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I am pretty sure that I came across a function that creates a vector of levels from a list but I just can't remember. Basically, I have something like > t <- list(A=c(4,1,4),B=c(3,7,9,2)) > t $A [1] 4 1 4 $B [1] 3 7 9 2 And I would like to get something like the following: t levels 4 1 1 1 4 1 3 2 7 2 9 2 2 2 I tried unlist without success. I also do remember that there is a

Dear R gurus, I have a very embarrassingly parallelizable job that I am trying to speed up with snow on our local cluster. Basically, I am doing ~50,000 t.test for a series of micro-array experiments, one gene at a time. Thus, I can easily spread the load across multiple processors and nodes. So, I have a master list object that tells me what rows to pick up for each genes to do the t.test from

Dear all, I am processing a very long and complicated list using lapply through a custom function and I would like to generate some sort of progress report. For instance, print a dot on the screen every time 1000 item have been process. Or even better, reporting the percent of the list that have been process every 10%. However, I can't seem to figure out a way to achieve that. For instance,

Dear all, I am running anova(lm()) on a series of different data frame and I am getting the following message Using dataFrame$levels as id variables 1. Why am I getting that message 2. How do I suppress it (or correct it). Thanks Marco -- Marco Blanchette, Ph.D. Assistant Investigator Stowers Institute for Medical Research 1000 East 50th St. Kansas City, MO 64110 Tel: 816-926-4071

Since I'm a SAS programmer, I'm used to creating command files in an editor for submission later. Is there a way to do this in R? I'd need to retain an ouput listing and a log to check for errors. _________________________________________________________________ Send e-mail faster without improving your typing skills. d_122008 [[alternative HTML version deleted]]

I all I am trying to combine columns from two dataframes to make a completely new dataframe consisting of one column of dates (ie a combination of dates from the two dataframes). >From the following dataframes a b 1 2008-07-27 1 2 2008-10-01 2 3 2008-08-15 3 4 2008-08-14 4 5 2008-08-14 5 6 2008-09-20 6 c d 1 2008-07-27 1 2 2008-10-01 2 I would like to get: z

Full_Name: Lixin Han Version: 2.4.0 OS: Windows 2000 Submission from: (NULL) (155.94.110.222) A character vector c('a','b') is supplied to rm(). As a result, 'c' is deleted unintentionally. > a <- 1:5 > b <- 'abc' > c <- letters > ls() [1] "a" "b" "c" > rm(c('a','b')) > ls() character(0)

Dears Sirs During my computational work I encountered unexpected behavior when calling "ar" function, namely # time series x<-ts(c(-0.2052083,-0.3764986,-0.3762448,0.3740089,0.2737568,2.8235722,- 1.7783313,0.2728676,-0.3273164),start=c(1978,3),frequency=4,end=c(1980,3)) # ar function res.ar<-ar(x,aic=TRUE,demean=F) # call "ar" again and ............

Running R.app on Mac OS X 10.4 > version _ platform powerpc-apple-darwin8.6.0 arch powerpc os darwin8.6.0 system powerpc, darwin8.6.0 status major 2 minor 3.1 year 2006

Running R.app on Mac OS X 10.4 > version _ platform powerpc-apple-darwin8.6.0 arch powerpc os darwin8.6.0 system powerpc, darwin8.6.0 status major 2 minor 3.1 year 2006

Hi all, Can anyone explain why the following use of the subset() function produces a different outcome than the use of the "[" extractor? The subset() function as used in density(subset(mydf, ht >= 150.0 & wt <= 150.0, select = c(age))) appears to me from documentation to be equivalent to density(mydf[mydf$ht >= 150.0 & mydf$wt <= 150.0, "age"])

Hi all, What are current methods people use in R to identify mis-spelled column names when selecting columns from a data frame? Alice Johnson recently tackled this issue (see [BioC] posting below). Due to a mis-spelled column name ("FileName" instead of "Filename") which produced no warning, Alice spent a fair amount of time tracking down this bug. With my fumbling fingers

Full_Name: Mike Ulrich Version: 2.9 OS: Mac OSX Submission from: (NULL) (69.169.178.34) The help documentation for loadings() lists more then one parameter. The function call only expects one parameter. The digits, cutoff, and sort parameters are not used in the function. ## S3 method for class 'loadings': print(x, digits = 3, cutoff = 0.1, sort = FALSE, ...) ## S3 method for class

Hi useRs, A recent coding infelicity along these lines yielded a corrupt data frame. foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4] > bar X1 X2 X3 X4 NewCol 1 1 4 7 10 <NA> 2 2 5 8 11

Dear list, I have a list of number sequences. Each number sequence has different numbers of elements. Is there a quick way (other than to iterate through the entire list) way to coerce list to matrix with NAs filling in the short sequences? An example of what I mean is this: A <- list(c(3,2,3),c(6,5)) I'd like to get A so that it is 3 2 3 6 5 NA Best, Ken

I recently received a Sigmaplot file (*.jnb) from a customer and would like to know if I can input it to a data frame and then manipulate the data in R.    I did a search on Google and on RSeek (www.rseek.org), but did not get any good hits.  Thank for any feedback and insight you can provide.   P.S. Love the flexibility of R and would love to keep using it.   Just wanting to know if this is

I'm trying to redirect where temporary files go under R to D:\temp\somerandomname from its default (C:\Documents and Settings\username\Temp\somerandomname) -- how do I go about doing this? --j -- Jonathan A. Greenberg, PhD Postdoctoral Scholar Center for Spatial Technologies and Remote Sensing (CSTARS) University of California, Davis One Shields Avenue The Barn, Room 250N Davis, CA 95616

Hi all, Hi Gavin, thx for your help i finally found out what i want to do and how to fix it. just needed to get some more level my cut level was too small... two question remain... a) can i somehow scale the twigs after cutting ? b) how can i label the nodes and how to label which one... thx !! -m.

Dear all, I have a dataframe with multiple observations and the levels as the last column, as in: d <- data.frame(A=sample(1:100,12),B=sample(1:100,12),levels=c(rep('A',4),rep('B',4),rep('C',4))) > d A B levels 1 77 40 A 2 14 18 A 3 56 7 A 4 46 27 A 5 63 35 B 6 80 21 B 7 3 54 B 8 93 76 B 9 5 46 C 10 16 53

I would like to figure out the equations for calculating "AIC" in both "step() function" and "AIC () function". They are different. Then I just type "step" in the R console, and found the "AIC" used in "step() function" is "extractAIC". I went to the R help, and found: "The criterion used is AIC = - 2*log L + k *