing
it without a loop too, but since you specifically asked with the
condition for(i in sep[-1]) I did it using a loop.
Best regards,
Josh
On Wed, Jun 2, 2010 at 7:24 AM, arnaud Gaboury
arnaud.gabo...@gmail.com wrote:
Dear group,
Here is my environment:
ls()
[1] DailyPL100419 DailyPL100420 DailyPL100421 dd
i
l PLglobal Pos100416 Pos100419
Pos100420
Pos100421 position
[13] resultsel Trad100416Trad100419
Trad100420Trad100421trade
With sel the following element :
sel -
c(100419, 100420, 100421)
DailyPL100419 , DailyPL100420,DailyPL100421 are all data
frames
with
same columns names. I want to rbind them with this condition :
for (i in sel[-1])
I have no idea how to write it.
TY for any help
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Senior in Psychology
University of California, Riverside
http://www.joshuawiley.com/
--
Joshua Wiley
Senior in Psychology
University of California, Riverside
http://www.joshuawiley.com/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] bind select data frames
Here we go :
dd-data.frame(do.call(rbind,
mget(paste(DailyPL,sel[-1],sep=),envir=.GlobalEnv)),row.names=NULL)
TY so much Joshua and Jorge.
-Original Message-
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
Sent: Wednesday, June 02, 2010 5:11 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] bind select data frames
On Wed, Jun 2, 2010 at 8:06 AM, arnaud Gaboury
arnaud.gabo...@gmail.com wrote:
I am working with something like this :
for (i in sel[-1]) {
dd-data.frame(do.call(rbind,
mget(paste(DailyPL,i,sep=),envir=.GlobalEnv)),row.names=NULL)
}
But the result is not good. In the case where sel[-1]-c(100420,
100421), dd is only equal to DailyPL100421
Yes, because it is in a loop, dd is reassigned for each value of i.
That is why in the first loop I sent you I included the old data in
the rbind, otherwise it is just overwritten everytime it loops.
Josh
-Original Message-
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
Sent: Wednesday, June 02, 2010 4:48 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] bind select data frames
Hello,
Does this do what you are looking for?
output - NULL
for(i in paste(DailyPL, sel, sep=)[-1]){
output - rbind(output, get(i))
}
If you just want to rbind all the data frames, there are ways of
doing
it without a loop too, but since you specifically asked with the
condition for(i in sep[-1]) I did it using a loop.
Best regards,
Josh
On Wed, Jun 2, 2010 at 7:24 AM, arnaud Gaboury
arnaud.gabo...@gmail.com wrote:
Dear group,
Here is my environment:
ls()
[1] DailyPL100419 DailyPL100420 DailyPL100421 dd
i
l PLglobal Pos100416 Pos100419
Pos100420
Pos100421 position
[13] resultsel Trad100416Trad100419
Trad100420Trad100421trade
With sel the following element :
sel -
c(100419, 100420, 100421)
DailyPL100419 , DailyPL100420,DailyPL100421 are all data
frames
with
same columns names. I want to rbind them with this condition :
for (i in sel[-1])
I have no idea how to write it.
TY for any help
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Senior in Psychology
University of California, Riverside
http://www.joshuawiley.com/
--
Joshua Wiley
Senior in Psychology
University of California, Riverside
http://www.joshuawiley.com/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] bind select data frames
Jorge,
Your line works and give the desired result. Now I need to be able to work
with i instead of 100419..., as I need to be able to change these numbers.
TY for your help
From: Jorge Ivan Velez [mailto:jorgeivanve...@gmail.com]
Sent: Wednesday, June 02, 2010 5:09 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] bind select data frames
Hi Arnaud,
Try the following (untested):
txt - paste('DailyPL',c(100419, 100420, 100421), sep = )
do.call(rbind, lapply(txt, get))
HTH,
Jorge
On Wed, Jun 2, 2010 at 10:24 AM, arnaud Gaboury wrote:
Dear group,
Here is my environment:
ls()
[1] DailyPL100419 DailyPL100420 DailyPL100421 dd i
l PLglobal Pos100416 Pos100419 Pos100420
Pos100421 position
[13] result sel Trad100416 Trad100419
Trad100420 Trad100421 trade
With sel the following element :
sel -
c(100419, 100420, 100421)
DailyPL100419 , DailyPL100420,DailyPL100421 are all data frames with
same columns names. I want to rbind them with this condition :
for (i in sel[-1])
I have no idea how to write it.
TY for any help
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] data frame manipulation with zero rows
Dear group, Here is the kind of data.frame I obtain every day with my function : futures - structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10, CORN Jul/10, CORN Jul/10, CORN Jul/10, LIVE CATTLE Aug/10, LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10 ), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406, 18407, 18408, 18406, 18407, 18407, 18407, 18407), class = Date), QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT = c(373.2500, 373.2500, 373.2500, 373.2500, 373.2500, 90.7750, 90.7750, 14.9200, 14.9200, 14.9200, 14.9200, 14.9200 )), .Names = c(DESCRIPTION, CREATED.DATE, QUANTITY, SETTLEMENT), row.names = c(NA, 12L), class = data.frame) I need then to apply to the df this following code line : PosFut=ddply(futures, c(DESCRIPTION,SETTLEMENT), summarise, POSITION= sum(QUANTITY))[,c(1,3,2)] It works perfectly in most of case, BUT I have a new problem: it can sometime occurs that my df futures is empty, with zero rows. futures - structure(list(DESCRIPTION = character(0), CREATED.DATE = structure(numeric(0), class = Date), QUANTITY = numeric(0), SETTLEMENT = character(0)), .Names = c(DESCRIPTION, CREATED.DATE, QUANTITY, SETTLEMENT), row.names = integer(0), class = data.frame) It is not the usual case, but it can happen. With this df, when I pass the above mentione line, I get an error : PosFut=ddply(futures, c(DESCRIPTION,SETTLEMENT), summarise, POSITION= sum(QUANTITY))[,c(1,3,2)] Error in tapply(1:nrow(data), splitv, list) : arguments must have same length How can I avoid this when my df is empty? Any help is appreciated __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation with zero rows
Brian, If I do understand correctly, I must use in my function something else than ddply() if I want to avoid any error each time my df has zero rows? Am I correct? TY -Original Message- From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk] Sent: Tuesday, June 01, 2010 9:47 AM To: arnaud Gaboury Subject: Re: [R] data frame manipulation with zero rows On Tue, 1 Jun 2010, arnaud Gaboury wrote: Dear group, Here is the kind of data.frame I obtain every day with my function : futures - structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10, CORN Jul/10, CORN Jul/10, CORN Jul/10, LIVE CATTLE Aug/10, LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10 ), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406, 18407, 18408, 18406, 18407, 18407, 18407, 18407), class = Date), QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT = c(373.2500, 373.2500, 373.2500, 373.2500, 373.2500, 90.7750, 90.7750, 14.9200, 14.9200, 14.9200, 14.9200, 14.9200 )), .Names = c(DESCRIPTION, CREATED.DATE, QUANTITY, SETTLEMENT), row.names = c(NA, 12L), class = data.frame) I need then to apply to the df this following code line : PosFut=ddply(futures, c(DESCRIPTION,SETTLEMENT), summarise, POSITION= sum(QUANTITY))[,c(1,3,2)] It works perfectly in most of case, BUT I have a new problem: it can sometime occurs that my df futures is empty, with zero rows. futures - structure(list(DESCRIPTION = character(0), CREATED.DATE = structure(numeric(0), class = Date), QUANTITY = numeric(0), SETTLEMENT = character(0)), .Names = c(DESCRIPTION, CREATED.DATE, QUANTITY, SETTLEMENT), row.names = integer(0), class = data.frame) It is not the usual case, but it can happen. With this df, when I pass the above mentione line, I get an error : PosFut=ddply(futures, c(DESCRIPTION,SETTLEMENT), summarise, POSITION= sum(QUANTITY))[,c(1,3,2)] Error in tapply(1:nrow(data), splitv, list) : arguments must have same length How can I avoid this when my df is empty? Ask the author of the (missing) function ddply() to correct the error of using 1:nrow(data) by replacing it by seq_len(nrow(data)). It's helpful to give example code, but much more helpful if you test it: yours cannot work without the function ddply() -- this is what 'self-contained' means in the footer here. Any help is appreciated __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data frame manipulation ddply
Dear group, Here is my data frame: futures - structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10, CORN Jul/10, CORN Jul/10, CORN Jul/10, LIVE CATTLE Aug/10, LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10 ), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406, 18407, 18408, 18406, 18407, 18407, 18407, 18407), class = Date), QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT = c(373.2500, 373.2500, 373.2500, 373.2500, 373.2500, 90.7750, 90.7750, 14.9200, 14.9200, 14.9200, 14.9200, 14.9200 )), .Names = c(DESCRIPTION, CREATED.DATE, QUANTITY, SETTLEMENT), row.names = c(NA, 12L), class = data.frame) Here is the line I pass : PosFut=ddply(futures, c(DESCRIPTION,SETTLEMENT), summarise, POSITION= sum(QUANTITY))[,c(1,3,2)] And here the result : PosFut - structure(list(DESCRIPTION = structure(1:3, .Label = c(CORN Jul/10, LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10), class = factor), POSITION = c(5, 4, 5), SETTLEMENT = structure(c(2L, 3L, 1L ), .Label = c(14.9200, 373.2500, 90.7750), class = factor)), .Names = c(DESCRIPTION, POSITION, SETTLEMENT), class = data.frame, row.names = c(NA, -3L)) I can no more use ddply, as this above command line is in a function, and this line should be able to work with a data frame with zero rows, and in this case ddply doesn't work. Any suggestion how to obtain the same result without ddply? TY for any help. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation ddply
Patrick, When apply to this following df : futures - structure(list(DESCRIPTION = character(0), CREATED.DATE = structure(numeric(0), class = Date), QUANTITY = numeric(0), SETTLEMENT = character(0)), .Names = c(DESCRIPTION, CREATED.DATE, QUANTITY, SETTLEMENT), row.names = integer(0), class = data.frame) PosFut - aggregate(futures$QUANTITY, list(DESCRIPTION = futures$DESCRIPTION,SETTLEMENT=futures$SETTLEMENT),sum)[,c(1,3,2)] Error in aggregate.data.frame(as.data.frame(x), ...) : no rows to aggregate -Original Message- From: Patrick Hausmann [mailto:patrick.hausm...@uni-bremen.de] Sent: Tuesday, June 01, 2010 11:38 AM To: arnaud Gaboury Subject: Re: [R] data frame manipulation ddply Hi Arnaud, maybe aggregate can help: PosFut - aggregate(futures$QUANTITY, list(DESCRIPTION = futures$DESCRIPTION, SETTLEMENT = futures$SETTLEMENT), sum)[, c(1,3,2)] HTH, Patrick Am 01.06.2010 11:02, schrieb arnaud Gaboury: Dear group, Here is my data frame: futures- structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10, CORN Jul/10, CORN Jul/10, CORN Jul/10, LIVE CATTLE Aug/10, LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10 ), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406, 18407, 18408, 18406, 18407, 18407, 18407, 18407), class = Date), QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT = c(373.2500, 373.2500, 373.2500, 373.2500, 373.2500, 90.7750, 90.7750, 14.9200, 14.9200, 14.9200, 14.9200, 14.9200 )), .Names = c(DESCRIPTION, CREATED.DATE, QUANTITY, SETTLEMENT), row.names = c(NA, 12L), class = data.frame) Here is the line I pass : PosFut=ddply(futures, c(DESCRIPTION,SETTLEMENT), summarise, POSITION= sum(QUANTITY))[,c(1,3,2)] And here the result : PosFut- structure(list(DESCRIPTION = structure(1:3, .Label = c(CORN Jul/10, LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10), class = factor), POSITION = c(5, 4, 5), SETTLEMENT = structure(c(2L, 3L, 1L ), .Label = c(14.9200, 373.2500, 90.7750), class = factor)), .Names = c(DESCRIPTION, POSITION, SETTLEMENT), class = data.frame, row.names = c(NA, -3L)) I can no more use ddply, as this above command line is in a function, and this line should be able to work with a data frame with zero rows, and in this case ddply doesn't work. Any suggestion how to obtain the same result without ddply? TY for any help. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation with zero rows
It is indeed ddply() from package plyr.
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: Tuesday, June 01, 2010 12:24 PM
To: Peter Ehlers
Cc: arnaud Gaboury; r-help@r-project.org
Subject: Re: [R] data frame manipulation with zero rows
On Tue, 1 Jun 2010, Peter Ehlers wrote:
On 2010-06-01 1:53, arnaud Gaboury wrote:
Brian,
If I do understand correctly, I must use in my function something
else than
ddply() if I want to avoid any error each time my df has zero rows?
Am I correct?
You could define a function to handle the zero-rows case:
f - function(x){
if(nrow(x) 1) out - x[, c(1,3,2)] # or whatever
else
out - ddply(x, c(DESCRIPTION,SETTLEMENT), summarise,
POSITION=sum(QUANTITY))[,c(1,3,2)]
out
}
f(futures)
Or simply fix ddply. We don't know what that is or what it should do
for the case of zero rows: it may or may not be the one in package
plyr.
-Peter Ehlers
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: Tuesday, June 01, 2010 9:47 AM
To: arnaud Gaboury
Subject: Re: [R] data frame manipulation with zero rows
On Tue, 1 Jun 2010, arnaud Gaboury wrote:
Dear group,
Here is the kind of data.frame I obtain every day with my function
:
futures-
structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10,
CORN Jul/10, CORN Jul/10, CORN Jul/10, LIVE CATTLE Aug/10,
LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10,
SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10
), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406,
18407, 18408, 18406, 18407, 18407, 18407, 18407), class = Date),
QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT =
c(373.2500,
373.2500, 373.2500, 373.2500, 373.2500, 90.7750,
90.7750, 14.9200, 14.9200, 14.9200, 14.9200,
14.9200
)), .Names = c(DESCRIPTION, CREATED.DATE, QUANTITY,
SETTLEMENT), row.names = c(NA, 12L), class = data.frame)
I need then to apply to the df this following code line :
PosFut=ddply(futures, c(DESCRIPTION,SETTLEMENT), summarise,
POSITION=
sum(QUANTITY))[,c(1,3,2)]
It works perfectly in most of case, BUT I have a new problem: it
can
sometime occurs that my df futures is empty, with zero rows.
futures-
structure(list(DESCRIPTION = character(0), CREATED.DATE =
structure(numeric(0), class = Date),
QUANTITY = numeric(0), SETTLEMENT = character(0)), .Names =
c(DESCRIPTION,
CREATED.DATE, QUANTITY, SETTLEMENT), row.names = integer(0),
class =
data.frame)
It is not the usual case, but it can happen. With this df, when I
pass the
above mentione line, I get an error :
PosFut=ddply(futures, c(DESCRIPTION,SETTLEMENT), summarise,
POSITION=
sum(QUANTITY))[,c(1,3,2)]
Error in tapply(1:nrow(data), splitv, list) :
arguments must have same length
How can I avoid this when my df is empty?
Ask the author of the (missing) function ddply() to correct the
error
of using 1:nrow(data) by replacing it by seq_len(nrow(data)).
It's helpful to give example code, but much more helpful if you
test
it: yours cannot work without the function ddply() -- this is what
'self-contained' means in the footer here.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] as.date
Dear group, Here is my df (obtained with a read.csv2()): df - structure(list(DESCRIPTION = c(COTTON NO.2 Jul/10, COTTON NO.2 Jul/10, PALLADIUM Jun/10, PALLADIUM Jun/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10), CREATED.DATE = c(13/05/2010, 13/05/2010, 14/05/2010, 14/05/2010, 10/05/2010, 10/05/2010), QUANITY = c(1, 1, -1, -1, 1, 1), CLOSING.PRICE = c(81.2000, 81.2000, 503.6000, 503.6000, 13.8900, 13.8900)), .Names = c(DESCRIPTION, CREATED.DATE, QUANITY, CLOSING.PRICE), row.names = c(NA, 6L), class = data.frame) str(df) 'data.frame': 6 obs. of 4 variables: $ DESCRIPTION : chr COTTON NO.2 Jul/10 COTTON NO.2 Jul/10 PALLADIUM Jun/10 PALLADIUM Jun/10 ... $ CREATED.DATE : chr 13/05/2010 13/05/2010 14/05/2010 14/05/2010 ... $ QUANITY : num 1 1 -1 -1 1 1 $ CLOSING.PRICE: chr 81.2000 81.2000 503.6000 503.6000 ... I want to change the class of df$CREATED.DATE from Chr to Date: pose$CREATED.DATE=as.Date(pose$CREATED.DATE,%d/%m/%y) Here is what I get : df - structure(list(DESCRIPTION = c(COTTON NO.2 Jul/10, COTTON NO.2 Jul/10, PALLADIUM Jun/10, PALLADIUM Jun/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10), CREATED.DATE = structure(c(18395, 18395, 18396, 18396, 18392, 18392), class = Date), QUANITY = c(1, 1, -1, -1, 1, 1), CLOSING.PRICE = c(81.2000, 81.2000, 503.6000, 503.6000, 13.8900, 13.8900)), .Names = c(DESCRIPTION, CREATED.DATE, QUANITY, CLOSING.PRICE), row.names = c(NA, 6L), class = data.frame) Where does the problem comes from?? Maybe from my sytem date ?? TY for any help __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] as.date
TY for the tip. The lower case is in fact the culprit. -Original Message- From: Erik Iverson [mailto:er...@ccbr.umn.edu] Sent: Tuesday, June 01, 2010 6:05 PM To: arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] as.date Where does the problem comes from?? Maybe from my sytem date ?? Simply from not reading the options carefully enough, from ?strptime, '%y' Year without century (00-99). If you use this on input, which century you get is system-specific. So don't! Most often values 00 to 68 are prefixed by 20 and 69 to 99 by 19 - that is the behaviour specified by the 2001 POSIX standard, but it does also say 'it is expected that in a future version of IEEE Std 1003.1-2001 the default century inferred from a 2-digit year will change'. '%Y' Year with century. You want %Y, not %y. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation with zero rows
Maybe not the cleanest way, but I create a fake data frame with one row so
ddply() is happy!!
if (nrow(futures)==0) futures-data.frame(...)
-Original Message-
From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Sent: Tuesday, June 01, 2010 12:07 PM
To: arnaud Gaboury
Cc: 'Prof Brian Ripley'; r-help@r-project.org
Subject: Re: [R] data frame manipulation with zero rows
On 2010-06-01 1:53, arnaud Gaboury wrote:
Brian,
If I do understand correctly, I must use in my function something
else than
ddply() if I want to avoid any error each time my df has zero rows?
Am I correct?
You could define a function to handle the zero-rows case:
f - function(x){
if(nrow(x) 1) out - x[, c(1,3,2)] # or whatever
else
out - ddply(x, c(DESCRIPTION,SETTLEMENT), summarise,
POSITION=sum(QUANTITY))[,c(1,3,2)]
out
}
f(futures)
-Peter Ehlers
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: Tuesday, June 01, 2010 9:47 AM
To: arnaud Gaboury
Subject: Re: [R] data frame manipulation with zero rows
On Tue, 1 Jun 2010, arnaud Gaboury wrote:
Dear group,
Here is the kind of data.frame I obtain every day with my function
:
futures-
structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10,
CORN Jul/10, CORN Jul/10, CORN Jul/10, LIVE CATTLE Aug/10,
LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10,
SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10
), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406,
18407, 18408, 18406, 18407, 18407, 18407, 18407), class = Date),
QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT =
c(373.2500,
373.2500, 373.2500, 373.2500, 373.2500, 90.7750,
90.7750, 14.9200, 14.9200, 14.9200, 14.9200,
14.9200
)), .Names = c(DESCRIPTION, CREATED.DATE, QUANTITY,
SETTLEMENT), row.names = c(NA, 12L), class = data.frame)
I need then to apply to the df this following code line :
PosFut=ddply(futures, c(DESCRIPTION,SETTLEMENT), summarise,
POSITION=
sum(QUANTITY))[,c(1,3,2)]
It works perfectly in most of case, BUT I have a new problem: it
can
sometime occurs that my df futures is empty, with zero rows.
futures-
structure(list(DESCRIPTION = character(0), CREATED.DATE =
structure(numeric(0), class = Date),
QUANTITY = numeric(0), SETTLEMENT = character(0)), .Names =
c(DESCRIPTION,
CREATED.DATE, QUANTITY, SETTLEMENT), row.names = integer(0),
class =
data.frame)
It is not the usual case, but it can happen. With this df, when I
pass the
above mentione line, I get an error :
PosFut=ddply(futures, c(DESCRIPTION,SETTLEMENT), summarise,
POSITION=
sum(QUANTITY))[,c(1,3,2)]
Error in tapply(1:nrow(data), splitv, list) :
arguments must have same length
How can I avoid this when my df is empty?
Ask the author of the (missing) function ddply() to correct the
error
of using 1:nrow(data) by replacing it by seq_len(nrow(data)).
It's helpful to give example code, but much more helpful if you test
it: yours cannot work without the function ddply() -- this is what
'self-contained' means in the footer here.
Any help is appreciated
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation change elements meeting criteria
Thank you for the answer.
Is there any way to combine if() and switch() in one line? In my case,
something like :
if(trade$Trade.Status==DEL)switch(.)
I would like to avoid the loop .
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Wednesday, May 26, 2010 9:15 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
see ?switch
X- rep(c(Buy,Sell,something else),each=5)
Y- rep(c(DEL,INS,DEL),5)
new.vect - X
for (i in which(Y==DEL)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy=Sell,
X[i])
}
cbind(new.vect,X,Y)
On Wed, May 26, 2010 at 7:43 PM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Dear group,
Here is my df :
trade -
structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name =
c(SUGAR NO.11,
CORN, CORN), Delivery.Prompt.Date = c(Jul/10, Jul/10,
Jul/10), Buy.Sell..Cleared. = c(Sell, Buy, Buy), Volume = c(1L,
2L, 1L), Price = c(15.2500, 368., 368.5000), Net.Charges..sum. =
c(4.01,
-8.64, -4.32)), .Names = c(Trade.Status, Instrument.Long.Name,
Delivery.Prompt.Date, Buy.Sell..Cleared., Volume, Price,
Net.Charges..sum.), row.names = c(NA, 3L), class = data.frame)
Here is what I want :
If trade$Trade.Status==DEL: then if trade$buy.Sell..Cleared==Sell , change
it to Buy, if trade$buy.Sell..Cleared==Buy, change it to Sell.
If trade$Trade.Status==INS, do nothing
I tried to work around with ifelse, but don't know how to deal with so many
conditions.
Any help is appreciated.
TY
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] switch function
Dear group, Here is my df : trades - structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name = c(SUGAR NO.11, CORN, CORN), Delivery.Prompt.Date = c(Jul/10, Jul/10, Jul/10), Buy.Sell..Cleared. = c(Sell, Buy, Buy), Volume = c(1L, 2L, 1L), Price = c(15.2500, 368., 368.5000), Net.Charges..sum. = c(4.01, -8.64, -4.32)), .Names = c(Trade.Status, Instrument.Long.Name, Delivery.Prompt.Date, Buy.Sell..Cleared., Volume, Price, Net.Charges..sum.), row.names = c(NA, 3L), class = data.frame) I want to replace Buy by Sell and Sell by Buy in column Buy.Sell..Cleared. when element in column Trade.Status is equal to DEL. I think I can write something like this : tradesnew-sapply(trades$Buy.Sell..Cleared[which(trades$Buy.Sell..Cleared== DEL),],switch,...) but I don't really know how to pass further arguments to the switch function. Any help is appreciated. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data frame change elements meeting criteria
If i pass this line :
tradesnew-sapply(trades[which(trades$Trade.Status==DEL),],switch,Sell=B
uy,Buy=Sell)
Here is what I get :
tradesnew
$Trade.Status
NULL
$Instrument.Long.Name
NULL
$Delivery.Prompt.Date
NULL
$Buy.Sell..Cleared.
[1] Buy
$Volume
[1] Buy
$Price
NULL
$Net.Charges..sum.
NULL
That's certainly not what I want.
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 8:43 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
The loop is due to the switch statement, not the condition. Without
condition it would become:
for (i in 1:length(Y)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy=Sell,
X[i])
}
You can make an sapply construct too off course :
new.vect - sapply(X[which(Y==DEL)],switch,Sell=Buy,Buy=Sell)
This will speed up things a little bit, but the effect is marginal.
Cheers
Joris
On Thu, May 27, 2010 at 8:33 AM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Thank you for the answer.
Is there any way to combine if() and switch() in one line? In my case,
something like :
if(trade$Trade.Status==DEL)switch(.)
I would like to avoid the loop .
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Wednesday, May 26, 2010 9:15 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
see ?switch
X- rep(c(Buy,Sell,something else),each=5)
Y- rep(c(DEL,INS,DEL),5)
new.vect - X
for (i in which(Y==DEL)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy=Sell,
X[i])
}
cbind(new.vect,X,Y)
On Wed, May 26, 2010 at 7:43 PM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Dear group,
Here is my df :
trade -
structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name =
c(SUGAR NO.11,
CORN, CORN), Delivery.Prompt.Date = c(Jul/10, Jul/10,
Jul/10), Buy.Sell..Cleared. = c(Sell, Buy, Buy), Volume = c(1L,
2L, 1L), Price = c(15.2500, 368., 368.5000), Net.Charges..sum. =
c(4.01,
-8.64, -4.32)), .Names = c(Trade.Status, Instrument.Long.Name,
Delivery.Prompt.Date, Buy.Sell..Cleared., Volume, Price,
Net.Charges..sum.), row.names = c(NA, 3L), class = data.frame)
Here is what I want :
If trade$Trade.Status==DEL: then if trade$buy.Sell..Cleared==Sell , change
it to Buy, if trade$buy.Sell..Cleared==Buy, change it to Sell.
If trade$Trade.Status==INS, do nothing
I tried to work around with ifelse, but don't know how to deal with so many
conditions.
Any help is appreciated.
TY
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation change elements meeting criteria
Joris,
If i pass this line :
tradesnew-sapply(trades[which(trades$Trade.Status==DEL),],switch,Sel
l=Buy,Buy=Sell)
Here is what I get :
tradesnew
$Trade.Status
NULL
$Instrument.Long.Name
NULL
$Delivery.Prompt.Date
NULL
$Buy.Sell..Cleared.
[1] Buy
$Volume
[1] Buy
$Price
NULL
$Net.Charges..sum.
NULL
That's certainly not what I want.
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 8:43 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
The loop is due to the switch statement, not the condition. Without
condition it would become:
for (i in 1:length(Y)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy=Sell,
X[i])
}
You can make an sapply construct too off course :
new.vect - sapply(X[which(Y==DEL)],switch,Sell=Buy,Buy=Sell)
This will speed up things a little bit, but the effect is marginal.
Cheers
Joris
On Thu, May 27, 2010 at 8:33 AM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Thank you for the answer.
Is there any way to combine if() and switch() in one line? In my case,
something like :
if(trade$Trade.Status==DEL)switch(.)
I would like to avoid the loop .
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Wednesday, May 26, 2010 9:15 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
see ?switch
X- rep(c(Buy,Sell,something else),each=5)
Y- rep(c(DEL,INS,DEL),5)
new.vect - X
for (i in which(Y==DEL)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy=Sell,
X[i])
}
cbind(new.vect,X,Y)
On Wed, May 26, 2010 at 7:43 PM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Dear group,
Here is my df :
trade -
structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name =
c(SUGAR NO.11,
CORN, CORN), Delivery.Prompt.Date = c(Jul/10, Jul/10,
Jul/10), Buy.Sell..Cleared. = c(Sell, Buy, Buy), Volume = c(1L,
2L, 1L), Price = c(15.2500, 368., 368.5000), Net.Charges..sum. =
c(4.01,
-8.64, -4.32)), .Names = c(Trade.Status, Instrument.Long.Name,
Delivery.Prompt.Date, Buy.Sell..Cleared., Volume, Price,
Net.Charges..sum.), row.names = c(NA, 3L), class = data.frame)
Here is what I want :
If trade$Trade.Status==DEL: then if trade$buy.Sell..Cleared==Sell , change
it to Buy, if trade$buy.Sell..Cleared==Buy, change it to Sell.
If trade$Trade.Status==INS, do nothing
I tried to work around with ifelse, but don't know how to deal with so many
conditions.
Any help is appreciated.
TY
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation change elements meeting criteria
Maybe should I be more precise. Here is what I have :
trades -
structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name =
c(SUGAR NO.11,
CORN, CORN), Delivery.Prompt.Date = c(Jul/10, Jul/10,
Jul/10), Buy.Sell..Cleared. = c(Sell, Buy, Buy), Volume = c(1L,
2L, 1L), Price = c(15.2500, 368., 368.5000), Net.Charges..sum. =
c(4.01,
-8.64, -4.32)), .Names = c(Trade.Status, Instrument.Long.Name,
Delivery.Prompt.Date, Buy.Sell..Cleared., Volume, Price,
Net.Charges..sum.), row.names = c(NA, 3L), class = data.frame)
Here is what I want :
tradesnew -
structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name =
c(SUGAR NO.11,
CORN, CORN), Delivery.Prompt.Date = c(Jul/10, Jul/10,
Jul/10), Buy.Sell..Cleared. = c(Buy, Buy, Buy), Volume = c(1L,
2L, 1L), Price = c(15.2500, 368., 368.5000), Net.Charges..sum. =
c(4.01,
-8.64, -4.32)), .Names = c(Trade.Status, Instrument.Long.Name,
Delivery.Prompt.Date, Buy.Sell..Cleared., Volume, Price,
Net.Charges..sum.), row.names = c(NA, 3L), class = data.frame)
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 10:38 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
Off course. You put in a matrix to sapply, but sapply is for vectors. You
want to apply the switch command on every entry of the vector
trades$Buy.Sell..Cleared for which trades$Trade.Status equals DEL. Why do
you try to put in a matrix with all variables for the observations where
status is DEL?
You should have done :
tradesnew-sapply(trades$Buy.Sell..Cleared[which(trades$Trade.Status==DEL)
],
switch,Sell=Buy,Buy=Sell)
Check the help files, and keep track of what goes in and out a function.
Cheers
Joris
On Thu, May 27, 2010 at 9:41 AM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Joris,
If i pass this line :
tradesnew-sapply(trades[which(trades$Trade.Status==DEL),],switch,Sel
l=Buy,Buy=Sell)
Here is what I get :
tradesnew
$Trade.Status
NULL
$Instrument.Long.Name
NULL
$Delivery.Prompt.Date
NULL
$Buy.Sell..Cleared.
[1] Buy
$Volume
[1] Buy
$Price
NULL
$Net.Charges..sum.
NULL
That's certainly not what I want.
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 8:43 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
The loop is due to the switch statement, not the condition. Without
condition it would become:
for (i in 1:length(Y)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy=Sell,
X[i])
}
You can make an sapply construct too off course :
new.vect - sapply(X[which(Y==DEL)],switch,Sell=Buy,Buy=Sell)
This will speed up things a little bit, but the effect is marginal.
Cheers
Joris
On Thu, May 27, 2010 at 8:33 AM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Thank you for the answer.
Is there any way to combine if() and switch() in one line? In my case,
something like :
if(trade$Trade.Status==DEL)switch(.)
I would like to avoid the loop .
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Wednesday, May 26, 2010 9:15 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
see ?switch
X- rep(c(Buy,Sell,something else),each=5)
Y- rep(c(DEL,INS,DEL),5)
new.vect - X
for (i in which(Y==DEL)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy=Sell,
X[i])
}
cbind(new.vect,X,Y)
On Wed, May 26, 2010 at 7:43 PM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Dear group,
Here is my df :
trade -
structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name =
c(SUGAR NO.11,
CORN, CORN), Delivery.Prompt.Date = c(Jul/10, Jul/10,
Jul/10), Buy.Sell..Cleared. = c(Sell, Buy, Buy), Volume = c(1L,
2L, 1L), Price = c(15.2500, 368., 368.5000), Net.Charges..sum. =
c(4.01,
-8.64, -4.32)), .Names = c(Trade.Status, Instrument.Long.Name,
Delivery.Prompt.Date, Buy.Sell..Cleared., Volume, Price,
Net.Charges..sum.), row.names = c(NA, 3L), class = data.frame)
Here is what I want :
If trade$Trade.Status==DEL: then if trade$buy.Sell..Cleared==Sell , change
it to Buy, if trade$buy.Sell..Cleared==Buy, change it to Sell.
If trade$Trade.Status==INS, do nothing
I tried to work around with ifelse, but don't know how to deal with so many
conditions.
Any help is appreciated.
TY
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be
Re: [R] data frame manipulation change elements meeting criteria
Sorry Joris, but I am totally lost on this issue!!
tradenews-sapply(trades$Buy.Sell..Cleared[which(trades$Trade.Status==DEL
)],switch,Sell=Buy,Buy=Sell)
tradenews
Sell
Buy
Not really what I want !!
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 10:38 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
Off course. You put in a matrix to sapply, but sapply is for vectors. You
want to apply the switch command on every entry of the vector
trades$Buy.Sell..Cleared for which trades$Trade.Status equals DEL. Why do
you try to put in a matrix with all variables for the observations where
status is DEL?
You should have done :
tradesnew-sapply(trades$Buy.Sell..Cleared[which(trades$Trade.Status==DEL)
],
switch,Sell=Buy,Buy=Sell)
Check the help files, and keep track of what goes in and out a function.
Cheers
Joris
On Thu, May 27, 2010 at 9:41 AM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Joris,
If i pass this line :
tradesnew-sapply(trades[which(trades$Trade.Status==DEL),],switch,Sel
l=Buy,Buy=Sell)
Here is what I get :
tradesnew
$Trade.Status
NULL
$Instrument.Long.Name
NULL
$Delivery.Prompt.Date
NULL
$Buy.Sell..Cleared.
[1] Buy
$Volume
[1] Buy
$Price
NULL
$Net.Charges..sum.
NULL
That's certainly not what I want.
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 8:43 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
The loop is due to the switch statement, not the condition. Without
condition it would become:
for (i in 1:length(Y)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy=Sell,
X[i])
}
You can make an sapply construct too off course :
new.vect - sapply(X[which(Y==DEL)],switch,Sell=Buy,Buy=Sell)
This will speed up things a little bit, but the effect is marginal.
Cheers
Joris
On Thu, May 27, 2010 at 8:33 AM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Thank you for the answer.
Is there any way to combine if() and switch() in one line? In my case,
something like :
if(trade$Trade.Status==DEL)switch(.)
I would like to avoid the loop .
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Wednesday, May 26, 2010 9:15 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
see ?switch
X- rep(c(Buy,Sell,something else),each=5)
Y- rep(c(DEL,INS,DEL),5)
new.vect - X
for (i in which(Y==DEL)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy=Sell,
X[i])
}
cbind(new.vect,X,Y)
On Wed, May 26, 2010 at 7:43 PM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Dear group,
Here is my df :
trade -
structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name =
c(SUGAR NO.11,
CORN, CORN), Delivery.Prompt.Date = c(Jul/10, Jul/10,
Jul/10), Buy.Sell..Cleared. = c(Sell, Buy, Buy), Volume = c(1L,
2L, 1L), Price = c(15.2500, 368., 368.5000), Net.Charges..sum. =
c(4.01,
-8.64, -4.32)), .Names = c(Trade.Status, Instrument.Long.Name,
Delivery.Prompt.Date, Buy.Sell..Cleared., Volume, Price,
Net.Charges..sum.), row.names = c(NA, 3L), class = data.frame)
Here is what I want :
If trade$Trade.Status==DEL: then if trade$buy.Sell..Cleared==Sell , change
it to Buy, if trade$buy.Sell..Cleared==Buy, change it to Sell.
If trade$Trade.Status==INS, do nothing
I tried to work around with ifelse, but don't know how to deal with so many
conditions.
Any help is appreciated.
TY
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
Re: [R] switch function
Maybe should I be more precise. Here is what I have :
trades -
structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name =
c(SUGAR NO.11, CORN, CORN), Delivery.Prompt.Date = c(Jul/10,
Jul/10, Jul/10), Buy.Sell..Cleared. = c(Sell, Buy, Buy), Volume =
c(1L, 2L, 1L), Price = c(15.2500, 368., 368.5000),
Net.Charges..sum. = c(4.01, -8.64, -4.32)), .Names = c(Trade.Status,
Instrument.Long.Name, Delivery.Prompt.Date, Buy.Sell..Cleared.,
Volume, Price, Net.Charges..sum.), row.names = c(NA, 3L), class =
data.frame)
Here is what I want :
tradesnew -
structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name =
c(SUGAR NO.11, CORN, CORN), Delivery.Prompt.Date = c(Jul/10,
Jul/10, Jul/10), Buy.Sell..Cleared. = c(Buy, Buy, Buy), Volume =
c(1L, 2L, 1L), Price = c(15.2500, 368., 368.5000),
Net.Charges..sum. = c(4.01, -8.64, -4.32)), .Names = c(Trade.Status,
Instrument.Long.Name, Delivery.Prompt.Date, Buy.Sell..Cleared.,
Volume, Price, Net.Charges..sum.), row.names = c(NA, 3L), class =
data.frame)
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 10:38 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
Off course. You put in a matrix to sapply, but sapply is for vectors. You
want to apply the switch command on every entry of the vector
trades$Buy.Sell..Cleared for which trades$Trade.Status equals DEL. Why do
you try to put in a matrix with all variables for the observations where
status is DEL?
You should have done :
tradesnew-sapply(trades$Buy.Sell..Cleared[which(trades$Trade.Status==DEL)
],
switch,Sell=Buy,Buy=Sell)
Check the help files, and keep track of what goes in and out a function.
Cheers
Joris
On Thu, May 27, 2010 at 9:41 AM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Joris,
If i pass this line :
tradesnew-sapply(trades[which(trades$Trade.Status==DEL),],switch,Sel
l=Buy,Buy=Sell)
Here is what I get :
tradesnew
$Trade.Status
NULL
$Instrument.Long.Name
NULL
$Delivery.Prompt.Date
NULL
$Buy.Sell..Cleared.
[1] Buy
$Volume
[1] Buy
$Price
NULL
$Net.Charges..sum.
NULL
That's certainly not what I want.
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 8:43 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
The loop is due to the switch statement, not the condition. Without
condition it would become:
for (i in 1:length(Y)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy=Sell,
X[i])
}
You can make an sapply construct too off course :
new.vect - sapply(X[which(Y==DEL)],switch,Sell=Buy,Buy=Sell)
This will speed up things a little bit, but the effect is marginal.
Cheers
Joris
On Thu, May 27, 2010 at 8:33 AM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Thank you for the answer.
Is there any way to combine if() and switch() in one line? In my case,
something like :
if(trade$Trade.Status==DEL)switch(.)
I would like to avoid the loop .
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Wednesday, May 26, 2010 9:15 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
see ?switch
X- rep(c(Buy,Sell,something else),each=5)
Y- rep(c(DEL,INS,DEL),5)
new.vect - X
for (i in which(Y==DEL)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy=Sell,
X[i])
}
cbind(new.vect,X,Y)
On Wed, May 26, 2010 at 7:43 PM, arnaud Gaboury arnaud.gabo...@gmail.com
wrote:
Dear group,
Here is my df :
trade -
structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name =
c(SUGAR NO.11,
CORN, CORN), Delivery.Prompt.Date = c(Jul/10, Jul/10,
Jul/10), Buy.Sell..Cleared. = c(Sell, Buy, Buy), Volume = c(1L,
2L, 1L), Price = c(15.2500, 368., 368.5000), Net.Charges..sum. =
c(4.01,
-8.64, -4.32)), .Names = c(Trade.Status, Instrument.Long.Name,
Delivery.Prompt.Date, Buy.Sell..Cleared., Volume, Price,
Net.Charges..sum.), row.names = c(NA, 3L), class = data.frame)
Here is what I want :
If trade$Trade.Status==DEL: then if trade$buy.Sell..Cleared==Sell , change
it to Buy, if trade$buy.Sell..Cleared==Buy, change it to Sell.
If trade$Trade.Status==INS, do nothing
I tried to work around with ifelse, but don't know how to deal with so many
conditions.
Any help is appreciated.
TY
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be
Re: [R] switch function
Thanks to Joris, this line does what I want : trades$Buy.Sell..Cleared.[which(trades$Trade.Status==DEL)] - sapply(trades$Buy.Sell..Cleared.[which(trades$Trade.Status==DEL)], switch,Sell=Buy,Buy=Sell) -Original Message- From: jim holtman [mailto:jholt...@gmail.com] Sent: Thursday, May 27, 2010 2:34 PM To: arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] switch function try this: toBuy - trades$Trade.Status == DEL trades$Buy.Sell..Cleared. == Sell toSell - trades$Trade.Status == DEL trades$Buy.Sell..Cleared. == Buy x - trades # make a copy to change x$Buy.Sell..Cleared.[toBuy] - Buy x$Buy.Sell..Cleared.[toSell] - Sell x Trade.Status Instrument.Long.Name Delivery.Prompt.Date Buy.Sell..Cleared. VolumePrice Net.Charges..sum. 1 DEL SUGAR NO.11 Jul/10 Buy 1 15.2500 4.01 2 INS CORN Jul/10 Buy 2 368. -8.64 3 INS CORN Jul/10 Buy 1 368.5000 -4.32 trades Trade.Status Instrument.Long.Name Delivery.Prompt.Date Buy.Sell..Cleared. VolumePrice Net.Charges..sum. 1 DEL SUGAR NO.11 Jul/10 Sell 1 15.2500 4.01 2 INS CORN Jul/10 Buy 2 368. -8.64 3 INS CORN Jul/10 Buy 1 368.5000 -4.32 On Thu, May 27, 2010 at 4:03 AM, arnaud Gaboury arnaud.gabo...@gmail.com wrote: Dear group, Here is my df : trades - structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name = c(SUGAR NO.11, CORN, CORN), Delivery.Prompt.Date = c(Jul/10, Jul/10, Jul/10), Buy.Sell..Cleared. = c(Sell, Buy, Buy), Volume = c(1L, 2L, 1L), Price = c(15.2500, 368., 368.5000), Net.Charges..sum. = c(4.01, -8.64, -4.32)), .Names = c(Trade.Status, Instrument.Long.Name, Delivery.Prompt.Date, Buy.Sell..Cleared., Volume, Price, Net.Charges..sum.), row.names = c(NA, 3L), class = data.frame) I want to replace Buy by Sell and Sell by Buy in column Buy.Sell..Cleared. when element in column Trade.Status is equal to DEL. I think I can write something like this : tradesnew- sapply(trades$Buy.Sell..Cleared[which(trades$Buy.Sell..Cleared== DEL),],switch,...) but I don't really know how to pass further arguments to the switch function. Any help is appreciated. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] writing function : can't find an object
Dear group,
Here is my function:
#return the daily PL for day y
PLDaily-function(x,y)
{
#find elements in my directory with LSCPos in the name, keep the numeric
part in the name and
#create a list
l-gsub(\\D,,dir()[grep(LSCPos,dir())])
#select in the list the desired elements
assign(sel,l[which(l==x):which(l==y)],envir=.GlobalEnv)
#here is another solution select-l[l %in% seq(x, y)]
#first we need to create the Pos and Trad elements
for (i in sel) {
assign(paste(Pos,i,sep=),position(i),envir=.GlobalEnv)
assign(paste(Trad,i,sep=),trade(i),envir=.GlobalEnv)
}
#access elements in my environment
posA-get(paste(c(Pos,x),collapse=))
posB-get(paste(c(Pos,y),collapse=))
av-get(paste(c(Trad,y),collapse=))
#apply some change on element columns then create only one data frame with
the three elements
allcon-ddply(rbind(av[,1:3],
transform(posA,prix=POSITION*SETTLEMENT,SETTLEMENT=NULL),
transform(posB, prix = -POSITION * SETTLEMENT, SETTLEMENT = NULL,
POSITION = POSITION * -1)),
DESCRIPTION,summarise,pl=sum(prix),quantity=sum(POSITION))
#remove the date in $DESCRPTION and add a new column $SHORTDESCRIPTION
allcon$SHORTDESCRIPTION-sub('
[a-z]{3}/[0-9]{2}','',allcon$DESCRIPTION,ignore.case=TRUE)
#read the contractvalue file
value-read.csv2(contractvalue.csv,sep=,,h=T,strip.white=T)
#merge value with allcon, change some columns, then merge with PosB,
replace NA by zero and assign the final result to element PL
zz-merge(transform(merge(value,allcon,all.y=T),SHORTDESCRIPTION=NULL,
VALUE=NULL,PL=-VALUE*pl,quantity=NULL),PosB,all.x=T,sort=F)
zz[is.na(zz)]-0
#PL-zz[c(1,3,4)]
assign(paste(DailyPL,y,sep=),zz[,c(1,3,4)],envir=.GlobalEnv)
}
Here is what I get :
PLDaily(100524,100525)
Error in as.data.frame(y) : object 'PosB' not found
ls()
[1] PLDailyPos100524 Pos100525 position sel
Trad100524 Trad100525 trade
Why R can't find PosB ?
TY for your help.
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] (no subject)
Dear group, Here is my data frame: dput(u) structure(list(DESCRIPTION = structure(c(2L, 5L, 6L, 7L, 9L, 11L, 12L, 15L, 14L, 16L, 1L, 10L, 3L, 4L, 13L, 8L, 17L), .Label = c(COFFEE C Jul/10, COPPER May/10, CORN Jul/10, CORN May/10, COTTON NO.2 Jul/10, CRUDE OIL miNY May/10, GOLD Jun/10, HENRY HUB NATURAL GAS May/10, ROBUSTA COFFEE (10) Jul/10, SILVER May/10, SOYBEANS Jul/10, SPCL HIGH GRADE ZINC USD, STANDARD LEAD USD, SUGAR NO.11 Jul/10, SUGAR NO.11 May/10, WHEAT Jul/10, WHEAT May/10), class = factor), PL = c(3500, -1874.999, -2612.503, -2169.998, -680, 425, 1025, 1008.000, -3057.599, 3212.5, -1781.251, -2265.0, 75, -387.5, 2950, 490.0013, 0), POSITION = c(-2, 3, 2, 2, 18, 3, -1, -1, 5, 5, 0, 0, 0, 0, 0, 0, 0)), .Names = c(DESCRIPTION, PL, POSITION ), class = data.frame, row.names = c(NA, -17L)) I want to give a warning message if one of the element of the POSITION column is different from zero. I tried using mapply with some line like this : mapply(if,u$POSITION,==0,print(WARNING:POSITIONS ARE WRONG,quote=F)) But it seems it is not the correct way to pass the various arguments. Any help is appreciated *** Arnaud Gaboury Mobile: +41 79 392 79 56 BBM: 255B488F __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] condition apply to elements of a data frame column
Oops, forgot to give a subject -Original Message- From: arnaud Gaboury [mailto:arnaud.gabo...@gmail.com] Sent: Wednesday, May 26, 2010 2:31 PM To: r-help@r-project.org Cc: 'arnaud Gaboury' Subject: Dear group, Here is my data frame: dput(u) structure(list(DESCRIPTION = structure(c(2L, 5L, 6L, 7L, 9L, 11L, 12L, 15L, 14L, 16L, 1L, 10L, 3L, 4L, 13L, 8L, 17L), .Label = c(COFFEE C Jul/10, COPPER May/10, CORN Jul/10, CORN May/10, COTTON NO.2 Jul/10, CRUDE OIL miNY May/10, GOLD Jun/10, HENRY HUB NATURAL GAS May/10, ROBUSTA COFFEE (10) Jul/10, SILVER May/10, SOYBEANS Jul/10, SPCL HIGH GRADE ZINC USD, STANDARD LEAD USD, SUGAR NO.11 Jul/10, SUGAR NO.11 May/10, WHEAT Jul/10, WHEAT May/10), class = factor), PL = c(3500, -1874.999, -2612.503, - 2169.998, -680, 425, 1025, 1008.000, -3057.599, 3212.5, -1781.251, -2265.0, 75, -387.5, 2950, 490.0013, 0), POSITION = c(-2, 3, 2, 2, 18, 3, -1, -1, 5, 5, 0, 0, 0, 0, 0, 0, 0)), .Names = c(DESCRIPTION, PL, POSITION ), class = data.frame, row.names = c(NA, -17L)) I want to give a warning message if one of the element of the POSITION column is different from zero. I tried using mapply with some line like this : mapply(if,u$POSITION,==0,print(WARNING:POSITIONS ARE WRONG,quote=F)) But it seems it is not the correct way to pass the various arguments. Any help is appreciated *** Arnaud Gaboury Mobile: +41 79 392 79 56 BBM: 255B488F *** __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] condition apply to elements of a data frame column
Joris, I want to add a line in a function with a print warning if one element of the column is 0. I could use if(sum(u$POSITION)0) as a condition, but I can imagine having one element equal to -2, and another one to 2. So in this case, sum=0, but the condition is false in fact (minimum of one element different from zero). From: Joris Meys [mailto:jorism...@gmail.com] Sent: Wednesday, May 26, 2010 2:48 PM To: arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] (no subject) What exactly are you trying to do? If you want to know which position is wrong, try : if (sum(u$POSITION==0)0) cat(WARNING:POSITION IS WRONG FOR ,which(u$POSITION==0),\n) or even : wrong - which(u$POSITION==0) if(length(wrong)0) cat(WARNING: POSITION IS WRONG FOR,u$DESCRIPTION[wrong],\n) Gives you the exact location of wrong positions. If you do that, make sure u$DESCRIPTION is a character vector and not a factor. Cheers Joris On Wed, May 26, 2010 at 2:31 PM, arnaud Gaboury arnaud.gabo...@gmail.com wrote: Dear group, Here is my data frame: dput(u) structure(list(DESCRIPTION = structure(c(2L, 5L, 6L, 7L, 9L, 11L, 12L, 15L, 14L, 16L, 1L, 10L, 3L, 4L, 13L, 8L, 17L), .Label = c(COFFEE C Jul/10, COPPER May/10, CORN Jul/10, CORN May/10, COTTON NO.2 Jul/10, CRUDE OIL miNY May/10, GOLD Jun/10, HENRY HUB NATURAL GAS May/10, ROBUSTA COFFEE (10) Jul/10, SILVER May/10, SOYBEANS Jul/10, SPCL HIGH GRADE ZINC USD, STANDARD LEAD USD, SUGAR NO.11 Jul/10, SUGAR NO.11 May/10, WHEAT Jul/10, WHEAT May/10), class = factor), PL = c(3500, -1874.999, -2612.503, -2169.998, -680, 425, 1025, 1008.000, -3057.599, 3212.5, -1781.251, -2265.0, 75, -387.5, 2950, 490.0013, 0), POSITION = c(-2, 3, 2, 2, 18, 3, -1, -1, 5, 5, 0, 0, 0, 0, 0, 0, 0)), .Names = c(DESCRIPTION, PL, POSITION ), class = data.frame, row.names = c(NA, -17L)) I want to give a warning message if one of the element of the POSITION column is different from zero. I tried using mapply with some line like this : mapply(if,u$POSITION,==0,print(WARNING:POSITIONS ARE WRONG,quote=F)) But it seems it is not the correct way to pass the various arguments. Any help is appreciated *** Arnaud Gaboury Mobile: +41 79 392 79 56 BBM: 255B488F __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joris Meys Statistical Consultant Ghent University Faculty of Bioscience Engineering Department of Applied mathematics, biometrics and process control Coupure Links 653 B-9000 Gent tel : +32 9 264 59 87 joris.m...@ugent.be --- Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data frame manipulation change elements meeting criteria
Dear group, Here is my df : trade - structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name = c(SUGAR NO.11, CORN, CORN), Delivery.Prompt.Date = c(Jul/10, Jul/10, Jul/10), Buy.Sell..Cleared. = c(Sell, Buy, Buy), Volume = c(1L, 2L, 1L), Price = c(15.2500, 368., 368.5000), Net.Charges..sum. = c(4.01, -8.64, -4.32)), .Names = c(Trade.Status, Instrument.Long.Name, Delivery.Prompt.Date, Buy.Sell..Cleared., Volume, Price, Net.Charges..sum.), row.names = c(NA, 3L), class = data.frame) Here is what I want : If trade$Trade.Status==DEL: then if trade$buy.Sell..Cleared==Sell , change it to Buy, if trade$buy.Sell..Cleared==Buy, change it to Sell. If trade$Trade.Status==INS, do nothing I tried to work around with ifelse, but don't know how to deal with so many conditions. Any help is appreciated. TY __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] writing function:loop and rbind
Dear group,
I have a function, let's call it myfun, wich give me a list of result:
R1,R2,R3...
There is a loop in this function to get my results. Here is the structure of
my function:
Myfun-function()
{
For (i in X ){
---instructions-
Ri
{
{
All Results (R1,R2...) are Data.frame. As a final result (call it Final),
I need to rbind all these dataframe. One solution could be to create another
loop, but I think I can avoid it. How can I add a line like this :
Final-rbind(R1,R2...) using the i parameter? Another solution could may be
to create a list of all my results, then apply rbind to the list?
Any idea would be appreciated.
TY in advance
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] writing function:loop and rbind
One solution is to create a list, then do.call.
Here is my environment:
ls()
[1] DailyPL100416 DailyPL100419 DailyPL100420 l ll
PLglobal Pos100415 Pos100416 Pos100419 Pos100420
position r
[13] resultsel selectTrad100415
Trad100416Trad100419Trad100420trade tt
w
l-list(grep(DailyPL,ls(),value=T)) #create a list of elements with
DailyPL in the name
l
[[1]]
[1] DailyPL100416 DailyPL100419 DailyPL100420
DF-do.call(rbind,l)
DF
[,1][,2][,3]
[1,] DailyPL100416 DailyPL100419 DailyPL100420
That's not what I want! I expect DF to be a data.frame binded by row.
I suspect there is an issue with get() or assign(), or something like that.
Any help is appreciated.
-Original Message-
From: arnaud Gaboury [mailto:arnaud.gabo...@gmail.com]
Sent: Monday, May 24, 2010 2:55 PM
To: r-help@r-project.org
Cc: 'arnaud Gaboury'
Subject: writing function:loop and rbind
Dear group,
I have a function, let's call it myfun, wich give me a list of result:
R1,R2,R3...
There is a loop in this function to get my results. Here is the
structure of
my function:
Myfun-function()
{
For (i in X ){
---instructions-
Ri
{
{
All Results (R1,R2...) are Data.frame. As a final result (call it
Final),
I need to rbind all these dataframe. One solution could be to create
another
loop, but I think I can avoid it. How can I add a line like this :
Final-rbind(R1,R2...) using the i parameter? Another solution could
may be
to create a list of all my results, then apply rbind to the list?
Any idea would be appreciated.
TY in advance
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] writing function:loop and rbind
Maybe is there a neater solution, but the function mget() does the trick. So
until further advice, I will work with your solution.
Thank you
-Original Message-
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
Sent: Monday, May 24, 2010 5:07 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] writing function:loop and rbind
On Mon, May 24, 2010 at 7:00 AM, arnaud Gaboury
arnaud.gabo...@gmail.com wrote:
One solution is to create a list, then do.call.
Here is my environment:
ls()
[1] DailyPL100416 DailyPL100419 DailyPL100420 l
ll
PLglobal Pos100415 Pos100416 Pos100419
Pos100420
position r
[13] resultsel selectTrad100415
Trad100416Trad100419Trad100420trade tt
w
l-list(grep(DailyPL,ls(),value=T)) #create a list of elements
with
DailyPL in the name
A list of their names, not the elements themselves.
l
[[1]]
[1] DailyPL100416 DailyPL100419 DailyPL100420
DF-do.call(rbind,l)
DF
[,1][,2][,3]
[1,] DailyPL100416 DailyPL100419 DailyPL100420
That's not what I want! I expect DF to be a data.frame binded by
row.
It is binded by row, but it is from the list you fed it, which is just
a list with a single element of character strings.
I suspect there is an issue with get() or assign(), or something like
that.
Any help is appreciated.
I am sure there is a neater solution, but something like this should
work where envir=whether the objects actually are.
do.call(rbind, mget(grep(DailyPL,ls(),value=TRUE),envir=.GlobalEnv))
Best regards,
Josh
-Original Message-
From: arnaud Gaboury [mailto:arnaud.gabo...@gmail.com]
Sent: Monday, May 24, 2010 2:55 PM
To: r-help@r-project.org
Cc: 'arnaud Gaboury'
Subject: writing function:loop and rbind
Dear group,
I have a function, let's call it myfun, wich give me a list of
result:
R1,R2,R3...
There is a loop in this function to get my results. Here is the
structure of
my function:
Myfun-function()
{
For (i in X ){
---instructions-
Ri
{
{
All Results (R1,R2...) are Data.frame. As a final result (call it
Final),
I need to rbind all these dataframe. One solution could be to create
another
loop, but I think I can avoid it. How can I add a line like this :
Final-rbind(R1,R2...) using the i parameter? Another solution could
may be
to create a list of all my results, then apply rbind to the list?
Any idea would be appreciated.
TY in advance
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Senior in Psychology
University of California, Riverside
http://www.joshuawiley.com/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] writing function:loop and rbind
Do you think there is a way to add somewhere the argument row.names=NULL ? Or
should I have to write another line to remove the row.names?
-Original Message-
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
Sent: Monday, May 24, 2010 5:07 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] writing function:loop and rbind
On Mon, May 24, 2010 at 7:00 AM, arnaud Gaboury
arnaud.gabo...@gmail.com wrote:
One solution is to create a list, then do.call.
Here is my environment:
ls()
[1] DailyPL100416 DailyPL100419 DailyPL100420 l
ll
PLglobal Pos100415 Pos100416 Pos100419
Pos100420
position r
[13] resultsel selectTrad100415
Trad100416Trad100419Trad100420trade tt
w
l-list(grep(DailyPL,ls(),value=T)) #create a list of elements
with
DailyPL in the name
A list of their names, not the elements themselves.
l
[[1]]
[1] DailyPL100416 DailyPL100419 DailyPL100420
DF-do.call(rbind,l)
DF
[,1][,2][,3]
[1,] DailyPL100416 DailyPL100419 DailyPL100420
That's not what I want! I expect DF to be a data.frame binded by
row.
It is binded by row, but it is from the list you fed it, which is just
a list with a single element of character strings.
I suspect there is an issue with get() or assign(), or something like
that.
Any help is appreciated.
I am sure there is a neater solution, but something like this should
work where envir=whether the objects actually are.
do.call(rbind, mget(grep(DailyPL,ls(),value=TRUE),envir=.GlobalEnv))
Best regards,
Josh
-Original Message-
From: arnaud Gaboury [mailto:arnaud.gabo...@gmail.com]
Sent: Monday, May 24, 2010 2:55 PM
To: r-help@r-project.org
Cc: 'arnaud Gaboury'
Subject: writing function:loop and rbind
Dear group,
I have a function, let's call it myfun, wich give me a list of
result:
R1,R2,R3...
There is a loop in this function to get my results. Here is the
structure of
my function:
Myfun-function()
{
For (i in X ){
---instructions-
Ri
{
{
All Results (R1,R2...) are Data.frame. As a final result (call it
Final),
I need to rbind all these dataframe. One solution could be to create
another
loop, but I think I can avoid it. How can I add a line like this :
Final-rbind(R1,R2...) using the i parameter? Another solution could
may be
to create a list of all my results, then apply rbind to the list?
Any idea would be appreciated.
TY in advance
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Senior in Psychology
University of California, Riverside
http://www.joshuawiley.com/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] writing function
Dear group, Here is my environment after I run a function, myfun() myfun() ls() [1] allconavprix16 DailyPL100416 DailyPL100419 DailyPL100420 l llmyl PL PLdaily PLglobal PLmonthly [13] Pos100415 Pos100416 Pos100419 Pos100420 pose15pose16position r result sel selectTrad100415 [25] Trad100416Trad100419Trad100420trade tt value w zz Elements DailyPL100416 DailyPL100419 DailyPL100420 are data frames created by a function, myfun(). If then I pass this line manually at the prompt: dd-data.frame(do.call(rbind, mget(grep(DailyPL,ls(),value=TRUE),envir=.GlobalEnv)),row.names=NULL) Here is what I get, wich is the expected result : dd - structure(list(DESCRIPTION = structure(c(2L, 5L, 6L, 7L, 9L, 11L, 12L, 15L, 14L, 16L, 1L, 10L, 3L, 4L, 13L, 8L, 17L, 2L, 3L, 5L, 7L, 9L, 11L, 12L, 13L, 14L, 16L, 18L, 8L, 6L, 10L, 15L, 5L, 19L, 9L, 10L, 11L, 12L, 13L, 14L, 16L, 2L, 3L, 20L, 18L, 7L, 21L), .Label = c(COFFEE C Jul/10, COPPER May/10, CORN Jul/10, CORN May/10, COTTON NO.2 Jul/10, CRUDE OIL miNY May/10, GOLD Jun/10, HENRY HUB NATURAL GAS May/10, ROBUSTA COFFEE (10) Jul/10, SILVER May/10, SOYBEANS Jul/10, SPCL HIGH GRADE ZINC USD, STANDARD LEAD USD, SUGAR NO.11 Jul/10, SUGAR NO.11 May/10, WHEAT Jul/10, WHEAT May/10, CRUDE OIL miNY Jun/10, HENRY HUB NATURAL GAS Jun/10, PALLADIUM Jun/10, PLATINUM Jul/10), class = factor), PL = c(3500, -1874.999, -2612.503, -2169.998, -680, 425, 1025, 1008.000, -3057.599, 3212.5, -1781.251, -2265.0, 75, -387.5, 2950, 490.0013, 0, 2612.498, -4162.5, 279.998, 589.9964, -3670, -1212.5, 887.5, -625, 3976.000, -7250, -1112.500, -289., -1414.999, 275, -526.4003, 6914.999, -19.99978, -2330, 54.99955, 1725, -3012.5, -5175, -1769.601, 3787.5, -537.5009, 175, -5330.001, 362.499, 590.0009, -2995.000)), .Names = c(DESCRIPTION, PL), row.names = c(NA, -47L), class = data.frame) If I add this same line at the end of my function , here is what I get: myfun() data frame with 0 columns and 0 rows #dd doesn't return What is wrong? Thank you for help. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] writing function
Thank you so much. You are totally right. -Original Message- From: Joshua Wiley [mailto:jwiley.ps...@gmail.com] Sent: Monday, May 24, 2010 6:33 PM To: arnaud Gaboury Cc: r-help@r-project.org Subject: Re: [R] writing function My guess is that either ls(), called inside grep() or mget() is looking in an environment you do not want it to. When you create a function, it has it's own environment. If you want the dataframes to be created inside the function call (which is what I think you were doing before), you should set ls() and mget() to use your function environment. If the dataframes are supposed to be in the global environment, I would try adding: mget(grep(DailyPL,ls(envir=.GlobalEnv),value=TRUE),envir=.GlobalEnv)) HTH, Josh On Mon, May 24, 2010 at 9:17 AM, arnaud Gaboury arnaud.gabo...@gmail.com wrote: Dear group, Here is my environment after I run a function, myfun() myfun() ls() [1] allconavprix16 DailyPL100416 DailyPL100419 DailyPL100420 l llmyl PL PLdaily PLglobal PLmonthly [13] Pos100415 Pos100416 Pos100419 Pos100420 pose15pose16position r result sel selectTrad100415 [25] Trad100416Trad100419Trad100420trade tt value w zz Elements DailyPL100416 DailyPL100419 DailyPL100420 are data frames created by a function, myfun(). If then I pass this line manually at the prompt: dd-data.frame(do.call(rbind, mget(grep(DailyPL,ls(),value=TRUE),envir=.GlobalEnv)),row.names=NULL) Here is what I get, wich is the expected result : dd - structure(list(DESCRIPTION = structure(c(2L, 5L, 6L, 7L, 9L, 11L, 12L, 15L, 14L, 16L, 1L, 10L, 3L, 4L, 13L, 8L, 17L, 2L, 3L, 5L, 7L, 9L, 11L, 12L, 13L, 14L, 16L, 18L, 8L, 6L, 10L, 15L, 5L, 19L, 9L, 10L, 11L, 12L, 13L, 14L, 16L, 2L, 3L, 20L, 18L, 7L, 21L), .Label = c(COFFEE C Jul/10, COPPER May/10, CORN Jul/10, CORN May/10, COTTON NO.2 Jul/10, CRUDE OIL miNY May/10, GOLD Jun/10, HENRY HUB NATURAL GAS May/10, ROBUSTA COFFEE (10) Jul/10, SILVER May/10, SOYBEANS Jul/10, SPCL HIGH GRADE ZINC USD, STANDARD LEAD USD, SUGAR NO.11 Jul/10, SUGAR NO.11 May/10, WHEAT Jul/10, WHEAT May/10, CRUDE OIL miNY Jun/10, HENRY HUB NATURAL GAS Jun/10, PALLADIUM Jun/10, PLATINUM Jul/10), class = factor), PL = c(3500, -1874.999, -2612.503, -2169.998, -680, 425, 1025, 1008.000, -3057.599, 3212.5, - 1781.251, -2265.0, 75, -387.5, 2950, 490.0013, 0, 2612.498, -4162.5, 279.998, 589.9964, -3670, -1212.5, 887.5, -625, 3976.000, -7250, -1112.500, -289., -1414.999, 275, -526.4003, 6914.999, -19.99978, -2330, 54.99955, 1725, -3012.5, -5175, -1769.601, 3787.5, -537.5009, 175, -5330.001, 362.499, 590.0009, -2995.000)), .Names = c(DESCRIPTION, PL), row.names = c(NA, -47L), class = data.frame) If I add this same line at the end of my function , here is what I get: myfun() data frame with 0 columns and 0 rows #dd doesn't return What is wrong? Thank you for help. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Senior in Psychology University of California, Riverside http://www.joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
