I suppose you could try quote=TRUE
Jim
On Sun, Sep 4, 2016 at 8:13 PM, Christofer Bogaso
wrote:
> Thanks Jim. But my data is like that and I have to live with that. Any
> idea on workaround. Thanks,
>
> On Sun, Sep 4, 2016 at 3:40 PM, Jim Lemon
Please go through an R tutorial or two before posting further here.
There are many good ones on the web. Some recommendations can be found
here:
https://www.rstudio.com/online-learning/
Your question:
"And my Year column is the first column in my csv file, which I
thought made it column 0. Am I
The "c" function creates vectors. Rows of data frames are data frames, not
vectors.
new_row <- data.frame( Prod_name = "Day_name", `Date 1`=1, `Date 2`=2,`Date
3`=3 )
data_may <- rbind( new_row, data_may )
Furthermore, data frames are NOT spreadsheets. "Day_num" looks suspiciously
UNlike a
Dear All,
I am relatively new to R and certainly new to the e-mailing list. I need
your help. I am working on a data frame, which looks like this:
Prod_name | Date 1 | Date 2 | Date 3 |
--|-||--|
Product 1| 3 | 4 |
Please use Reply-all to keep the mailing list in the loop. I cannot provide
private assistance, and others may provide valuable input or respond faster
than I can.
It is very common that people cannot provide the original data. That means more
work for YOU, though, not for us. It is up to
Hi,
I would like to use the C spline functions if R for a FORTRAN subroutine. What
header file should I refer to?
Regards
Filippo Monari
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
Didnt work getting unused argument error.
On Sun, Sep 4, 2016 at 4:47 PM, Jim Lemon wrote:
> I suppose you could try quote=TRUE
>
> Jim
>
>
> On Sun, Sep 4, 2016 at 8:13 PM, Christofer Bogaso
> wrote:
>> Thanks Jim. But my data is like that
Hi Bailey,
Treat it as a guess, but try this:
for (i in c(1:3)){
y<-mydata[,i]
x <- mblm(y ~ Year, mydata, repeated = FALSE)
print(x)
}
I'm not sure that you can mix indexed columns with column names. Also,
Year is column 4, no?
Jim
On Sun, Sep 4, 2016 at 11:43 AM, Bailey Hewitt
A useful rule is to fix the first error you understand and hope that the
others go away.
On 04/09/2016 04:05, Tamar Michaeli wrote:
Any help in resolving the following errors will be appreciated:
pollutantmean <- function(directory, pollutant, id=1:332)
+ file_inc <- list.files("specdata",
Shouldn't get that with write.csv.
Jim
On Sun, Sep 4, 2016 at 9:29 PM, Christofer Bogaso
wrote:
> Didnt work getting unused argument error.
>
> On Sun, Sep 4, 2016 at 4:47 PM, Jim Lemon wrote:
>> I suppose you could try quote=TRUE
>>
>>
Hi Michael,
On top of all suggestions, I can mention the following packages for linear
programming problems:
https://cran.r-project.org/web/packages/linprog/linprog.pdf
https://cran.r-project.org/web/packages/lpSolve/lpSolve.pdf
https://cran.r-project.org/web/packages/clue/clue.pdf
(see
Hi again,
I was trying to read a subset of Data from a CSV file using below code
as example :
library(sqldf)
Dat = structure(list(`col 1` = c(133261L, 133261L, 133262L, 133262L
), `col 2` = structure(1:4, .Label = c("aaa1", "aaa2", "bbb1, bbb",
"bbb3, bbb"), class = "factor"), `col 3` =
Hi Christofer,
You have embedded commas in your data structure. This is guaranteed to
mess up a CSV read.
Jim
On Sun, Sep 4, 2016 at 5:54 PM, Christofer Bogaso
wrote:
> Hi again,
>
> I was trying to read a subset of Data from a CSV file using below code
> as
Thanks Jim. But my data is like that and I have to live with that. Any
idea on workaround. Thanks,
On Sun, Sep 4, 2016 at 3:40 PM, Jim Lemon wrote:
> Hi Christofer,
> You have embedded commas in your data structure. This is guaranteed to
> mess up a CSV read.
>
> Jim
>
>
>
Any help in resolving the following errors will be appreciated:
> pollutantmean <- function(directory, pollutant, id=1:332)
+ file_inc <- list.files("specdata", full.names=TRUE)
> dat <- data.frame()
> for(i in 1:10) {
+ dat <- rbind(dat, read.csv(file_inc[i]))
+ }
Error in read.table(file =
Dear all,
I believe that this will be a more helpful way to put the problem:
structure(list(Prod_name = c("Banana", "Apple", "Orange", "Yoghurt",
"Eggs", "Milk", "Day_num"), X1.1.2000 = c("1", "0", "4", "3",
"6", "2", "1"), X2.1.2000 = c("2", "4", "1", "5", "3", "0", "2"
), X3.1.2000 = c("1",
My suggested approach:
dta <- structure(list(Prod_name = c("Banana", "Apple", "Orange",
"Yoghurt",
"Eggs", "Milk", "Day_num"), X1.1.2000 = c("1", "0", "4", "3",
"6", "2", "1"), X2.1.2000 = c("2", "4", "1", "5", "3", "0", "2"
), X3.1.2000 = c("1", "5", "2", "3", "0", "4", "3"), X4.1.2000 =
pretty sure you just missed the `{` at the beginning of the `function`
definition block.
On Sun, Sep 4, 2016 at 7:38 AM, Michael Dewey
wrote:
> A useful rule is to fix the first error you understand and hope that the
> others go away.
>
> On 04/09/2016 04:05, Tamar
> On Sep 3, 2016, at 8:05 PM, Tamar Michaeli wrote:
>
> Any help in resolving the following errors will be appreciated:
>
>> pollutantmean <- function(directory, pollutant, id=1:332)
> + file_inc <- list.files("specdata", full.names=TRUE)
>> dat <- data.frame()
>> for(i
Time for an R tutorial or two to learn how to use the "apply" family
in R. I think what you want is:
merged_list <- lapply(merging, get)
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in
Dear list,
I have a vector of strings that cannot be described by one pattern. So
let's say I construct a vector of patterns in the same length as the vector
of strings, can I do the element wise pattern recognition and string
substitution.
For example,
pattern1 <-
Hello,
I have a vector of characters that I know will be object names and I'd like
to treat this vector as a series of names to create a list. But, for the
life of me, I cannot figure out how to treat a vector of characters as a
vector of object names when creating a list.
For example, this does
Hi Ryan,
How about:
names(merged.parameters)<-merging
Jim
On Mon, Sep 5, 2016 at 7:57 AM, Ryan Utz wrote:
> Hello,
>
> I have a vector of characters that I know will be object names and I'd like
> to treat this vector as a series of names to create a list. But, for the
>
On 09/05/2016 12:07 AM, Bert Gunter wrote:
Time for an R tutorial or two to learn how to use the "apply" family
in R. I think what you want is:
merged_list <- lapply(merging, get)
Or even:
named_merged_list <- mget(merging)
Anyway, probably you could arrive to a list of parameters
Thank you, Dénes. Better yet.
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sun, Sep 4, 2016 at 4:13 PM, Dénes Tóth wrote:
>
Dear R users,
I am trying to simulate a pseudo R console, evaluating commands. This is a
simple example of a series of command lines which get evaluated:
foo <- function(x) {
print(x)
}
print)foo)
foo(2)
If I copied and pasted this example in an R console, this is the result to
replicate
You might want to look at the "evaluate" package.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sun, Sep 4, 2016 at 4:47 PM, Adrian Dușa
Your opening assertion is false.
Provide a reproducible example and someone will demonstrate.
--
Sent from my phone. Please excuse my brevity.
On September 4, 2016 9:06:59 PM PDT, Jun Shen wrote:
>Dear list,
>
>I have a vector of strings that cannot be described by one
Well, he did provide an example, and...
> z <- c('TX.WT.CUT.mean','mg.tx.cv')
> sub("^.+?\\.(.+)\\.[^.]+$","\\1",z)
[1] "WT.CUT" "tx"
## seems to do what was requested.
Jeff would have to amplify on his initial statement however: do you
mean that separate patterns can always be combined via
29 matches
Mail list logo