Re: [R] Job scheduling in R
Cron still works, but launchd/launchctl seems to be preferred by some if you're on Mac OS X. J. On Fri, Nov 19, 2010 at 7:09 PM, Steve Lianoglou mailinglist.honey...@gmail.com wrote: Hi, On Fri, Nov 19, 2010 at 10:09 AM, amit jain budd...@indiatimes.com wrote: Hi All, Can anyone point to any package/resouce to schedule a job in R which runs a .R file at a specified time ? I couldn't find anything useful in R Reference manual or RSiteSearch. I am sure its there but i am unable to locate. If you're on a *nix-type machine, look into cron: http://en.wikipedia.org/wiki/Cron -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ 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] assignment operator saving factor level as number
Perhaps this will help: test1 - test2 - data.frame(col1=factor(c(1:3), labels=c(a, b, c))) test3 - data.frame(col1 = 1:3) Now: test2[2,1] - test1$col1[1] test2$col1 [1] a a c Levels: a b c vs test3[2,1] - test1$col1[1] test3$col1 [1] 1 1 3 Because test3's first column, col1, is a vector of numeric, and each element of a vector must have the same data type (numeric, factor, etc), it will coerce the data coming in to have the same data type (if it can). In this case, the data type is numeric. Had it been a character coming in, because it can't coerce a character to a numeric, it would have made the entire vector a vector of characters: test3[2,1] - 'b' test3$col1 [1] 1 b 3 Hope that demonstrates what's probably going on, Jeff. On Fri, Nov 5, 2010 at 3:54 PM, Wade Wall wade.w...@gmail.com wrote: Hi all, I have a dataframe (df1) that I am trying to select values from to a second dataframe that at the current time is only for the selected items from df1 (df2). The values that I am trying to save from df1 are factors with alphanumeric names df1 looks like this: 'data.frame': 3014 obs. of 13 variables: $ Num : int 1 1 1 2 2 2 3 3 3 4 ... $ Tag_Num : int 1195 1195 1195 1162 1162 1162 1106 1106 1106 1173 ... $ Site : Factor w/ 25 levels PYBR002A,PYBR003B,..: 1 1 1 1 1 1 1 1 1 1 ... $ Site_IndNum : Factor w/ 1044 levels PYBR002A_001,..: 1 1 1 2 2 2 3 3 3 4 ... ... $ Area : num 463.3 29.5 101.8 152.9 34.6 ... However, whenever I try to assign values, like this df2[j,1]-df2$Site[i] the values are changed from alphanumeric (e.g. PYBR003A) to numerals (e.g. 1). Does anyone know why this is happening and how I can assign the actual values from df1 to df2? Thanks in advance, Wade [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] assignment operator saving factor level as number
Glad you figured it out, but just be aware that if you set one value of the column to be a character, it will make the whole vector characters. This could cause issues for analysis if you need numerics or factors. If the column is supposed to be a factor to begin with, set it to be so; if you have two data frames, one with a column of factors (dat2) and one with what should be a column of factors (dat1), you can use something like this: dat1$columnThatShouldBeFactors - as.factor( dat1$columnThatShouldBeFactors, levels=levels(dat2$columnThatIsAlreadyFactors) ) Cheers, Jeff. On Fri, Nov 5, 2010 at 6:03 PM, Wade Wall wade.w...@gmail.com wrote: Hi all, Thanks for the help. Jeffrey was right; my initial dataframe did not have the columns defined for factors. I solved it using Jorge's example of using as.character. Sorry for not being more clear before. Wade On Fri, Nov 5, 2010 at 4:12 PM, Jeffrey Spies jsp...@virginia.edu wrote: Perhaps this will help: test1 - test2 - data.frame(col1=factor(c(1:3), labels=c(a, b, c))) test3 - data.frame(col1 = 1:3) Now: test2[2,1] - test1$col1[1] test2$col1 [1] a a c Levels: a b c vs test3[2,1] - test1$col1[1] test3$col1 [1] 1 1 3 Because test3's first column, col1, is a vector of numeric, and each element of a vector must have the same data type (numeric, factor, etc), it will coerce the data coming in to have the same data type (if it can). In this case, the data type is numeric. Had it been a character coming in, because it can't coerce a character to a numeric, it would have made the entire vector a vector of characters: test3[2,1] - 'b' test3$col1 [1] 1 b 3 Hope that demonstrates what's probably going on, Jeff. On Fri, Nov 5, 2010 at 3:54 PM, Wade Wall wade.w...@gmail.com wrote: Hi all, I have a dataframe (df1) that I am trying to select values from to a second dataframe that at the current time is only for the selected items from df1 (df2). The values that I am trying to save from df1 are factors with alphanumeric names df1 looks like this: 'data.frame': 3014 obs. of 13 variables: $ Num : int 1 1 1 2 2 2 3 3 3 4 ... $ Tag_Num : int 1195 1195 1195 1162 1162 1162 1106 1106 1106 1173 ... $ Site : Factor w/ 25 levels PYBR002A,PYBR003B,..: 1 1 1 1 1 1 1 1 1 1 ... $ Site_IndNum : Factor w/ 1044 levels PYBR002A_001,..: 1 1 1 2 2 2 3 3 3 4 ... ... $ Area : num 463.3 29.5 101.8 152.9 34.6 ... However, whenever I try to assign values, like this df2[j,1]-df2$Site[i] the values are changed from alphanumeric (e.g. PYBR003A) to numerals (e.g. 1). Does anyone know why this is happening and how I can assign the actual values from df1 to df2? Thanks in advance, Wade [[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.
Re: [R] parallel for loop
Take a look at the parallel computing section of: http://cran.r-project.org/web/views/HighPerformanceComputing.html specifically the line concerning the foreach package. Jeff. On Sun, Oct 31, 2010 at 6:38 PM, sachinthaka.abeyward...@allianz.com.au wrote: Hi all, Just following on from a previous thread (for loop). Is there a parallel 'for' loop like matlab (parfor maybe?). I know there was a Nvidia GPU version for blas somewhere. But is there a CPU or a GPU version of the for loop? Thanks, Sachin p.s. sorry about the corporate notice below: cant get rid of it. Don't have other mail client access at the office :( --- Please consider the environment before printing this email --- Allianz - Best General Insurance Company of the Year 2010* Allianz - General Insurance Company of the Year 2009+ * Australian Banking and Finance Insurance Awards + Australia and New Zealand Insurance Industry Awards This email and any attachments has been sent by Allianz Australia Insurance Limited (ABN 15 000 122 850) and is intended solely for the addressee. It is confidential, may contain personal information and may be subject to legal professional privilege. Unauthorised use is strictly prohibited and may be unlawful. If you have received this by mistake, confidentiality and any legal privilege are not waived or lost and we ask that you contact the sender and delete and destroy this and any other copies. In relation to any legal use you may make of the contents of this email, you must ensure that you comply with the Privacy Act (Cth) 1988 and you should note that the contents may be subject to copyright and therefore may not be reproduced, communicated or adapted without the express consent of the owner of the copyright. Allianz will not be liable in connection with any data corruption, interruption, delay, computer virus or unauthorised access or amendment to the contents of this email. If this email is a commercial electronic message and you would prefer not to receive further commercial electronic messages from Allianz, please forward a copy of this email to unsubscr...@allianz.com.au with the word unsubscribe in the subject header. __ 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] ggplot2: how to label lines?
Hi, all, Let's say I have some time series data--10 subjects measured 20 times--that I plot as follows: library(ggplot2) dat - data.frame(subject=as.factor(rep(1:10, each=20)), time=rep(1:20, 10), measure=as.vector(replicate(10, rnorm(20, mean=runif(1, 0, 15), sd=runif(1, 1, 3) p - qplot(time, measure, data=dat, colour=subject, geom=line) p What would be the preferred way to add a single label to every line? For instance, labels might be most readable at the beginning, end, or peak (max value) of every line. I could do: p + geom_text(aes(label=subject)) But this gets messy when the labels are longer than single digits; instead, I want a single label per line. I suppose a dataset could be put together composed of single points and labels and then layered atop p, but perhaps there is a more direct method. Any help would be greatly appreciated, Jeff. __ 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] Text wrapping in R
I would demonstrate one of the many LaTeX table functions. Off hand, packages xtable, hmisc, and quantreg all have functions that convert R objects to LaTeX tables. If they're unwilling to work in LaTeX, you can use something like LaTeXiT or Laeqed to create PDFs or PNGs of the tables for insertion into whatever report tool they use. Note that the latter will require a change to the preamble to not constantly be in math mode. Mac: http://www.chachatelier.fr/programmation/latexit_en.php Windows: http://www.thrysoee.dk/laeqed/ Hope that helps, Jeff. On Mon, Oct 25, 2010 at 12:59 AM, Johannes Huesing johan...@huesing.name wrote: I am about to give an introduction to R to some clinical data managers used to SAS. There is already a lot of material in printed form and on the web that paves the way. What I haven't found so far are text wrapping capabilities in setting tables in raw text as in SAS PROC REPORT. At the moment i would direct them at producing HTML output from R and pipe the result through lynx. Coming from SAS they may not be prepared to walk the Unix way of choosing the best tool for the right job. Have I overlooked a package that does something similar to SAS PROC REPORT? -- Johannes Hüsing There is something fascinating about science. One gets such wholesale returns of conjecture mailto:johan...@huesing.name from such a trifling investment of fact. http://derwisch.wikidot.com (Mark Twain, Life on the Mississippi) __ 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] Help reading table rows into lists
To get just the list you wanted, Gabor's solution is more elegant, but
here's another using the apply family. First, your data:
dat - scan(file=/g/bork8/waller/test_COGtoPath.txt,what=character,sep=\n)
I expect dat to be a vector of strings where each string is a line of
values separated by tabs, which I think, by looking at your other
code, is what you get.
sapply(dat, function(x){
tmp-unlist(strsplit(x, '\t', fixed=T))
out - list(tmp[seq_along(tmp)[-1]])
names(out) - tmp[1]
out
}, USE.NAMES=F)
The one difference between the two is that if you have a COG with no
pathways (might not be realistic or that big of a deal), this solution
will have the COG name in the list with a value of character(0) where
Gabor's will omit the COG completely. Again, probably not a big deal.
Cheers,
Jeff.
On Sun, Oct 10, 2010 at 11:40 AM, Alison Waller alison.wal...@embl.de wrote:
Hi all,
I have a large table mapping thousands of COGs(groups of genes) to pathways.
# Ex
COG0001 patha pathb pathc
COG0002 pathd pathe
COG0003 pathe pathf pathg pathh
##
I would like to combine this information into a big list such as below
COG2PATHWAY-list(COG0001=c(patha,pathb,pathc),COG0002=c(pathd,pathe),COG0003=c(pathf,pathg,pathh))
I am stuck and have tried various methods involving (probably mangled)
versions of lappy and loops.
Any suggestions on the most efficient way to do this would be great.
Thanks,
Alison
Here is my latest attempt.
#
line_num-length(scan(file=/g/bork8/waller/test_COGtoPath.txt,what=character,sep=\n))
COG2Path-vector(list,line_num)
COG2Path-lapply(1:(line_num-1),function(x)
scan(file=/g/bork8/waller/test_COGtopath.txt,skip=x,nlines=1,quiet=T,what='character',sep=\t))
#
I am getting an error
#
COG2Path-lapply(1:(line_num-1),function(x)
scan(file=/g/bork8/waller/test_COGtopath.txt,skip=x,nlines=1,quiet=T,what='character',sep=\t))
Error in file(file, r) : cannot open the connection
In addition: Warning message:
In file(file, r) :
But if I do scan alone I don't get an error
# then I suppose it looks like the easiest wasy to name the list variables
is using unix to cut the first column out and then read that in.
names(COG2Path)-scan(file=/g/bork8/waller/test_col_names.txt,sep=\t,what=character)
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Re: [R] StrSplit
Jim's solution is the ideal way to read in the data: using the sep=; argument in read.table. However, if you do for some reason have a vector of strings like the following (maybe someone gives you an Rdata file instead of the raw data file): MF_Data - c(106506;AIG India Liquid Fund-Institutional Plan-Daily Dividend Option;1001.;1001.;1001.;02-Oct-2010,106511;AIG India Liquid Fund-Institutional Plan-Growth Option;1210.4612;1210.4612;1210.4612;02-Oct-2010) Then you can use this to get a data frame: as.data.frame(do.call(rbind, lapply(MF_Data, function(x) unlist(strsplit(x, ';') Cheers, Jeff. On Sat, Oct 9, 2010 at 12:30 PM, jim holtman jholt...@gmail.com wrote: Is this what you are after: x - c(Scheme Code;Scheme Name;Net Asset Value;Repurchase Price;Sale Price;Date + , + ,Open Ended Schemes ( Liquid ) + , + , + , AIG Global Investment Group Mutual Fund + , 106506;AIG India Liquid Fund-Institutional Plan-Daily Dividend Option;1001.;1001.;1001.;02-Oct-2010 + , 106511;AIG India Liquid Fund-Institutional Plan-Growth Option;1210.4612;1210.4612;1210.4612;02-Oct-2010 + , 106507;AIG India Liquid Fund-Institutional Plan-Weekly Dividend Option;1001.8765;1001.8765;1001.8765;02-Oct-2010 + , 106503;AIG India Liquid Fund-Retail Plan-DailyDividend Option;1001.;1001.;1001.;02-Oct-2010) myData - read.table(textConnection(x[7:10]), sep=';') closeAllConnections() str(myData) 'data.frame': 4 obs. of 6 variables: $ V1: int 106506 106511 106507 106503 $ V2: Factor w/ 4 levels AIG India Liquid Fund-Institutional Plan-Daily Dividend Option,..: 1 2 3 4 $ V3: num 1001 1210 1002 1001 $ V4: num 1001 1210 1002 1001 $ V5: num 1001 1210 1002 1001 $ V6: Factor w/ 1 level 02-Oct-2010: 1 1 1 1 myData V1 V2 V3 V4 V5 V6 1 106506 AIG India Liquid Fund-Institutional Plan-Daily Dividend Option 1001.000 1001.000 1001.000 02-Oct-2010 2 106511 AIG India Liquid Fund-Institutional Plan-Growth Option 1210.461 1210.461 1210.461 02-Oct-2010 3 106507 AIG India Liquid Fund-Institutional Plan-Weekly Dividend Option 1001.876 1001.876 1001.876 02-Oct-2010 4 106503 AIG India Liquid Fund-Retail Plan-DailyDividend Option 1001.000 1001.000 1001.000 02-Oct-2010 On Sat, Oct 9, 2010 at 12:18 PM, Santosh Srinivas santosh.srini...@gmail.com wrote: Newbie question ... I am looking something equivalent to read.delim but which accepts a text line as parameter instead of a file input. Below is my problem, I'm unable to get the exact output which is a simple data frame of the data where the delimiter exists ... coming quite close though I have a data frame with 10 lines called MF_Data MF_Data [1:10] [1] Scheme Code;Scheme Name;Net Asset Value;Repurchase Price;Sale Price;Date [2] [3] Open Ended Schemes ( Liquid ) [4] [5] [6] AIG Global Investment Group Mutual Fund [7] 106506;AIG India Liquid Fund-Institutional Plan-Daily Dividend Option;1001.;1001.;1001.;02-Oct-2010 [8] 106511;AIG India Liquid Fund-Institutional Plan-Growth Option;1210.4612;1210.4612;1210.4612;02-Oct-2010 [9] 106507;AIG India Liquid Fund-Institutional Plan-Weekly Dividend Option;1001.8765;1001.8765;1001.8765;02-Oct-2010 [10] 106503;AIG India Liquid Fund-Retail Plan-DailyDividend Option;1001.;1001.;1001.;02-Oct-2010 Now for the lines below .. they are delimted by ; ... I am using tempTxt - MF_Data[7] MF_Data_F - unlist(strsplit(tempTxt,;, fixed = TRUE)) tempTxt - MF_Data[8] MF_Data_F1 - unlist(strsplit(tempTxt,;, fixed = TRUE)) MF_Data_F - rbind(MF_Data_F,MF_Data_F1) But MF_Data_F is not a simple 2X6 data frame which is what I want __ 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-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] Counting unique items in a list of matrices
If you just want a list of matrices and their counts, you can use
Peter's list of matrices, L, and then:
With plyr:
require(plyr)
count(unlist(lapply(L, toString)))
Without plyr:
as.data.frame(table(unlist(lapply(L, toString
Cheers,
Jeff.
On Sat, Oct 9, 2010 at 12:44 PM, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2010-10-07 10:10, Jim Silverton wrote:
Hello,
I gave a list of 2 x 2 matrices called matlist. I have about 5000 2 x 2
matrices. I would like to count how many of each 2 x 2 unique matrix I
have.
So I am thinking that I need a list of the unique 2 x 2 matrices and their
counts. Can anyone help.
Here's one way, using the plyr package:
require(plyr)
## make a list of 2X2 matrices
L - vector('list', 5000)
set.seed(4321)
for(i in 1:5000) L[[i]] - matrix(round(runif(4), 1), 2, 2)
## convert each matrix to a string of 4 numbers, then
## form dataframe
dL - ldply(L, function(.x) toString(unlist(.x)))
## add an index vector
dL$ind - seq_len(5000)
## count unique strings; return string, frequency, indeces
result - ddply(dL, .(V1), summarize,
freq=length(V1),
idx=toString(ind))
-Peter Ehlers
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Re: [R] StrSplit
Obviously Jim's solution does work, and I did not intend to imply it didn't. In fact, his read.table solution would work both if the OP had a semi-colon delimited file to begin with (which I was trying to say was ideal from a workflow standpoint) or a vector of strings (for use when paired with textConnections). Using strsplit is merely another solution for the latter situation. I thought the OP might appreciate seeing how to use the function that they indicated they were having problems with. Plus, I have a penchant for R-ishly unreadble code. ;) Thanks for clarifying, Jeff. On Sat, Oct 9, 2010 at 1:04 PM, David Winsemius dwinsem...@comcast.net wrote: On Oct 9, 2010, at 12:46 PM, Jeffrey Spies wrote: Jim's solution is the ideal way to read in the data: using the sep=; argument in read.table. However, if you do for some reason have a vector of strings like the following (maybe someone gives you an Rdata file instead of the raw data file): MF_Data - c(106506;AIG India Liquid Fund-Institutional Plan-Daily Dividend Option;1001.;1001.;1001.;02-Oct-2010,106511;AIG India Liquid Fund-Institutional Plan-Growth Option;1210.4612;1210.4612;1210.4612;02-Oct-2010) Then you can use this to get a data frame: as.data.frame(do.call(rbind, lapply(MF_Data, function(x) unlist(strsplit(x, ';') If you are suggesting that Jim's solution would not work here, then I would disagree and suggest you try offering your vector (without the cr's inserted by our mail clients) to his code. It should work just fine and be far more readable. On the other hand if you were offering this with an explanation that strsplit's split argument is more flexible than the sep argument in the read functions because it accepts regular expressions and so can handle situations where multiple separators exist in the same line, then I would applaud you. -- David. Cheers, Jeff. On Sat, Oct 9, 2010 at 12:30 PM, jim holtman jholt...@gmail.com wrote: Is this what you are after: x - c(Scheme Code;Scheme Name;Net Asset Value;Repurchase Price;Sale Price;Date + , + ,Open Ended Schemes ( Liquid ) + , + , + , AIG Global Investment Group Mutual Fund + , 106506;AIG India Liquid Fund-Institutional Plan-Daily Dividend Option;1001.;1001.;1001.;02-Oct-2010 + , 106511;AIG India Liquid Fund-Institutional Plan-Growth Option;1210.4612;1210.4612;1210.4612;02-Oct-2010 + , 106507;AIG India Liquid Fund-Institutional Plan-Weekly Dividend Option;1001.8765;1001.8765;1001.8765;02-Oct-2010 + , 106503;AIG India Liquid Fund-Retail Plan-DailyDividend Option;1001.;1001.;1001.;02-Oct-2010) myData - read.table(textConnection(x[7:10]), sep=';') closeAllConnections() str(myData) 'data.frame': 4 obs. of 6 variables: $ V1: int 106506 106511 106507 106503 $ V2: Factor w/ 4 levels AIG India Liquid Fund-Institutional Plan-Daily Dividend Option,..: 1 2 3 4 $ V3: num 1001 1210 1002 1001 $ V4: num 1001 1210 1002 1001 $ V5: num 1001 1210 1002 1001 $ V6: Factor w/ 1 level 02-Oct-2010: 1 1 1 1 myData V1 V2 V3 V4 V5 V6 1 106506 AIG India Liquid Fund-Institutional Plan-Daily Dividend Option 1001.000 1001.000 1001.000 02-Oct-2010 2 106511 AIG India Liquid Fund-Institutional Plan-Growth Option 1210.461 1210.461 1210.461 02-Oct-2010 3 106507 AIG India Liquid Fund-Institutional Plan-Weekly Dividend Option 1001.876 1001.876 1001.876 02-Oct-2010 4 106503 AIG India Liquid Fund-Retail Plan-DailyDividend Option 1001.000 1001.000 1001.000 02-Oct-2010 On Sat, Oct 9, 2010 at 12:18 PM, Santosh Srinivas santosh.srini...@gmail.com wrote: Newbie question ... I am looking something equivalent to read.delim but which accepts a text line as parameter instead of a file input. Below is my problem, I'm unable to get the exact output which is a simple data frame of the data where the delimiter exists ... coming quite close though I have a data frame with 10 lines called MF_Data MF_Data [1:10] [1] Scheme Code;Scheme Name;Net Asset Value;Repurchase Price;Sale Price;Date [2] [3] Open Ended Schemes ( Liquid ) [4] [5] [6] AIG Global Investment Group Mutual Fund [7] 106506;AIG India Liquid Fund-Institutional Plan-Daily Dividend Option;1001.;1001.;1001.;02-Oct-2010 [8] 106511;AIG India Liquid Fund-Institutional Plan-Growth Option;1210.4612;1210.4612;1210.4612;02-Oct-2010 [9] 106507;AIG India Liquid Fund-Institutional Plan-Weekly Dividend Option;1001.8765;1001.8765;1001.8765;02-Oct-2010 [10] 106503;AIG India Liquid Fund-Retail Plan-DailyDividend Option;1001.;1001.;1001.;02-Oct-2010 Now for the lines below .. they are delimted by ; ... I am using tempTxt - MF_Data[7] MF_Data_F - unlist(strsplit(tempTxt,;, fixed = TRUE)) tempTxt - MF_Data[8] MF_Data_F1 - unlist(strsplit(tempTxt,;, fixed = TRUE)) MF_Data_F - rbind(MF_Data_F,MF_Data_F1) But MF_Data_F
Re: [R] function using values separated by a comma
Here's another method without using any external regular expression libraries:
dat - read.table(tc - textConnection(
'0,1 1,3 40,10 0,0
20,5 4,2 10,40 10,0
0,11 1,2 120,10 0,0'), sep=)
mat - apply(dat, c(1,2), function(x){
temp - as.numeric(unlist(strsplit(x, ',')))
min(temp)/sum(temp)
})
For mat[2,4], I get 0 (as did the other solutions), and you get 1, so
check on that. If you want the divide-by-0 NaNs to be 0, you can check
that by replacing
min(temp)/sum(temp)
with:
ifelse(is.nan(val-min(temp)/sum(temp)), 0, val)
This has an advantage over:
mat[is.na(mat)] - 0
in that you might have true missingness in your data and is.na won't
be able to distinguish it.
Cheers,
Jeff.
On Fri, Oct 8, 2010 at 1:19 AM, burgundy saub...@yahoo.com wrote:
Hello,
I have a dataframe (tab separated file) which looks like the example below -
two values separated by a comma, and tab separation between each of these.
[,1] [,2] [,3] [ ,4]
[1,] 0,1 1,3 40,10 0,0
[2,] 20,5 4,2 10,40 10,0
[3,] 0,11 1,2 120,10 0,0
I would like to calculate the percentage of the smallest number separated by
the comma by:
1) summing the values e.g. for [1,3] where 40,10, 40+10 = 50
2) taking the first value and dividing it by the total e.g. for [1,3], 40/50
= 0.8
3) where the value generated by 2) is 0.5, print 1-value, otherwise, leave
value e.g. for [1,3], where value is 0.8, print 1-0.8 = 0.2
plan to generate file like:
[,1] [,2] [,3] [,4]
[1,] 1 0.25 0.2 0
[2,] 0.2 0.33 0.2 1
[3,] 1 0.33 0.08 0
Apologies, I know this is very complex. Any help, even just some pointers on
how to write a general function where values are separated by a comma, is
realy very much appreciated!
Thank you
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Re: [R] Create variable by name
An alternative to Peter's solution:
createVariable - function(name) {assign(name, NULL, envir=.GlobalEnv)}
Jeff.
On Wed, Oct 6, 2010 at 12:32 PM, Ralf B ralf.bie...@gmail.com wrote:
Can one create a variable through a function by name
createVariable - function(name) {
outputVariable = name
name - NULL
}
after calling
createVariable(myVar)
I would like to have a variable myVar initialized with NULL in my
environment. Is this possible?
Ralf
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Re: [R] Empty data frame does not maintain column type
This should do it: df - data.frame(a=character(0), b=character(0), stringsAsFactors=F) because: typeof(factor(0)) is integer while: typeof(character(0)) is character. Cheers, Jeff. On Wed, Oct 6, 2010 at 1:00 PM, N David Brown hubd...@gmail.com wrote: Does anyone know why a data frame created with empty character columns converts them to integer columns? df-data.frame(a=character(0),b=character(0)) df-rbind(df,c(a,a)) typeof(df[1,1]) [1] integer AsIs doesn't help: df-data.frame(a=I(character(0)),b=I(character(0))) df-rbind(df,I(c(a,a))) typeof(df[1,1]) [1] integer Any suggestions on how to overcome this would be appreciated. Best wishes, David __ 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] R: Tools for thinking about data analysis and graphics
Hi, Michael, When I teach/preach on R, I emphasize the language's focus on data, both in its objects and operations. It might seems basic, but it's fundamental to most of the features you and others have mentioned. As a statistical programming language, what we intend to do with R is often very naturally accomplished using vector operations on tabular data, where columns represent variables of the same data type and rows represent observations of these variables for a given member of the dataset. Fortunately, these are core components of R. For instance, we can easily perform complex selections of variables and/or members, which, more often than not, serve as input to or power the functions that generate the statistics and graphics we care about. Unfortunately, vector operations seem to be difficult for people to learn how to use properly, and there are penalties for not using them, but as they say: no pain, no gain. :) If you'd be willing to share the materials you create for your talk, I'd be interested in seeing them. Cheers, Jeff. On Wed, Oct 6, 2010 at 5:05 PM, Michael Friendly frien...@yorku.ca wrote: I'm giving a talk about some aspects of language and conceptual tools for thinking about how to solve problems in several programming languages for statistical computing and graphics. I'm particularly interested in language features that relate to: o expressive power: ease of translating what you want to do into the results you want o elegance: how well does the code provide a simple human-readable description of what is done? o extensibility: ease of generalizing a method to wider scope o learnability: your learning curve (rate, asymptote) For R, some things to cite are (a) data and function objects, (b) object-oriented methods (S3 S4); (c) function mapping over data with *apply methods and plyr. What other language features of R should be on this list? I would welcome suggestions (and brief illustrative examples). -Michael -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele Street Web: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ 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] R-help
As Joshua said, in your example sim isn't be declared anywhere
(neither in the environment nor as an argument in a function), but you
might try something more R-ish:
prop.doubles - function(sim){
sum(sample(1:6, sim,replace=T)==sample(1:6,sim,replace=T))/sim
}
prop.doubles(1000)
Cheers,
Jeff.
On Tue, Oct 5, 2010 at 2:30 AM, Joshua Wiley jwiley.ps...@gmail.com wrote:
On Mon, Oct 4, 2010 at 9:13 PM, Lemarian WallaceIII tott...@yahoo.com wrote:
Im trying to simulate the rolling of a pair of dice
this is my function:
#function to simulate tosses of a pair of dice
#from the simulation, the program returns the empirical probability of
#observing a double
count - 0
for(j in 1:sim){#begin loop
die1 - sample(1:6,1)
print(die1)
die2 - sample(1:6,1)
print(die2)
count - ifelse(die1 == die2, count + 1, count)
}#end loop
emprob - count/sim
return(count,emprob)
} #end program
these are the errors that keep coming up:
Error in 1:sim : 'sim' is missing
You need to define an object called 'sim', otherwise you are telling R
to go look up the value of a non-existent variable. Does your
function have a first part? All I am seeing is the body and the end.
Josh
How do I correct this?
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--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
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Re: [R] Sampling from data set
We'll probably need much more info, but this should get you started: nameOfDataSet[sample(1:1, 100),] You can replace the 1 with dim(nameOfDataSet)[1] to make it more dynamic. Jeff. On Tue, Oct 5, 2010 at 3:07 AM, Jumlong Vongprasert jumlong.u...@gmail.com wrote: Dear all. I have data with 2 variable x,y size 1. I want to sampling from this data with size 100. How I can do it. THANK. -- Jumlong Vongprasert Institute of Research and Development Ubon Ratchathani Rajabhat University Ubon Ratchathani THAILAND 34000 [[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.
Re: [R] Sampling from data set
If poproh.3 was your dataset as a data.frame (an object with row and column dimensions), you need a comma following the row selection (sample(...)) to indicate that you want to select those rows and all columns: newsample -poprho.3[sample(1:1,100),] # note the last comma in the brackets General use is: my.data.frame[rows,columns] Where either rows or columns (or both) can be left blank to indicate that you want all of them. Similarly, a selection of the first column would have been (comma followed by column number): newsample -poprho.3[sample(1:1,100),1] That's why your: newsample -as.matrix(nameofdataset[sample(1:1,100),]) worked; the as.matrix wasn't necessary to simply sample the data. Cheers, Jeff. On Tue, Oct 5, 2010 at 3:54 AM, Jumlong Vongprasert jumlong.u...@gmail.com wrote: Dear Jeffrey. I used newsample -as.matrix(nameofdataset[sample(1:1,100),]). Now it include all 2 variable. Thank you for your answer to inspire. Jumlong 2010/10/5 Jeffrey Spies jsp...@virginia.edu We'll probably need much more info, but this should get you started: nameOfDataSet[sample(1:1, 100),] You can replace the 1 with dim(nameOfDataSet)[1] to make it more dynamic. Jeff. On Tue, Oct 5, 2010 at 3:07 AM, Jumlong Vongprasert jumlong.u...@gmail.com wrote: Dear all. I have data with 2 variable x,y size 1. I want to sampling from this data with size 100. How I can do it. THANK. -- Jumlong Vongprasert Institute of Research and Development Ubon Ratchathani Rajabhat University Ubon Ratchathani THAILAND 34000 [[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. -- Jumlong Vongprasert Institute of Research and Development Ubon Ratchathani Rajabhat University Ubon Ratchathani THAILAND 34000 __ 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] For help on Open .tar.gz file in R under Windows
read.table('path.to.file.tar.gz')
or any of its derivatives (read.csv, read.csv2, etc.). Of course
you'll need to set the arguments appropriately. For example:
read.table('path.to.comma.delimited.tar.gz', sep=,, header=T)
Cheers,
Jeff.
On Mon, Oct 4, 2010 at 6:48 PM, wenyue sun wenyue...@gmail.com wrote:
Hey All,
I am new in R. Need help on code that can open the .tar.gz file in R under
Windows. Can any one help.
Thanks in Advance.
Wayne
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Re: [R] Modifying a data.frame
You should examine what is being looped over when you use a for loop
with the i in dataframe syntax:
j-1; for(i in ex){ cat('step', j, i, sep= , fill=T); j-j+1}
As you can see, each column in ex is being set to i for each step of
the for loop. Instead, it seems that you want to step over every
row--a change made in the first line:
for (i in 1:dim(ex)[1]) {
if(ex[i,3]==A|| ex[i,3]==C){
ex[i,4]- -
}else {
ex[i,4]-10
}
}
1:dim(ex)[1] is then a vector of row index values that is looped over.
A more R-ish version of this might be:
ex[,4] - ifelse(ex$eff == 'A' | ex$eff == 'C', -, 10)
I'm not sure this is the case, but if - is supposed to represent
missingness, missing values are represented by `NA`s in R.
ex[,4] - ifelse(ex$eff == 'A' | ex$eff == 'C', NA, 10)
?NA for more info. Note: those are not single quotes, but instead back-ticks.
Hope that helps,
Jeff.
On Sun, Oct 3, 2010 at 4:58 AM, Bapst Beat beat.ba...@braunvieh.ch wrote:
Hello list members
I have a problem with modifying a data.frame.
As an example given is a data.frame called ex :
ex-data.frame(id=c(1,2,3,4,5,6),obs=c(14,9,20,36,55,47),eff=c(A,A,B,C,C,C))
After that I would like to modify the object ex with the following short
script:
for (i in ex) {
if(ex[i,3]==A|| ex[i,3]==C){
ex[i,4]--
}
else {
ex[i,4]-10
}
}
This script is creating an error message:
Fehler in if (ex[i, 3] == A || ex[i, 3] == C) { :
Fehlender Wert, wo TRUE/FALSE nötig ist
Why this script doesn't work properly?
Thanks a lot for your hints
Beat
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Re: [R] A problem about nomogram--thank you for you help
Firstly, `*` is the multiplication operator in R. Secondly, you'll need to convert your factors to numerics: L-0.559*as.numeric(T.Grade)-0.896*as.numeric(Smoking)+0.92*as.numeric(Sex)-1.338 Cheers, Jeff. 2010/10/3 笑啸 dingdongl...@126.com: dear professor: I am a doctor of urinary,and I am developing a nomogram of bladder tumor.Now I have a problem about this. I have got the result like this through analysing the dataset exp11.sav through multinominal logistic regression by SPSS 17.0.(the Sig. is high,that is good ,it is just aexperimental data ) Parameter Estimates Ya B Std. Error Wald df Sig. Exp(B) 95% Confidence Interval for Exp(B) Lower Bound Upper Bound 1 Intercept -1.338 .595 5.059 1 .024 T.Grade .559 .319 3.076 1 .079 1.749 .936 3.265 Sex .920 .553 2.766 1 .096 2.511 .849 7.428 Smoking -.896 .474 3.580 1 .058 .408 .161 1.033 a. The reference category is: 0. And after that,I want to develop the nomogram through R-Project. And I load the package rms T.Grade-factor(0:3,labels=c(G0, G1, G2,G3)) Sex-factor(0:1,labels=c(F,M)) Smoking-factor(0:1,labels=c(No,yes)) L-0.559T.Grade-0.896Smoking+0.92Sex-1.338 # error (错误: 不适用于非函数;error:it is not fit the non-function) The R-project index that the last program error. can you tell me where is the mistake.and how to get the correct equation . thank you for you help! And I an sorry about my poor english! truly yours __ 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] Ranked Set Sampling
This is certainly not my area of expertise, but like Peter mentioned, Jeff Terpstra published this: http://www.jstatsoft.org/v14/i07 which has R code listed as supplements. Joe McKean seems to keep an updated version of that code here: http://www.stat.wmich.edu/red5328/WWest/ And Brent Johnson has extended that code for variable selection/regression: http://userwww.service.emory.edu/~bajohn3/software.html Maybe that's a start; if you find more by following Peter's suggestion of privately contacting authors, please follow-up with the list. Cheers, Jeff. On Sun, Oct 3, 2010 at 1:38 PM, Peter Dalgaard pda...@gmail.com wrote: On 10/03/2010 06:32 PM, David Winsemius wrote: Ahmed Albatineh wrote: Are you aware of any package that calculates Ranked Set Sample? If you have a code that you are willing to share, I will acknowledge that in my work. Thanks much Ahmed I wonder if this is a phrase that is uniformly understood? One possibility is that you are asking to sample elements of a set based on some ranking function. In that case you may need to describe in more detail how you want to handle ties and whether this function is supposed to deal with multivariate strata. (There are many base functions that handle univariate situations and there are packages that provide support for more complex ones.) You are also requested (in the Posting Guide) to provide an example that can be cut and pasted and desired results against which responder can judge the degree to which their efforts agree with your hopes. It's a fairly well-defined concept. I.e., you can google for it... On the other hand, same search points to authors like Jeff Terpstra who explicitly says that he codes in R, so maybe ask him instead of the the world at large? -- Peter Dalgaard Center for Statistics, Copenhagen Business School 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-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] can't find and install reshape2??
The first argument in download.packages should be of type character or
a vector of characters.
This worked for me:
install.packages('reshape2')
as did:
download.packages('reshape2', '~/Downloads/')
Cheers,
Jeff.
On Sun, Oct 3, 2010 at 8:57 PM, Chris Howden
ch...@trickysolutions.com.au wrote:
Hi everyone,
I’m trying to install reshape2.
But when I click on “install package” it’s not coming up!?!?! I’m getting
reshape, but no reshape2?
I’ve also tried download.packages(reshape2, destdir=c:\\)
download.packages(Reshape2, destdir=c:\\)…but no luck!!!
Does anyone have any ideas what could be going on?
Chris Howden
Founding Partner
Tricky Solutions
Tricky Solutions 4 Tricky Problems
Evidence Based Strategic Development, IP development, Data Analysis,
Modelling, and Training
(mobile) 0410 689 945
(fax / office) (+618) 8952 7878
ch...@trickysolutions.com.au
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Re: [R] R data opening problem
If I understand your problem correctly, I think you need to be doing: summary(data.name) The functions read.dta and read.spss both return things (data frames, if you use the to.data.frame=T argument with read.spss). So whatever variable you set is what you should be doing a summary on. In this case, you are setting data.name to be the data frame returned by read.dta: data.name - read.dta(file.choose()) If you'd rather, you could use: kelleya - read.dta(file.choose()) summary(kelleya) Hope that helps, Jeff. On Sat, Oct 2, 2010 at 10:01 PM, Alla Manukyan allama...@yahoo.com wrote: Dear Sir/Madam, I have just installed R for Windows. I am trying to open a stata file and I have a problem. I have used the following commands: install.packages(foreign) library(foreign) data.name - read.dta(file.choose()) # then I choose a kelleya.dta file and click on open and then I use the following command: summary(kelleya) Error in object[[i]] : object of type 'closure' is not subsettable I have also used opening an spss file using read.spss command but I get the same error message. Can you help me please? Alla __ 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] Need to incorporate the use of na.rm into custom function
You can use na.omit on x after it is passed into your apply function if na.rm == T, or simply pass your na.rm to functions that use it, such as sum. Hope that helps, Jeff. On Fri, Oct 1, 2010 at 11:07 AM, Ochsner, Scott A sochs...@bcm.edu wrote: Hi, Take a matrix with missing values: X = matrix(rnorm(10), ncol = 5) X[2,4]=NA X [,1] [,2] [,3] [,4] [,5] [1,] -0.1566427 -0.7382232 -1.0564624 -0.8412139 0.9370319 [2,] -1.0289865 -0.8452054 -0.1349459 NA -0.1749113 I want to apply a custom function over the rows such that the NAs are ignored in a similar fashion as to how the following works. means-apply(X,1,mean,na.rm=TRUE) Custom function: liptak-function (x,df) 2*(pnorm(abs(sum(x*df)/sqrt(sum(df^2))),lower.tail=FALSE)) I want to be able to do the following: rslt-apply(X,1,liptak,na.rm=TRUE) Can someone point me in the right direction on how to incorporate the use of na.rm into my function? Scott sessionInfo() R version 2.11.0 (2010-04-22) i386-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.11.0 Scott A. Ochsner, PhD NURSA Bioinformatics Baylor College of Medicine One Baylor Plaza Mail Stop: BCM-130 Houston, TX 77030 Voice: (713) 798-6227 Fax: (713) 790-1275 __ 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] [Help]:How to use loop to achieve this aim?
Correcting Karena's solution:
for(i in 1:22) {
chrn - paste(chr,i,sep=)
chrn=MEDIPS.readAlignedSeqences(BSgenome=hg19, file=chrn,numrows=
?) # no quotes around chrn
chrn=MEDIPS.genomeVector(data=chrn, bin_size=50,extend=250)
frames=? # I don't know how you get this variable
...
out_file - paste(frames.chr, i, .meth.txt, sep=) # dynamic file name
write.table(frames, file=out_file, sep=\t, quote=F, col.names=T,
row.names=F) # use var you just made
}
Replace the ???s.
Jeff.
On Fri, Oct 1, 2010 at 11:27 AM, qcshare qcsh...@gmail.com wrote:
It doesn't work...
where should change?
Thanks.
--
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[R] LR Decomposition?
Hi all, Is there an LR decomposition function in R and, if not, how can we get the non-compact representation of Q from QR decomposition? Thanks, Jeff. -- View this message in context: http://www.nabble.com/LR-Decomposition--tp18072588p18072588.html Sent from the R help mailing list archive at Nabble.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] S4 pass-by-value work-around?
Howdy all,
I have a problem that I'd like some advice/help in solving---it has to do
with R's pass-by-value system. I understand the issue, but am wondering if
anyone has found a working solution in dealing with it for cases when one
wants to modify an object inside of a method, specifically when working with
S4. I'm aware that R.oo is able to deal with this using S3, but I'd really
rather stick to S4.
The basics of what I would like to do are coded below:
setClass(MyMatrix,
representation(
parameters=matrix,
uniqueCount=numeric
),
prototype(
parameters=matrix(numeric(0),0,0),
uniqueCount=1
)
)
setGeneric(createUniqueName, function(object)
standardGeneric(createUniqueName))
setMethod(createUniqueName, MyMatrix, function(object){
retval - paste(unique_, [EMAIL PROTECTED], sep=)
[EMAIL PROTECTED] - [EMAIL PROTECTED] + 1
return(retval)
})
x - new(MyMatrix, parameters=matrix(0, 2, 2))
createUniqueName(x) # returns unique_1
x # [EMAIL PROTECTED] is still 1
I understand why this is happening, but am wondering how people in the
community have dealt with it, specifically when using S4. Any advice would
be appreciated. Also, I am aware that this is somewhat of a silly example,
but it should allow you to see what I'm trying to accomplish.
Thank you,
Jeff.
--
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Re: [R] S4 pass-by-value work-around?
Thanks for the response, Martin. While the solutions offered update the
object appropriately, we wouldn't get the desired return value (a string
followed by the counter, unique_1) when the methods are called. Do you
know a way of dealing with this?
Jeff.
Martin Morgan wrote:
Hi Jeff --
two different scenarios are to overwrite the current object, along the
lines of
y - uniquify(y)
where uniquify is a method like createUniqueName but returns the
(modified) instance rather than unique name
setMethod('uniquify', 'MyMatrix', function(x) {
[EMAIL PROTECTED] - # something unique
x
})
The second is a replacement method, along the lines of
setGeneric(uniqueCount-,
function(x, ..., value) standardGeneric(uniqueCount-))
setReplaceMethod(uniqueCount,
signature=c(x=MyMatrix, value=numeric),
function(x, ..., value) {
[EMAIL PROTECTED] - value
x
})
uniqueCount(x) - uniqueCount(x) + 1
x # now modified
This is untested psuedo-code, so I hope it's right enough to get you
going.
Martin
Jeffrey Spies [EMAIL PROTECTED] writes:
Howdy all,
I have a problem that I'd like some advice/help in solving---it has to do
with R's pass-by-value system. I understand the issue, but am wondering
if
anyone has found a working solution in dealing with it for cases when one
wants to modify an object inside of a method, specifically when working
with
S4. I'm aware that R.oo is able to deal with this using S3, but I'd
really
rather stick to S4.
The basics of what I would like to do are coded below:
setClass(MyMatrix,
representation(
parameters=matrix,
uniqueCount=numeric
),
prototype(
parameters=matrix(numeric(0),0,0),
uniqueCount=1
)
)
setGeneric(createUniqueName, function(object)
standardGeneric(createUniqueName))
setMethod(createUniqueName, MyMatrix, function(object){
retval - paste(unique_, [EMAIL PROTECTED], sep=)
[EMAIL PROTECTED] - [EMAIL PROTECTED] + 1
return(retval)
})
x - new(MyMatrix, parameters=matrix(0, 2, 2))
createUniqueName(x) # returns unique_1
x # [EMAIL PROTECTED] is still 1
I understand why this is happening, but am wondering how people in the
community have dealt with it, specifically when using S4. Any advice
would
be appreciated. Also, I am aware that this is somewhat of a silly
example,
but it should allow you to see what I'm trying to accomplish.
Thank you,
Jeff.
--
View this message in context:
http://www.nabble.com/S4-pass-by-value-work-around--tp17997553p18012246.html
Sent from the R help mailing list archive at Nabble.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.
