[R] Density or Boxplot with median and mean
Hi there, I would like to be able to draw a density plot or a box plot where the median and the median and the mean would be visible. If I decide a density plot I need to put two big marks one for the median and one for the mean, which I do not know how I can achieve to put marks in a density plot. For that I am using plot(density(myVector)) while on the boxplot median is already visible but mean not. To have the mean there I would have to add one more line on each boxplot, perhaps of different color but I am not sure if that is possible in R. boxplot(myVector) I am using where myVector can be something like myVector-seq(1,200) I would like to thank you all in advance Regards Alex [[alternative HTML version deleted]] __ 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] Density or Boxplot with median and mean
On 01/22/2014 07:37 PM, Alaios wrote: Hi there, I would like to be able to draw a density plot or a box plot where the median and the median and the mean would be visible. If I decide a density plot I need to put two big marks one for the median and one for the mean, which I do not know how I can achieve to put marks in a density plot. For that I am using plot(density(myVector)) while on the boxplot median is already visible but mean not. To have the mean there I would have to add one more line on each boxplot, perhaps of different color but I am not sure if that is possible in R. boxplot(myVector) I am using where myVector can be something like myVector-seq(1,200) Hi Alex, On a density plot you can use abline: abline(v=mean(myVector),col=red) abline(v=median(myVector),col=green) I don't know of any boxplot function that will plot two measures of central tendency, but box.heresy in plotrix will plot any one measure that you like. Jim __ 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] collapsing records
Hi
That is great!
Thanks
On Mon, Jan 20, 2014 at 12:10 PM, Jim Lemon j...@bitwrit.com.au wrote:
On 01/20/2014 11:44 AM, Bill wrote:
I am trying to read a csv file with a date-time field. There are many rows
with the same date but different times. I first want to clear the times so
that rows from the same day have the same date-time field (called Date).
There is another field called Text and I want to collapse all the records
with the same date so that there is only one record for this date and with
a text field that contains all the strings from all the corresponding text
fields. At the same time I want to create a new field that has the count
of
how many records were collapsed for each date. There is a third field
called Tw.ID and I was trying to use tapply on this field to do this.
Later
I will create a DocumentTermMatrix with the tm package on this dataframe.
In the code below I have not figured out how to collapse the data so that
there is only one record for each date and I don't really have a good way
to add in a count field. Can anyone make any suggestions?
Thanks.
install.packages(c(tm))
library(tm)
y.df=read.csv(YHOO3000.csv, header=TRUE)
y.df$Date= as.POSIXlt( y.df$Date)
ysub14.df=y.df
ysub14.df$Date=y.df$Date -14*3600 #I pushed the record times back a little
here.
ysub14.df$Date=as.Date(ysub14.df$Date, %Y-%m-%d)
# might want to use groups-
unstack(data.frame(ysub14.df$Text,ysub14.df$Date))
# to put all the tweets for one day into a group. This makes a list
# I think, with the name of the list being the Date and
# the tweets for that date being stored in a vector.
countgroup2=tapply(ysub14.df$Tw.ID,ysub14.df$Date,length)
Hi Bill,
Here is one way:
# get some date-time strings
dates-paste(2014-01-,10:15, ,sample(0:23,20),
:,sample(0:60,20),:,sample(0:60,20),sep=)
# function to return stupid text
sillytext-function(n) {
return(paste(sample(letters[1:26],n),sep=,collapse=))
}
# get the stupid text
ttext-sapply(rep(10,20),sillytext)
# make the data frame
y.df-data.frame(dates,ttext)
# convert the date-time strings to dates
y.df$dates-
as.Date(format(as.Date(dates,%Y-%m-%d %H:%M:%S),
Y-%m-%d),Y-%m-%d)
library(prettyR)
# stretch out all the text strings for each day
y2.df-stretch_df(y.df,dates,ttext)
# get the dimension of the resulting data frame
ydim-dim(y2.df)
# function to count the NAs
nna-function(x) return(sum(is.na(x)))
# add a column with a count of _not_ NAs
y2.df$nrec-
(ydim[2]-1)-apply(as.matrix(y2.df[,2:ydim[2]]),1,nna)
Jim
[[alternative HTML version deleted]]
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Re: [R] My problem with R
Hi First of all without posting some of your data it is really difficult to understand what you really want. Just select only a small part for 2 your data frames and post an output from dput. e.g. dput(data3[1:10, 1:7]) dput(data4[1:10, 1:7]) Most probably resulting timestamp is in POSIXlt mode which has list structure. If you want to work with it as with vector you need to transfer it to POSIXct timestamp -as.POSIXct(timestamp) and only after that to do rbinding. However I am not sure that rbind preserves time/date format. see below test[100:110] [1] 2014-01-08 15:00:00 CET 2014-01-08 15:00:00 CET [3] 2014-01-08 15:00:00 CET 2014-01-08 15:00:00 CET [5] 2014-01-08 15:00:00 CET 2014-01-08 15:00:00 CET [7] 2014-01-08 15:00:00 CET 2014-01-08 15:00:00 CET [9] 2014-01-08 15:00:00 CET 2014-01-08 15:00:00 CET [11] 2014-01-08 15:00:00 CET rbind(test[100:110], test[110:120]) [,1] [,2] [,3] [,4] [,5] [,6] [1,] 1389189600 1389189600 1389189600 1389189600 1389189600 1389189600 [2,] 1389189600 1389189600 1389189600 1389189600 1389189600 1389189600 [,7] [,8] [,9] [,10] [,11] [1,] 1389189600 1389189600 1389189600 1389189600 1389189600 [2,] 1389189600 1389189600 1389189600 1389189600 1389189600 Resulting number are seconds from 1.1.1970 UTC. Look at Examples in POSIX help page. I am still just guessing, without actual data it is difficult to understand where is the problem. I still believe that merge(data1, data2, all=T) is the most efficient way. Petr From: Lee Marine [mailto:marine1...@gmail.com] Sent: Wednesday, January 22, 2014 9:16 AM To: PIKAL Petr Subject: Re: [R] My problem with R Thanks for your re-email. I success merge function(really appreciate you) but I have one problem. That is,,, time data rbind... I can not solved it. I want to see 2013-10-04 04:42:00 2013-10-04 04:42:01 2013-10-04 04:42:02 ... timestamp(as you know) is really good, timestamp3-strptime(paste(data3[,1],data3[,2], sep=),format=%Y-%m-%d %H:%M:%S) timestamp4-strptime(paste(data4[,1],data4[,2], sep=),format=%Y-%m-%d %H:%M:%S) ... I can not use merge because I need row binding, not column binding. but It did not work. Time-rbind(rbind(timestamp3,timestamp4,timestamp5,timestamp6,timestamp7,timestamp8,timestamp9,timestamp10,timestamp11,timestamp12,timestamp13,timestamp14,timestamp15,timestamp16) str(Time) 'data.frame': 14 obs. of 9 variables: $ sec :List of 14 ..$ timestamp3 : num 0 1 2 3 4 5 6 7 8 9 ... ..$ timestamp4 : num 0 1 2 3 4 5 6 7 8 9 ... ..$ timestamp5 : num 0 1 2 3 4 5 6 7 8 9 ... ..$ timestamp6 : num 0 1 2 3 4 5 6 7 8 9 ... ..$ timestamp7 : num 0 1 2 3 4 5 6 7 8 9 ... ..$ timestamp8 : num 0 1 2 3 4 5 6 7 8 9 ... ..$ timestamp9 : num 0 1 2 3 4 5 6 7 8 9 ... ..$ timestamp10: num 0 1 2 3 4 5 6 7 8 9 ... ..$ timestamp11: num 0 1 2 3 4 5 6 7 8 9 ... ..$ timestamp12: num 0 1 2 4 5 6 7 8 9 10 ... ..$ timestamp13: num 0 1 2 3 4 5 6 7 8 9 ... ..$ timestamp14: num 0 1 2 3 4 5 6 7 8 9 ... ..$ timestamp15: num 1 2 3 4 5 6 7 8 9 10 ... ..$ timestamp16: num 0 1 2 3 4 5 6 7 8 9 ... $ min :List of 14 ..$ timestamp3 : int 42 42 42 42 42 42 42 42 42 42 ... ..$ timestamp4 : int 0 0 0 0 0 0 0 0 0 0 ... summary(Time) sec.Length sec.Class sec.Mode min.Length min.Class min.Mode 36568-none- numeric 36568-none- numeric 85206-none- numeric 85206-none- numeric 85207-none- numeric 85207-none- numeric 85206-none- numeric 85206-none- numeric ... hour.Length hour.Class hour.Mode mday.Length mday.Class mday.Mode 36568-none- numeric 36568-none- numeric 85206-none- numeric 85206-none- numeric 85207-none- numeric 85207-none- numeric ... mon.Length mon.Class mon.Mode year.Length year.Class year.Mode 36568-none- numeric 36568-none- numeric 85206-none- numeric 85206-none- numeric ... wday.Length wday.Class wday.Mode yday.Length yday.Class yday.Mode 36568-none- numeric 36568-none- numeric 85206-none- numeric 85206-none- numeric ... isdst.Length isdst.Class isdst.Mode 36568-none- numeric 85206-none- numeric I want to result just like that;;; [1] 2013-10-04 04:42:00 2013-10-04 04:42:01 2013-10-04 04:42:02 [4] 2013-10-04 04:42:03 2013-10-04 04:42:04 2013-10-04 04:42:05 ... I tried to find many ways mid-rbind(timestamp3=timestamp3, timestamp4=timestamp4) mid-merge(timestamp3, timestamp4) mid-rbind(as.POSIXct(timestamp3,timestamp4)) ... etc all fail...T^ T What should I do? Best Regards, Marine 2014/1/22 PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz Hi From: Lee Marine [mailto:marine1...@gmail.commailto:marine1...@gmail.com] Sent: Wednesday, January 22, 2014 2:01 AM To: PIKAL Petr Subject: Re: [R] My problem with R ps.
Re: [R] Density or Boxplot with median and mean
Thanks Jim.. once again your rock On Wednesday, January 22, 2014 9:51 AM, Jim Lemon j...@bitwrit.com.au wrote: On 01/22/2014 07:37 PM, Alaios wrote: Hi there, I would like to be able to draw a density plot or a box plot where the median and the median and the mean would be visible. If I decide a density plot I need to put two big marks one for the median and one for the mean, which I do not know how I can achieve to put marks in a density plot. For that I am using plot(density(myVector)) while on the boxplot median is already visible but mean not. To have the mean there I would have to add one more line on each boxplot, perhaps of different color but I am not sure if that is possible in R. boxplot(myVector) I am using where myVector can be something like myVector-seq(1,200) Hi Alex, On a density plot you can use abline: abline(v=mean(myVector),col=red) abline(v=median(myVector),col=green) I don't know of any boxplot function that will plot two measures of central tendency, but box.heresy in plotrix will plot any one measure that you like. Jim [[alternative HTML version deleted]] __ 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] complicated IF
Hello. I am trying to work out some complicated if() logic. I thought of using which() and if() but cannot get it. I have a dataframe that looks like this: head(deleteFridayTest) Date nrec 1 2011-07-17 667 2 2011-07-18 266 3 2009-10-29 29 4 2009-10-30 211 5 2009-10-31 237 6 2009-11-01 898 I want to take the values in nrec for consecutive Friday, Saturday and Sundays and average them and replace Sundays value with that average. I came up with this: deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Sun,]$nrec - (deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Sun,]$nrec + deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Sat,]$nrec + deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Fri,]$nrec)/3 but this won't work for my data because sometimes one or more of the days of data may be missing. For example Friday's data could be missing, or Friday and Saturday, or Sunday may be missing, or they all may be missing, etc. The rule I want to implement is that if any of Friday, Saturday, or Sunday is available then I want to have an entry for Sunday (call it 'X'). If all 3 days are missing then nothing should be done and there will be no entry for X. If any of the days Fri, Sat, Sun are available then X should be the average of those values (e.g. if two days are available then sum and divide by 2, if just one day is available then just use that value for X). Can anyone suggest how to go about this? Thank you. [[alternative HTML version deleted]] __ 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] Setting up an R server.
Hi John,
A server it means a computer that has an operating system, where you can
run R. For example, a Linux OS can be run in a computer connected to
your Local Area Network at home. There you can install R and you can
communicate with it via batch mode or interactive. The simplest is you
become root of your system and use a software yum to install R (yum
install R). As Linux OS I use Fedora (free software), but one can use a
number of other flavors of Linux OS. This software (yum) will place the
executable of R in one of the bin dirs which is reachable from any
directory you enter in the system. What you need after it is an ssh
(secure shell) from the apps of ipad (apple), or if windows (x-win
software). At home, as long as you have connected your server with your
router, you can reach the server with no problem, by learning local
addresses or defining a name to the system, while remotely (distant from
your home) you need to know a fixed system address (probably you need to
buy it) from your internet company. If more than one server you can do R
parallel computing at home by using software such as gridware to
distribute jobs.
Hope this gives an idea about what you are planning to do :-) .
Best,
Aldi
On 1/20/2014 10:53 AM, R. Michael Weylandt wrote:
Perhaps http://www.rstudio.com/ide/docs/server/getting_started
Michael
On Mon, Jan 20, 2014 at 9:12 AM, John Sorkin
jsor...@grecc.umaryland.edu wrote:
Can someone provide suggestions about how to best set up an R server? I would
like to be able to run R on my IPad. It sounds like the only way to do this is
to have the IPad access an R server. The server will be at my home, connected
to the internet via my cable company (comcast). I don't yet know if the server
will be a linux box or a windows box. I would appreciate advice about setting
up both kinds of servers.
Thank you,
John
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)
Confidentiality Statement:
This email message, including any attachments, is for ...{{dropped:14}}
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--
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[R] R in remote mode
Dear all, I have written a simulation in R that has a significant running time (probably 60-80 hours). While I can run the code on my laptop, it tends to slow things down to a significant extent and it leads to a very high CPU temperature overall. Is there an easy and convenient way to run R remotely on some outside server or PC? Any services that you are aware off? I know that there is a way to run R on Amazon EC but I'm wondering whether there is something even simpler. Ideally I am looking for a remote access to a PC where R is already installed and where I can simply copy-paste my code and run it. Please let me know in case you have any ideas, Thanks in advance, Michael Michael Haenlein Professor of Marketing ESCP Europe Paris, France [[alternative HTML version deleted]] __ 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] complicated IF
Here's one way of doing it. Does not use complicated IFs; just
splits the data and works on it.
x - read.table(text =Date nrec
+
+ 1 2011-07-17 667
+
+ 2 2011-07-18 266
+
+ 3 2009-10-29 29
+
+ 4 2009-10-30 211
+
+ 5 2009-10-31 237
+
+ 6 2009-11-01 898, header = TRUE, as.is = TRUE)
# convert to Date
x$Date - as.Date(x$Date)
# add week of year
x$week - format(x$Date, %Y%W)
# add the day of week
x$day - format(x$Date, %w)
# process each week, substituting the mean if Sunday exists
result - do.call(rbind
+ , lapply(split(x, x$week), function(.week){
+ means - mean(.week$nrec[.week$day %in% c('0', '5', '6')])
+ .week$nrec[.week$day == '0'] - means
+ .week
+ })
+ )
result
Date nrec week day
200943.3 2009-10-29 29. 200943 4
200943.4 2009-10-30 211. 200943 5
200943.5 2009-10-31 237. 200943 6
200943.6 2009-11-01 448.6667 200943 0
201128 2011-07-17 667. 201128 0
201129 2011-07-18 266. 201129 1
x
Date nrec week day
1 2011-07-17 667 201128 0
2 2011-07-18 266 201129 1
3 2009-10-29 29 200943 4
4 2009-10-30 211 200943 5
5 2009-10-31 237 200943 6
6 2009-11-01 898 200943 0
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Wed, Jan 22, 2014 at 6:33 AM, Bill william...@gmail.com wrote:
Hello. I am trying to work out some complicated if() logic.
I thought of using which() and if() but cannot get it.
I have a dataframe that looks like this:
head(deleteFridayTest)
Date nrec
1 2011-07-17 667
2 2011-07-18 266
3 2009-10-29 29
4 2009-10-30 211
5 2009-10-31 237
6 2009-11-01 898
I want to take the values in nrec for consecutive Friday, Saturday and
Sundays and average them and replace Sundays value with that average.
I came up with this:
deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Sun,]$nrec -
(deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Sun,]$nrec +
deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Sat,]$nrec +
deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Fri,]$nrec)/3
but this won't work for my data because sometimes one or more of the days
of data may be missing. For example Friday's data could be missing, or
Friday and Saturday, or Sunday may be missing, or they all may be missing,
etc.
The rule I want to implement is that
if any of Friday, Saturday, or Sunday is available then I want to have an
entry for Sunday (call it 'X'). If all 3 days are missing then nothing
should be done and there will be no entry for X. If any of the days Fri,
Sat, Sun are available then X should be the average of those values (e.g.
if two days are available then sum and divide by 2, if just one day is
available then just use that value for X).
Can anyone suggest how to go about this?
Thank you.
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] complicated IF
Hello Jim,
Thanks for this. I will study it. One thing, you wrote # process each
week, substituting the mean if Sunday exists. Even if Sunday's data does
not exist, I need an entry for Sunday if Friday or Saturday (or both)
exist. I don't yet understand what you wrote so I am not sure if that is
the case.
Bill
On Wed, Jan 22, 2014 at 10:04 PM, jim holtman jholt...@gmail.com wrote:
Here's one way of doing it. Does not use complicated IFs; just
splits the data and works on it.
x - read.table(text =Date nrec
+
+ 1 2011-07-17 667
+
+ 2 2011-07-18 266
+
+ 3 2009-10-29 29
+
+ 4 2009-10-30 211
+
+ 5 2009-10-31 237
+
+ 6 2009-11-01 898, header = TRUE, as.is = TRUE)
# convert to Date
x$Date - as.Date(x$Date)
# add week of year
x$week - format(x$Date, %Y%W)
# add the day of week
x$day - format(x$Date, %w)
# process each week, substituting the mean if Sunday exists
result - do.call(rbind
+ , lapply(split(x, x$week), function(.week){
+ means - mean(.week$nrec[.week$day %in% c('0', '5', '6')])
+ .week$nrec[.week$day == '0'] - means
+ .week
+ })
+ )
result
Date nrec week day
200943.3 2009-10-29 29. 200943 4
200943.4 2009-10-30 211. 200943 5
200943.5 2009-10-31 237. 200943 6
200943.6 2009-11-01 448.6667 200943 0
201128 2011-07-17 667. 201128 0
201129 2011-07-18 266. 201129 1
x
Date nrec week day
1 2011-07-17 667 201128 0
2 2011-07-18 266 201129 1
3 2009-10-29 29 200943 4
4 2009-10-30 211 200943 5
5 2009-10-31 237 200943 6
6 2009-11-01 898 200943 0
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.
On Wed, Jan 22, 2014 at 6:33 AM, Bill william...@gmail.com wrote:
Hello. I am trying to work out some complicated if() logic.
I thought of using which() and if() but cannot get it.
I have a dataframe that looks like this:
head(deleteFridayTest)
Date nrec
1 2011-07-17 667
2 2011-07-18 266
3 2009-10-29 29
4 2009-10-30 211
5 2009-10-31 237
6 2009-11-01 898
I want to take the values in nrec for consecutive Friday, Saturday and
Sundays and average them and replace Sundays value with that average.
I came up with this:
deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Sun,]$nrec -
(deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Sun,]$nrec +
deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Sat,]$nrec +
deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Fri,]$nrec)/3
but this won't work for my data because sometimes one or more of the days
of data may be missing. For example Friday's data could be missing, or
Friday and Saturday, or Sunday may be missing, or they all may be
missing,
etc.
The rule I want to implement is that
if any of Friday, Saturday, or Sunday is available then I want to have an
entry for Sunday (call it 'X'). If all 3 days are missing then nothing
should be done and there will be no entry for X. If any of the days Fri,
Sat, Sun are available then X should be the average of those values
(e.g.
if two days are available then sum and divide by 2, if just one day is
available then just use that value for X).
Can anyone suggest how to go about this?
Thank you.
[[alternative HTML version deleted]]
__
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.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] complicated IF
Here is the change to create a Sunday in a week if it does not exist.
I took out the Sunday (2009-11-01) for testing and you will notice
that week 201129 did not have a Sunday, so it has NaN as the result.
x - read.table(text =Date nrec
+
+ 1 2011-07-17 667
+
+ 2 2011-07-18 266
+
+ 3 2009-10-29 29
+
+ 4 2009-10-30 211
+
+ 5 2009-10-31 237, header = TRUE, as.is = TRUE)
# convert to Date
x$Date - as.Date(x$Date)
# add week of year
x$week - format(x$Date, %Y%W)
# add the day of week
x$day - format(x$Date, %w)
# process each week, substituting the mean if Sunday exists
result - do.call(rbind
+ , lapply(split(x, x$week), function(.week){
+ means - mean(.week$nrec[.week$day %in% c('0', '5', '6')])
+ # check if Sunday exists; if not, create it
+ if (!any(.week$day == '0')){
+ # create a new entry for Sunday
+ .week - rbind(.week[1, ], .week) # new entry in row 1
+ # convert date to Sunday by backing off the days of the week
+ .week$Date[1L] - .week$Date[1L] - as.numeric(.week$day[1L]) + 7
+ .week$day[1L] - '0' # make it a Sunday
+ }
+ .week$nrec[.week$day == '0'] - means
+ .week
+ })
+ )
result
Date nrec week day
200943.3 2009-11-01 224 200943 0 # added
200943.31 2009-10-29 29 200943 4
200943.4 2009-10-30 211 200943 5
200943.5 2009-10-31 237 200943 6
2011282011-07-17 667 201128 0
201129.2 2011-07-24 NaN 201129 0 # no other days to average
201129.21 2011-07-18 266 201129 1
x
Date nrec week day
1 2011-07-17 667 201128 0
2 2011-07-18 266 201129 1
3 2009-10-29 29 200943 4
4 2009-10-30 211 200943 5
5 2009-10-31 237 200943 6
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.
On Wed, Jan 22, 2014 at 8:25 AM, Bill william...@gmail.com wrote:
Hello Jim,
Thanks for this. I will study it. One thing, you wrote # process each week,
substituting the mean if Sunday exists. Even if Sunday's data does not
exist, I need an entry for Sunday if Friday or Saturday (or both) exist. I
don't yet understand what you wrote so I am not sure if that is the case.
Bill
On Wed, Jan 22, 2014 at 10:04 PM, jim holtman jholt...@gmail.com wrote:
Here's one way of doing it. Does not use complicated IFs; just
splits the data and works on it.
x - read.table(text =Date nrec
+
+ 1 2011-07-17 667
+
+ 2 2011-07-18 266
+
+ 3 2009-10-29 29
+
+ 4 2009-10-30 211
+
+ 5 2009-10-31 237
+
+ 6 2009-11-01 898, header = TRUE, as.is = TRUE)
# convert to Date
x$Date - as.Date(x$Date)
# add week of year
x$week - format(x$Date, %Y%W)
# add the day of week
x$day - format(x$Date, %w)
# process each week, substituting the mean if Sunday exists
result - do.call(rbind
+ , lapply(split(x, x$week), function(.week){
+ means - mean(.week$nrec[.week$day %in% c('0', '5', '6')])
+ .week$nrec[.week$day == '0'] - means
+ .week
+ })
+ )
result
Date nrec week day
200943.3 2009-10-29 29. 200943 4
200943.4 2009-10-30 211. 200943 5
200943.5 2009-10-31 237. 200943 6
200943.6 2009-11-01 448.6667 200943 0
201128 2011-07-17 667. 201128 0
201129 2011-07-18 266. 201129 1
x
Date nrec week day
1 2011-07-17 667 201128 0
2 2011-07-18 266 201129 1
3 2009-10-29 29 200943 4
4 2009-10-30 211 200943 5
5 2009-10-31 237 200943 6
6 2009-11-01 898 200943 0
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Wed, Jan 22, 2014 at 6:33 AM, Bill william...@gmail.com wrote:
Hello. I am trying to work out some complicated if() logic.
I thought of using which() and if() but cannot get it.
I have a dataframe that looks like this:
head(deleteFridayTest)
Date nrec
1 2011-07-17 667
2 2011-07-18 266
3 2009-10-29 29
4 2009-10-30 211
5 2009-10-31 237
6 2009-11-01 898
I want to take the values in nrec for consecutive Friday, Saturday and
Sundays and average them and replace Sundays value with that average.
I came up with this:
deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Sun,]$nrec -
(deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Sun,]$nrec +
deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Sat,]$nrec +
deleteFridayTest[dayOfWeek(deleteFridayTest$Date)==Fri,]$nrec)/3
but this won't work for my data because sometimes one or more of the
days
of data may be missing. For example Friday's data could be missing, or
Friday and Saturday, or Sunday may be missing, or they all may be
missing,
etc.
The rule I want to implement is that
if any of Friday, Saturday, or Sunday is available then I want to have
[R] ETAS-Help
Hello, My name is Katerina, i am new to R and i am working with the ETAS package. My goal is to fit the spatiotemporal etas model to an aftershock sequence ( atach file example.csv).I have installed the packages: spatstat, SAPP and ETAS. By reading the ETAS package manual i saw the data must be in class ppx. Could you please help me on how to convert my data (example.csv) into class ppx ? Thank you in advance,Katerina __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to get the numbers of factors in a matrix
sapply(X, function(m){nlevels(factor(m$latitudes))})
I think that length(unique(x)) is a more direct and easier to remember
way of determining the number of unique values in the vector x,
rather than nlevels(factor(x)).
Bill Dunlap
TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of PIKAL Petr
Sent: Tuesday, January 21, 2014 11:29 PM
To: 张以春
Cc: r-help@r-project.org
Subject: Re: [R] how to get the numbers of factors in a matrix
Hi
elaborating answers you already got
sapply(X, function(m){nlevels(factor(m$latitudes))})
tapply(N$latitudes, N$species, function(x) nlevels(factor(x)))
shall do the trick
Petr
From: 张以春 [mailto:yczh...@nigpas.ac.cn]
Sent: Tuesday, January 21, 2014 2:58 PM
To: PIKAL Petr
Cc: r-help@r-project.org
Subject: Re: RE: [R] how to get the numbers of factors in a matrix
Dear Pikal,
Thank you very much for your answer.
I think your example is just the problem I have.
In the following example you gave to me,
ff-factor(letters[1:5])
levels(ff[1:2])
[1] a b c d e
fff-ff[1:2]
nlevels(fff)
[1] 5
fff
[1] a b
Levels: a b c d e
In my understanding, fff is a subset of ff. Why fff's levels is not a, b
but a,b,c,d,e.
My problem is quite similar to the example. I just want to split the matrix
into many
subsets and calculate the levels of every subset. Can you tell me how to do?
Thank you
very much!
Best regards,
Yichun
-原始邮件-
发件人: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
发送时间: 2014年1月21日 星期二
收件人: 张以春 yczh...@nigpas.ac.cnmailto:yczh...@nigpas.ac.cn, r-
h...@r-project.orgmailto:r-help@r-project.org
r-help@r-project.orgmailto:r-
h...@r-project.org
抄送:
主题: RE: [R] how to get the numbers of factors in a matrix
Hi
It is rather difficult to understand what problem you have.
post some data e.g. by
dput(head(bigmatrix))
Maybe your problem is in a factor feature that it preserves also empty
levels until you
specifically drop them.
ff-factor(letters[1:5])
levels(ff[1:2])
[1] a b c d e
fff-ff[1:2]
nlevels(fff)
[1] 5
fff
[1] a b
Levels: a b c d e
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org
[mailto:r-help-bounces@r-
project.orgmailto:r-help-bounces@r-%0b%3e %3e project.org] On Behalf Of
???
Sent: Tuesday, January 21, 2014 7:36 AM
To: r-help@r-project.orgmailto:r-help@r-project.org
Subject: [R] how to get the numbers of factors in a matrix
Dear friends,
I have a question do not know how to resolve.
I have a big matrix composed of different columns (I use N here). A
column is species and another one is latitudes. Now, I want to know
how I can get the number of different latitudes for every species.
I have tried to split the matrix according to species (X-split(N,
N$species) and then use sapply(X, function(m){nlevels(m$latitudes)}) to
get that. But the result shows the total factor numbers of latitudes
but not the factor numbers of every species I splitted. Also, I have
tried to use tapply(N$latitudes, N$species, nlevels) to do this. The
result is the same. I am confused about this. Can someone help me with
that? Thank you very much!
Best regards,
Yichun
[[alternative HTML version deleted]]
__
R-help@r-project.orgmailto:R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou
určeny pouze
jeho adresátům.
Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně
jeho
odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze svého
systému.
Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email
jakkoliv
užívat, rozšiřovat, kopírovat či zveřejňovat.
Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či
zpožděním přenosu e-mailu.
V případě, že je tento e-mail součástí obchodního jednání:
- vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření
smlouvy, a to z
jakéhokoliv důvodu i bez uvedení důvodu.
- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout;
Odesílatel
tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce s
dodatkem či
odchylkou.
- trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným
dosažením
shody na všech jejích náležitostech.
- odesílatel tohoto emailu informuje, že není oprávněn uzavírat za
společnost žádné
smlouvy
[R] a problem with table() and duplicates
Dear List, I have a data.frame like this: name religion neighbor religion.neighbor pippo a minnie a pluto a mickey a paperino b donald a paperino b minnie b when I table(dataframe$religion) my data.frame, I get a b 2 2 of course, paperino is cited twice but should be counted once. Is there anything I can do in order to keep the data.frame the way it is but tell R to count values once if they are repeated? the point is that each row represent a relation, thus I cannot simply remove duplicates... any help more than welcome! Best regards, Simone __ 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] a problem with table() and duplicates
try: table(dataframe$religion[!duplicated(dataframe$name)]) 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. On Wed, Jan 22, 2014 at 11:04 AM, Simone Gabbriellini simone.gabbriell...@gmail.com wrote: Dear List, I have a data.frame like this: name religion neighbor religion.neighbor pippo a minnie a pluto a mickey a paperino b donald a paperino b minnie b when I table(dataframe$religion) my data.frame, I get a b 2 2 of course, paperino is cited twice but should be counted once. Is there anything I can do in order to keep the data.frame the way it is but tell R to count values once if they are repeated? the point is that each row represent a relation, thus I cannot simply remove duplicates... any help more than welcome! Best regards, Simone __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] a problem with table() and duplicates
that is awesome, thank you Jim! 2014/1/22 jim holtman jholt...@gmail.com: try: table(dataframe$religion[!duplicated(dataframe$name)]) Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. On Wed, Jan 22, 2014 at 11:04 AM, Simone Gabbriellini simone.gabbriell...@gmail.com wrote: Dear List, I have a data.frame like this: name religion neighbor religion.neighbor pippo a minnie a pluto a mickey a paperino b donald a paperino b minnie b when I table(dataframe$religion) my data.frame, I get a b 2 2 of course, paperino is cited twice but should be counted once. Is there anything I can do in order to keep the data.frame the way it is but tell R to count values once if they are repeated? the point is that each row represent a relation, thus I cannot simply remove duplicates... any help more than welcome! Best regards, Simone __ 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. -- - Simone Gabbriellini, PhD Post-doctoral Researcher ANR founded research project DIFFCERAM GEMASS, CNRS Paris-Sorbonne. mobile: +39 340 39 75 626 email: simone.gabbriell...@cnrs.fr __ 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] New version of document on R programming - with videos [French]
I hereby announce the availability of the Fourth edition of my document «Introduction à la programmation en R» (in French) in the contributed documentation section of CRAN. The document is now accompanied by a set of short videos on more challenging topics like creation and indexing of arrays, the order() function, etc. In my highly biased opinion, the illustration are pretty good. I might consider making English versions if there is sufficient interest. The videos are in the YouTube channel http://www.youtube.com/user/VincentGouletIntroR Please note that I don't currently monitor r-help regularly, so do not hesitate to write to me directly. *** [The rest of this message is in French for the target audience] La quatrième édition de mon document «Introduction à la programmation en R» est maintenant disponible dans la section de la documentation par les tiers (Contributed Documentation) de CRAN: http://cran.r-project.org/other-docs.html L’ouvrage est basé sur des notes de cours et des exercices utilisés à l’École d’actuariat de l’Université Laval. L’enseignement du langage R est axé sur l’exposition à un maximum de code — que nous avons la prétention de croire bien écrit — et sur la pratique de la programmation. C’est pourquoi les chapitres sont rédigés de manière synthétique et qu’ils comportent peu d’exemples au fil du texte. En revanche, le lecteur est appelé à lire et à exécuter le code informatique se trouvant dans les sections d’exemples à la fin de chacun des chapitres. Ce code et les commentaires qui l’accompagnent reviennent sur l’essentiel des concepts du chapitre et les complémentent souvent. Nous considérons l’exercice d’«étude active» consistant à exécuter du code et à voir ses effet comme essentielle à l’apprentissage du langage R. Le code des sections d’exemples est disponible en format électronique au même endroit que le document. Cette quatrième édition de l’ouvrage se distingue principalement de la précédente par l’ajout de liens vers des vidéos réalisées par l’auteur qui reviennent sur certains sujets plus délicats. Ces vidéos sont disponibles dans la chaîne YouTube http://www.youtube.com/user/VincentGouletIntroR Le document est publié sous licence Creative Commons. Vincent Goulet, Ph.D. Professor of Actuarial Science Directeur général de la formation continue Université Laval __ 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] how to get the numbers of factors in a matrix
On 22 Jan 2014, at 15:51 , William Dunlap wdun...@tibco.com wrote:
sapply(X, function(m){nlevels(factor(m$latitudes))})
I think that length(unique(x)) is a more direct and easier to remember
way of determining the number of unique values in the vector x,
rather than nlevels(factor(x)).
However, it may make you forget the possibility of NA:
length(unique(factor(c(1,2,NA
[1] 3
nlevels(factor(c(1,2,NA)))
[1] 2
-pd
Bill Dunlap
TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of PIKAL Petr
Sent: Tuesday, January 21, 2014 11:29 PM
To: 张以春
Cc: r-help@r-project.org
Subject: Re: [R] how to get the numbers of factors in a matrix
Hi
elaborating answers you already got
sapply(X, function(m){nlevels(factor(m$latitudes))})
tapply(N$latitudes, N$species, function(x) nlevels(factor(x)))
shall do the trick
Petr
From: 张以春 [mailto:yczh...@nigpas.ac.cn]
Sent: Tuesday, January 21, 2014 2:58 PM
To: PIKAL Petr
Cc: r-help@r-project.org
Subject: Re: RE: [R] how to get the numbers of factors in a matrix
Dear Pikal,
Thank you very much for your answer.
I think your example is just the problem I have.
In the following example you gave to me,
ff-factor(letters[1:5])
levels(ff[1:2])
[1] a b c d e
fff-ff[1:2]
nlevels(fff)
[1] 5
fff
[1] a b
Levels: a b c d e
In my understanding, fff is a subset of ff. Why fff's levels is not a, b
but a,b,c,d,e.
My problem is quite similar to the example. I just want to split the matrix
into many
subsets and calculate the levels of every subset. Can you tell me how to do?
Thank you
very much!
Best regards,
Yichun
-原始邮件-
发件人: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
发送时间: 2014年1月21日 星期二
收件人: 张以春 yczh...@nigpas.ac.cnmailto:yczh...@nigpas.ac.cn, r-
h...@r-project.orgmailto:r-help@r-project.org
r-help@r-project.orgmailto:r-
h...@r-project.org
抄送:
主题: RE: [R] how to get the numbers of factors in a matrix
Hi
It is rather difficult to understand what problem you have.
post some data e.g. by
dput(head(bigmatrix))
Maybe your problem is in a factor feature that it preserves also empty
levels until you
specifically drop them.
ff-factor(letters[1:5])
levels(ff[1:2])
[1] a b c d e
fff-ff[1:2]
nlevels(fff)
[1] 5
fff
[1] a b
Levels: a b c d e
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org
[mailto:r-help-bounces@r-
project.orgmailto:r-help-bounces@r-%0b%3e %3e project.org] On Behalf Of
???
Sent: Tuesday, January 21, 2014 7:36 AM
To: r-help@r-project.orgmailto:r-help@r-project.org
Subject: [R] how to get the numbers of factors in a matrix
Dear friends,
I have a question do not know how to resolve.
I have a big matrix composed of different columns (I use N here). A
column is species and another one is latitudes. Now, I want to know
how I can get the number of different latitudes for every species.
I have tried to split the matrix according to species (X-split(N,
N$species) and then use sapply(X, function(m){nlevels(m$latitudes)}) to
get that. But the result shows the total factor numbers of latitudes
but not the factor numbers of every species I splitted. Also, I have
tried to use tapply(N$latitudes, N$species, nlevels) to do this. The
result is the same. I am confused about this. Can someone help me with
that? Thank you very much!
Best regards,
Yichun
[[alternative HTML version deleted]]
__
R-help@r-project.orgmailto: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.
Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou
určeny pouze
jeho adresátům.
Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně
jeho
odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze svého
systému.
Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email
jakkoliv
užívat, rozšiřovat, kopírovat či zveřejňovat.
Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či
zpožděním přenosu e-mailu.
V případě, že je tento e-mail součástí obchodního jednání:
- vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření
smlouvy, a to z
jakéhokoliv důvodu i bez uvedení důvodu.
- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout;
Odesílatel
tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce s
dodatkem či
odchylkou.
- trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným
dosažením
shody na všech jejích náležitostech.
-
[R] Variance analysis
Dear R-helpers... I've be trying to run a variance analysis to compare means between various lines in various treatments. I have 10 genotypes (GEN), tested in 2 environments (ENV) and in each environment there are 3 repetitions (REP). Several traits were recoded (yield, flowering, plant height...) First I checked whether the residuals were normally distributed and then the homogeneity of variances. For those which satisfied the assumptions for ANOVA, I performed aov. I tested two models, one simple (GEN and ENV being fixed effects) and the other mixed effects (REP) aov1 - aov (Y~GEN*ENV, data=mydata) aov2 - aov (Y~GEN*ENV+Error(REP/ENV, data=mydata) When I wanted to compare the likelihood of these models, I failed performing the extractAIC for the mixed model (aov2). Is there any reason why extractAIC doesn't work in models including a random effect? As for other traits the assumption of homoscedasticity was violated I ran a lmer When I ran the following model lmer1 - lmer(Y~GEN*ENV + (1|REP), data=mydata) the following error message came Error in `contrasts-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : contrasts can be applied only to factors with 2 or more levels Could you please help me with this? Thanks Cecile [[alternative HTML version deleted]] __ 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] Problems with a R-packages
Hallo, Trust, that my request is one that can be posted under this mailing list Would like to install lossDev, Version in the Repository 3.0.0-4 and I did it under CRAN(sources). get as response versuche URL 'http://cran.at.r-project.org/src/contrib/lossDev_3.0.0-4.tar.gz' Content type 'application/x-gzip' length 1539914 bytes (1.5 Mb) URL geöffnet == downloaded 1.5 Mb ERROR: dependencies rjags, logspline are not available for package lossDev * removing /Library/Frameworks/R.framework/Versions/3.0/Resources/library/lossDev Die heruntergeladenen Quellpakete sind in /private/var/folders/1d/7y5t0hy57q5bphrlsrgkq43hgp/T/RtmpipUlzT/downloaded_packages Question: Was this download successful? Because in the R-packages-installation-window, there appears no remark in the column installed version, it is still blank. Next I tried to install rjags and logspline. Same result regarding download. Even if this question is simple, any help highly appreciated. I am available via Skype too, to share my screen. Using MacBook Pro. Regards, Andreas [[alternative HTML version deleted]] __ 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] reduce space between factors groups in a graph
Dear all, I am preparing a graph in which values derived from 2 variable are displayed using the stripchart function. I have applied the factor function to separate the 2 variables in two groups, although I noticed that the graph works anyway even without the factorisation of the variables. However there is a lot of space between the two factors, thus most of the graph is empty. It is possible to reduce the space between the factors? The function I have implemented is more or less like this: stripchart ( Y ~ factor(X), method = stack , offset=1/3, vertical = TRUE, las=1, pch=19, ylim=Y, xlim=c(1, 2), par(mai=c(1,1,0.5,0.1)) ) Best regards Luigi [[alternative HTML version deleted]] __ 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] install ggplot2
Hi Dear helper, I installed Rx643.0.2 on my windows 7 Enterprise computer, and I installed all the packages as well. However, when I tried to use ggplot2 package with the commend library(ggplot2, lib.loc=C:/Users/JXD043/Documents/R/win-library/3.0), I got the following error message: Error in library(ggplot2, lib.loc = C:/Users/JXD043/Documents/R/win-library/3.0) : there is no package called 'ggplot2' Then I tried to reinstall the package with install.packages(ggplot2) The message I got is : Installing package into 'C:/Users/JXD043/Documents/R/win-library/3.0' (as 'lib' is unspecified) trying URL 'http://cran.rstudio.com/bin/windows/contrib/3.0/ggplot2_0.9.3.1.zip' Content type 'application/zip' length 2657708 bytes (2.5 Mb) opened URL downloaded 2.5 Mb package 'ggplot2' successfully unpacked and MD5 sums checked Warning in install.packages : cannot remove prior installation of package 'ggplot2' The downloaded binary packages are in C:\Users\JXD043\AppData\Local\Temp\RtmpuIVA2s\downloaded_packages So I extracted the files into the folder, however, I still can't get the library work. Please help me and let me know what can I do to fix it. Thanks! Jie DAI [[alternative HTML version deleted]] __ 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] Multiple corrgrams or joining jpg/png
Hello world, I have a database with time series of concentration of nutrients for several lakes. I wanted *one* corrgram for each nutrient in all lakes (correlation of a single nutrient content of all lakes in different years). The single corrgram works pretty well, but I cannot create a page with all nutrients on one page, i.e. several corrgrams on one page. The usual mfrow and layout commands do not work (splom has the same problem). I wonder if anyone already has a solution. A workaround would be to write a single jpg/png for each corrgram and join them. Is there a possibility to do this *automatically* or even *within* R? (I do not want to align 300 figures manually :-) Merci and greetings, GEorg -- Georg Hoermann, Department of Hydrology and Water Resources Management Kiel University, Germany +49/431/2190916, mo: +49/176/64335754, icq:348340729, skype: ghoermann __ 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] geo_bar x= and y= warnings and error help
Any insight on issues leading to the following error modes would be appreciated. #Version_1 CALL alphaDivOTU - ggplot(data=alphaDivOTU_pt1to5, aes(y = Num.OTUs,x = Patient,fill = Timepoint)) + geom_bar(position = position_dodge) + theme(text = element_text(family = 'Helvetica-Narrow',size = 18.0)) + scale_fill_manual(guide = guide_legend(),values = c(forestgreen,gray44,dodgerblue2,royalblue2,royalblue4,blue3)) + scale_y_continuous(breaks = pretty_breaks(n = 10.0,min.n = 5.0)) ggsave(plot=alphaDivOTU, filename='alphaDivOTU.png', scale=1, dpi=300, width=10, height=10, units=c(cm)) #Version_1 Error modes Mapping a variable to y and also using stat=bin. With stat=bin, it will attempt to set the y value to the count of cases in each group. This can result in unexpected behavior and will not be allowed in a future version of ggplot2. If you want y to represent counts of cases, use stat=bin and don't map a variable to y. If you want y to represent values in the data, use stat=identity. See ?geom_bar for examples. (Deprecated; last used in version 0.9.2) Error in .$position$adjust : object of type 'closure' is not subsettable #Version_2 CALL alphaDivOTU - ggplot(data=alphaDivOTU_pt1to5, aes(y = Num.OTUs,x = Patient,fill = Timepoint)) + geom_bar(position = position_dodge, stat = identity) + theme(text = element_text(family = 'Helvetica-Narrow',size = 18.0)) + scale_fill_manual(guide = guide_legend(),values = c(forestgreen,gray44,dodgerblue2,royalblue2,royalblue4,blue3)) + scale_y_continuous(breaks = pretty_breaks(n = 10.0,min.n = 5.0)) ggsave(plot=alphaDivOTU, filename='alphaDivOTU.png', scale=1, dpi=300, width=10, height=10, units=c(cm)) #For Version_2 I get the error: Error in stat$parameters : object of type 'closure' is not subsettable [[alternative HTML version deleted]] __ 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] Multiple corrgrams or joining jpg/png
1. When using a package from CRAN, you usually want to copy the package author on the question. (In this case, me.) 2. The corrgram function is basically a wrapper around the pairs() function. What you want to do doesn't seem to be possible based on this discussion: https://stat.ethz.ch/pipermail/r-help/2004-December/063112.html Kevin On Wed, Jan 22, 2014 at 3:40 PM, Georg Hörmann ghoerm...@hydrology.uni-kiel.de wrote: Hello world, I have a database with time series of concentration of nutrients for several lakes. I wanted *one* corrgram for each nutrient in all lakes (correlation of a single nutrient content of all lakes in different years). The single corrgram works pretty well, but I cannot create a page with all nutrients on one page, i.e. several corrgrams on one page. The usual mfrow and layout commands do not work (splom has the same problem). I wonder if anyone already has a solution. A workaround would be to write a single jpg/png for each corrgram and join them. Is there a possibility to do this *automatically* or even *within* R? (I do not want to align 300 figures manually :-) Merci and greetings, GEorg -- Georg Hoermann, Department of Hydrology and Water Resources Management Kiel University, Germany +49/431/2190916, mo: +49/176/64335754, icq:348340729, skype: ghoermann __ 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. -- Kevin Wright [[alternative HTML version deleted]] __ 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] xyplots in lattice - strange behaviour, possible bug?
Hello everyone,
I'm very green on R, I'm following a Coursera course about it when I
hit a problem when I rewrote the same code the professor use in the
lecture.
I'm running Win 7 x64, R 3.0.2 x64 and the last version of Rstudio IDE
I put up this script:
library(lattice)
x - rnorm (100)
z - x + rnorm(100)
f - gl(2,50,labels =c(Groups 1 , Groups 2))
xyplot (z ~ x | f,
panel = function (x, z, ...) {
panel.xyplot(x,z, ...)
panel.abline(h = median(z),
lty=2
)})
In my box don't work, it give no error in the terminal, the plotting
windonw will be opened, the graphbox drawed with all the ticks and the
titles as intended but instead of the actual data plot inside the
graph I have this error Error using packet x argment z is
missing, with no default where x is 1 or 2 as the script draw two
graphs.
I reported this behaviour in the lecture forum and someone replicated it.
I replicated this behaviour even with R alone running the above script
with the same results.
If I call traceback() no value is given, there is no traceback.
Apparently not everyone could replicate this behaviour for some reason.
As you could see the code must work but didn't.
A similar thing happens if I change the part after the function with
another like:
... same code of above...
panel= function(x,y, ...) {
panel.xyplot(x,z, ...)
fit - lm(y~x)
panel.abline(fit)
})
but don't happens if I call a xyplot without calling a function in it.
Have any ideas?
__
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] xyplots in lattice - strange behaviour, possible bug?
Well, if the professor wrote that, it wouldn't have run for him
either! You need to take better notes.
What's going on: You need to distinguish between formal and actual arguments.
?panel.xyplot
tells you that the formal arguments for this function are x,**y** ,...
(emphasis added) and NOT x,**z**,...
The **actual** argument for y passed to the function will be z. So
change your z to a y in your function call and it will run:
library(lattice)
x - rnorm (100)
z - x + rnorm(100)
f - gl(2,50,labels =c(Groups 1 , Groups 2))
xyplot (z ~ x | f,
panel = function (x, y, ...) {
panel.xyplot(x,y, ...)
panel.abline(h = median(y),
lty=2
)})
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
H. Gilbert Welch
On Wed, Jan 22, 2014 at 2:10 PM, Manlio Calvi manlio.ca...@gmail.com wrote:
Hello everyone,
I'm very green on R, I'm following a Coursera course about it when I
hit a problem when I rewrote the same code the professor use in the
lecture.
I'm running Win 7 x64, R 3.0.2 x64 and the last version of Rstudio IDE
I put up this script:
library(lattice)
x - rnorm (100)
z - x + rnorm(100)
f - gl(2,50,labels =c(Groups 1 , Groups 2))
xyplot (z ~ x | f,
panel = function (x, z, ...) {
panel.xyplot(x,z, ...)
panel.abline(h = median(z),
lty=2
)})
In my box don't work, it give no error in the terminal, the plotting
windonw will be opened, the graphbox drawed with all the ticks and the
titles as intended but instead of the actual data plot inside the
graph I have this error Error using packet x argment z is
missing, with no default where x is 1 or 2 as the script draw two
graphs.
I reported this behaviour in the lecture forum and someone replicated it.
I replicated this behaviour even with R alone running the above script
with the same results.
If I call traceback() no value is given, there is no traceback.
Apparently not everyone could replicate this behaviour for some reason.
As you could see the code must work but didn't.
A similar thing happens if I change the part after the function with
another like:
... same code of above...
panel= function(x,y, ...) {
panel.xyplot(x,z, ...)
fit - lm(y~x)
panel.abline(fit)
})
but don't happens if I call a xyplot without calling a function in it.
Have any ideas?
__
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] xyplots in lattice - strange behaviour, possible bug?
On Jan 22, 2014, at 3:02 PM, Bert Gunter gunter.ber...@gene.com wrote: Well, if the professor wrote that, it wouldn't have run for him either! Fortune? Or just a great line? Don McKenzie Research Ecologist Pacific Wildland Fire Science Lab US Forest Service Affiliate Professor School of Environmental and Forest Sciences University of Washington d...@uw.edu __ 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] xyplots in lattice - strange behaviour, possible bug?
... and I should have added (more complexity!) that the formula method
of xyplot parses the formula and passes down what's on the left hand
side of ~ to the y argument of the panel function.
And if all else fails, read the docs! -- in this case for ?xyplot --
where it explicitly says:
... A panel function appropriate for the functions described here
would usually expect arguments named x and y, which would be provided
by the conditioning process
And please oh please do not suggest as a newbie that your confusion is
due to bugs in long used and extensively tested R code. That just
seems arrogant to me (I didn't get it so the software must be
buggy).
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
H. Gilbert Welch
On Wed, Jan 22, 2014 at 3:02 PM, Bert Gunter bgun...@gene.com wrote:
Well, if the professor wrote that, it wouldn't have run for him
either! You need to take better notes.
What's going on: You need to distinguish between formal and actual arguments.
?panel.xyplot
tells you that the formal arguments for this function are x,**y** ,...
(emphasis added) and NOT x,**z**,...
The **actual** argument for y passed to the function will be z. So
change your z to a y in your function call and it will run:
library(lattice)
x - rnorm (100)
z - x + rnorm(100)
f - gl(2,50,labels =c(Groups 1 , Groups 2))
xyplot (z ~ x | f,
panel = function (x, y, ...) {
panel.xyplot(x,y, ...)
panel.abline(h = median(y),
lty=2
)})
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
H. Gilbert Welch
On Wed, Jan 22, 2014 at 2:10 PM, Manlio Calvi manlio.ca...@gmail.com wrote:
Hello everyone,
I'm very green on R, I'm following a Coursera course about it when I
hit a problem when I rewrote the same code the professor use in the
lecture.
I'm running Win 7 x64, R 3.0.2 x64 and the last version of Rstudio IDE
I put up this script:
library(lattice)
x - rnorm (100)
z - x + rnorm(100)
f - gl(2,50,labels =c(Groups 1 , Groups 2))
xyplot (z ~ x | f,
panel = function (x, z, ...) {
panel.xyplot(x,z, ...)
panel.abline(h = median(z),
lty=2
)})
In my box don't work, it give no error in the terminal, the plotting
windonw will be opened, the graphbox drawed with all the ticks and the
titles as intended but instead of the actual data plot inside the
graph I have this error Error using packet x argment z is
missing, with no default where x is 1 or 2 as the script draw two
graphs.
I reported this behaviour in the lecture forum and someone replicated it.
I replicated this behaviour even with R alone running the above script
with the same results.
If I call traceback() no value is given, there is no traceback.
Apparently not everyone could replicate this behaviour for some reason.
As you could see the code must work but didn't.
A similar thing happens if I change the part after the function with
another like:
... same code of above...
panel= function(x,y, ...) {
panel.xyplot(x,z, ...)
fit - lm(y~x)
panel.abline(fit)
})
but don't happens if I call a xyplot without calling a function in it.
Have any ideas?
__
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] ETAS-Help
On 01/22/2014 10:03 PM, katerina stavrianaki wrote: Hello, My name is Katerina, i am new to R and i am working with the ETAS package. My goal is to fit the spatiotemporal etas model to an aftershock sequence ( atach file example.csv).I have installed the packages: spatstat, SAPP and ETAS. By reading the ETAS package manual i saw the data must be in class ppx. Could you please help me on how to convert my data (example.csv) into class ppx ? Thank you in advance,Katerina Hi Katerina, Your example data didn't make it through to the list, but it seems that what you want to do is to read that file into a data frame (see read.csv) and then pass that data frame to the ppx function in spatstat. Read the help page for ppx to see how this is done, I think that all you need is the data frame and a vector of data types in the ccord.type argument. It looks like you should only pass those columns of the data frame that are spatial, temporal or local coordinates, and marks, so make sure that you know how to specify a subset of columns (e.g. df[,c(2,4,5,6)]) if you have other information in the data frame. Jim __ 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] subset and na.rm not really suppressing NA values
I have a dataset mydf with a field EMAIL_ADDRESS. When importing, I specified: mydf - read.csv(file = extract, header = TRUE, stringsAsFactors = FALSE, na.strings=c(NA,)) I've also tried setting na.strings= c(NA,,NA) but I don't know if it's appropriate to put NA there. I'm running a - subset(mydf, VALID_EMAIL == FALSE, na.rm = TRUE, select = EMAIL_ADDRESS) dput(head(a,5)) structure(list(EMAIL_ADDRESS = c(NA_character_, NA_character_, NA_character_, NA_character_, NA_character_)), .Names = EMAIL_ADDRESS, row.names = c(17L, 22L, 23L, 24L, 30L), class = data.frame) The results show a lot of NA values on screen and in the dput statement. I don't quite understand why it is doing that. I would have expected it to exclude those since I had the na.rm = TRUE statement. Do you have any suggestions? Thanks! -- Jeff [[alternative HTML version deleted]] __ 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] subset and na.rm not really suppressing NA values
I don't think na.rm is a valid at parameter for the subset function. I would normally use the is.na function to logically test for NA values. I also don't know where your VALID_EMAIL variable is coming from. a - subset(mydf, !is.na(EMAIL_ADDRESS)) The na.strings argument to read.csv and friends is used to help recognise strings in the input that should be treated as NA. If you don't see NA in your input file then it will have no effect on the data import. --- 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. Jeff Johnson mrjeffto...@gmail.com wrote: I have a dataset mydf with a field EMAIL_ADDRESS. When importing, I specified: mydf - read.csv(file = extract, header = TRUE, stringsAsFactors = FALSE, na.strings=c(NA,)) I've also tried setting na.strings= c(NA,,NA) but I don't know if it's appropriate to put NA there. I'm running a - subset(mydf, VALID_EMAIL == FALSE, na.rm = TRUE, select = EMAIL_ADDRESS) dput(head(a,5)) structure(list(EMAIL_ADDRESS = c(NA_character_, NA_character_, NA_character_, NA_character_, NA_character_)), .Names = EMAIL_ADDRESS, row.names = c(17L, 22L, 23L, 24L, 30L), class = data.frame) The results show a lot of NA values on screen and in the dput statement. I don't quite understand why it is doing that. I would have expected it to exclude those since I had the na.rm = TRUE statement. Do you have any suggestions? Thanks! __ 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] unable to carry through object in a nested function
Hi There,
I am having trouble carrying through an object listed in the outer function
into the inner function of a nested pair.
This is my code:
FindGreaterThanProportion- function (y, SSThreshold) {
DenomCells-length(y)
NumerCells-subset(y,ySSThreshold)
PropCells-length(NumerCells)/DenomCells
return (PropCells)
}
GetPropPlot- function (PrePropData, direction, SSThreshold, cell, stim) {
# Split up the dataframe with ddply and apply the function
print(direction)
print (SSThreshold)
if (direction==) {
PropData-ddply(PrePropData,.(SUBJECT, STIM, CELL, SIGNAL, DAY),
summarise,
PROP=FindGreaterThanProportion(SIMSCORE, SSThreshold))
}
PlotSubset - subset(PropData, PropData$STIM==stim PropData$CELL == cell)
PropPlot-ggplot(data=PlotSubset, aes(x=DAY, y=PROP)) +
geom_line() +
geom_point() +
facet_grid(SIGNAL~SUBJECT)
return(FinalPropPlot)
}
When I run the code, I get this error:
[1]
[1] 2
Error in subset.default(y, y SSThreshold) :
object 'SSThreshold' not found
It seems as if SSThreshold is not being passed into FindGreaterThanProportion.
Any help would be appreciated to determine what I am doing incorrectly.
Thanks in advance.
George Chen
This email message may contain legally privileged and/or confidential
information. If you are not the intended recipient(s), or the employee or
agent responsible for the delivery of this message to the intended
recipient(s), you are hereby notified that any disclosure, copying,
distribution, or use of this email message is prohibited. If you have received
this message in error, please notify the sender immediately by e-mail and
delete this email message from your computer. Thank you.
__
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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.
Re: [R] unable to carry through object in a nested function
Hi,
This is a resend of a previous message reproduced below but with sample data to
run.
Thanks.
George Chen
Hi There,
I am having trouble carrying through an object listed in the outer function
into the inner function of a nested pair.
sample data below -
library(plyr)
SUBJECT-c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)
STIM-c(No,No,No,No,No,No,No,No,No,No,No,No,No,No,No,No,No,No,No,No)
CELL-c(CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4)
SIGNAL-c(ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC)
DAY-c(7,7,7,7,7,7,7,7,7,7,1,1,1,1,1,1,1,1,1,1)
SIMSCORE-c(2,3,4,5,2,3,4,5,2,3,4,5,2,3,4,5,2,3,4,5)
PrePropData-data.frame(SUBJECT, STIM, CELL, SIGNAL, DAY, SIMSCORE)
This is my code:
FindGreaterThanProportion- function (y, SSThreshold) {
DenomCells-length(y)
NumerCells-subset(y,ySSThreshold)
PropCells-length(NumerCells)/DenomCells
return (PropCells)
}
GetPropPlot- function (PrePropData, direction, SSThreshold, cell, stim) {
# Split up the dataframe with ddply and apply the function
print(direction)
print (SSThreshold)
if (direction==) {
PropData-ddply(PrePropData,.(SUBJECT, STIM, CELL, SIGNAL, DAY),
summarise,
PROP=FindGreaterThanProportion(SIMSCORE, SSThreshold))
}
PlotSubset - subset(PropData, PropData$STIM==stim PropData$CELL == cell)
PropPlot-ggplot(data=PlotSubset, aes(x=DAY, y=PROP)) +
geom_line() +
geom_point() +
facet_grid(SIGNAL~SUBJECT)
return(FinalPropPlot)
}
GetPropPlot(PrePropData, direction=, SSThreshold=3, CD4, No)
When I run the code, I get this error:
[1]
[1] 3
Error in subset.default(y, y SSThreshold) :
object 'SSThreshold' not found
It seems as if SSThreshold is not being passed into FindGreaterThanProportion.
Any help would be appreciated to determine what I am doing incorrectly.
Thanks in advance.
George Chen
This email message may contain legally privileged and/or confidential
information. If you are not the intended recipient(s), or the employee or
agent responsible for the delivery of this message to the intended
recipient(s), you are hereby notified that any disclosure, copying,
distribution, or use of this email message is prohibited. If you have received
this message in error, please notify the sender immediately by e-mail and
delete this email message from your computer. Thank you.
__
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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.
Re: [R] unable to carry through object in a nested function
Hi,
Try this:
GetPropPlot- function (PrePropData, direction, SSThreshold, cell, stim) {
if (direction== ) {
PropData-ddply(PrePropData,.(SUBJECT, STIM, CELL, SIGNAL, DAY),
here(summarise),
PROP=FindGreaterThanProportion(SIMSCORE, SSThreshold))
}
PlotSubset - subset(PropData, STIM==stim CELL == cell)
FinalPropPlot-ggplot(data=PlotSubset, aes(x=DAY, y=PROP)) +
geom_line() +
geom_point() +
facet_grid(SIGNAL~SUBJECT)
return(FinalPropPlot)
}
GetPropPlot(PrePropData, direction=, SSThreshold=3, CD4, No)
A.K.
On Wednesday, January 22, 2014 10:51 PM, Chen, George
george.c...@roswellpark.org wrote:
Hi,
This is a resend of a previous message reproduced below but with sample data to
run.
Thanks.
George Chen
Hi There,
I am having trouble carrying through an object listed in the outer function
into the inner function of a nested pair.
sample data below -
library(plyr)
SUBJECT-c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)
STIM-c(No,No,No,No,No,No,No,No,No,No,No,No,No,No,No,No,No,No,No,No)
CELL-c(CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4,CD4)
SIGNAL-c(ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC,ABC)
DAY-c(7,7,7,7,7,7,7,7,7,7,1,1,1,1,1,1,1,1,1,1)
SIMSCORE-c(2,3,4,5,2,3,4,5,2,3,4,5,2,3,4,5,2,3,4,5)
PrePropData-data.frame(SUBJECT, STIM, CELL, SIGNAL, DAY, SIMSCORE)
This is my code:
FindGreaterThanProportion- function (y, SSThreshold) {
DenomCells-length(y)
NumerCells-subset(y,ySSThreshold)
PropCells-length(NumerCells)/DenomCells
return (PropCells)
}
GetPropPlot- function (PrePropData, direction, SSThreshold, cell, stim) {
# Split up the dataframe with ddply and apply the function
print(direction)
print (SSThreshold)
if (direction==) {
PropData-ddply(PrePropData,.(SUBJECT, STIM, CELL, SIGNAL, DAY),
summarise,
PROP=FindGreaterThanProportion(SIMSCORE, SSThreshold))
}
PlotSubset - subset(PropData, PropData$STIM==stim PropData$CELL == cell)
PropPlot-ggplot(data=PlotSubset, aes(x=DAY, y=PROP)) +
geom_line() +
geom_point() +
facet_grid(SIGNAL~SUBJECT)
return(FinalPropPlot)
}
GetPropPlot(PrePropData, direction=, SSThreshold=3, CD4, No)
When I run the code, I get this error:
[1]
[1] 3
Error in subset.default(y, y SSThreshold) :
object 'SSThreshold' not found
It seems as if SSThreshold is not being passed into FindGreaterThanProportion.
Any help would be appreciated to determine what I am doing incorrectly.
Thanks in advance.
George Chen
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Re: [R] install ggplot2
Hard to say. Read the Posting Guide for useful suggestions on getting help, including providing the output of sessionInfo() and avoiding HTML formatted emails because what you see is not what we see. I suggest you delete the ggplot2 subdirectory in your win-library directory and try the install.packages(ggplot2) command again. If you have installed packages while running R As Administrator then you may have permissions problems on your win-library directory and need to delete it (As Administrator) and reinstall R. --- 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. Dai, Jie j...@amfam.com wrote: Hi Dear helper, I installed Rx643.0.2 on my windows 7 Enterprise computer, and I installed all the packages as well. However, when I tried to use ggplot2 package with the commend library(ggplot2, lib.loc=C:/Users/JXD043/Documents/R/win-library/3.0), I got the following error message: Error in library(ggplot2, lib.loc = C:/Users/JXD043/Documents/R/win-library/3.0) : there is no package called 'ggplot2' Then I tried to reinstall the package with install.packages(ggplot2) The message I got is : Installing package into 'C:/Users/JXD043/Documents/R/win-library/3.0' (as 'lib' is unspecified) trying URL 'http://cran.rstudio.com/bin/windows/contrib/3.0/ggplot2_0.9.3.1.zip' Content type 'application/zip' length 2657708 bytes (2.5 Mb) opened URL downloaded 2.5 Mb package 'ggplot2' successfully unpacked and MD5 sums checked Warning in install.packages : cannot remove prior installation of package 'ggplot2' The downloaded binary packages are in C:\Users\JXD043\AppData\Local\Temp\RtmpuIVA2s\downloaded_packages So I extracted the files into the folder, however, I still can't get the library work. Please help me and let me know what can I do to fix it. Thanks! Jie DAI [[alternative HTML version deleted]] __ 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.
