similar to: string split at xth position

Dear all, I want to concatenate the elements of two vectors such as a<-c("a1","a2") b<-c("b1","b2") and obtain "a1b1", "a1b2","a2b1","a2b2" I tried the paste and paste0 functions, but they yielded elementwise concatenation such as "a1b1","a2b2" I am wondering that is there an efficient

I have a data frame in wide format that I'd like to convert to long format. For example, in wide format I have: id A1B1 A1B2 A2B1 A2B2 1 1 400 475 420 510 2 2 390 500 470 472 3 3 428 512 555 610 4 4 703 787 801 822 5 5 611 634 721 705 6 6 543 522 612 788 7 7 411 488 506 623 8 8 654 644 711 795 A is one repeated-measures variable with levels 1 and 2. B is a second repeated-measures variable

I have design with two repeated-measures factor, and no grouping factor. I can analyze the dataset successfully in other software, including my legacy DOS version BMDP, and R's 'aov' function. I would like to use 'Anova' in 'car' in order to obtain the sphericity tests and the H-F corrected p-values. I do not believe the data are truly deficient in rank. I

To whom it may concern I made some analysis with R using the command Anova. However, I found some problmes with the output obtained by selecting type II o type III sum of squares. Briefly, I have to do a 2x3 mixed model anova, wherein the first factor is a between factor and the second factor is a within factor. I use the command Anova in the list below, because I want to obtain also the sum

Hi all, In preparation for teaching a class next week, I've been reviewing R's standard modelling algebra. I've used it for a long time and have a pretty good intuitive feel for how it works, but would like to understand more of the technical details. The best (online) reference I've found so far is the section in "An Introduction to R"

Hi folks, Suppose I have a series of cases each with categorical factors A, B. What is the best way to "glue" A and B together into a single factor? For example, given A0 B1 ... A1 B1 ... A0 B2 ... A1 B0 ... A0 B0 ... A1 B2 ... then I'd like to end up with a single factor with levels A0B0, A0B1, A0B2, A1B0, A1B1, A1B2 according to all the combinations which actually occur in

Hi, I am using colsplit (package = reshape) to split all strings in a column according to the same patterns. Here an example: library(reshape2) df1 <- data.frame(x=c("str1_name2", "str3_name5")) df2 <- data.frame(df1, colsplit(df1$x, pattern = "_", names=c("str","name"))) This is nearly what I want but I want to remove the words

Hi, I've got a dataframe having a code as column name. Addtionally I have another dataframe with a two columns (and lots of rows), the first containing the code and the second some Text (real name). Now I'd like to use the information (pairs of code and name) of the second dataframe to rename all the columnnames in the first dataframe. How is it possible to achieve that? Here a small

Hi, I would like to split dataframe based on one colum and want to connect the two dataframes by rows (like rbind). Here a small example: # The orgininal dataframe df1 <- data.frame(col1 = c("A","A","B","B"),col2 = c(1:4), col3 = c(1:4)) # The datafame how it could look like df2 <- data.frame(A.col2 = c(1,2), A.col3 = c(1,2), B.col2 = c(3,4), B.col3

Hi, I am trying to calculated various summary statistics using the dcast function of reshape2. This works perfectly for getting the mean, sum, length, sd. But when I want to calculate the median I get an error. I tried it with and without removing NAs: my_median <- function(x) median(x, na.rm = FALSE) median_df <- dcast(patch_stats_dfm,formula=species~input+barriers,my_median) Error in

Hi, I'd like to get mean values for the margins of my casted data.frame. For the casting I am using dcast() from reshape2. However, when I set the margins parameter (margins=c("grand\_row")) I get following error concerning an unrecognized escape character '\_'. So what is the correct command to get the outermost margins only in reshape2? /johannes [[alternative HTML

Hi, is there a way to color the branches or text label of the branches of dendrograms e.g. from hclust() according to a grouping variable. Here I have something in mind like: http://www.sigmaaldrich.com/content/dam/sigma-aldrich/life-science/biowire/biowire-fall-2010/proteome-figure-1.Par.0001.Image.gif (ectodermal, endodermal, mesodermal). /johannes

I am wondering how to obtain SE estimates for fixed effects from a nonlinear mixed effects model? I have fixed effects corresponding to three factors A, B and C with 2, 3 and 3 levels respectively. I have fit a model of the following general form: nlme1<-nlme(y~ SasympOrig(x, Asym, lrc), data=df, fixed=list(Asym~A*B*C, lrc~A*B*C), start=c(fixef(ETR.nlme)[1], rep(0,17), fixef(ETR.nlme)[2],

Hi, I'd like to use Latin Hypercube Sampling (LHC) in the the context of uncertainty / sensitivity analysis of a complex model with approximately 10 input variables. With the LHC approach I'd like to generate parameter combinations for my model input variables. Therefore I came across an simple example here on the mailing list (

Hi there. I just switched from icecast1 to icecast2. Right now I'm not sure if that was a very good idea. I do like the ogg-support, but I lack a good amount of documentation (Hey, I'd be glad to help you out, but how can there be consistent documentation for a software that is only available through cvs(-snapshots)?) and good linux source clients. I used xmms-liveice or muse before, but

Dear R users, I would like to compare two dataframes, actually their categorical variables (as factors) only (there are 12, from column 1 to 12). The reason I do that is that I got 2 datasets from two different methods and I would like to be sure that each method used the same data (3D images) to extract 2 different sets of 3D parameters. Is it clear so far? So I thought about using

Suppose I have a vector of strings: c("A1B2","A3C4","B5","C6A7B8") [1] "A1B2" "A3C4" "B5" "C6A7B8" where each string is a sequence of <column><value> pairs (fixed width, in this example both value and name are 1 character, in reality the column name is 6 chars and value is 2 digits). I need to

The entries in xenstore have permission attributes. The attributes can be easily altered by xenstore-chmod, however, I cannot find a easy way to see them. I''ve modified xenstore_client.c to raise a new utility. The utility checks the permission and makes an easy-look output. Please tell me any suggestions. Thanks. Signed-off-by: Frank Pan <frankpzh@gmail.com> ---

Hola a tod en s ?alquien sabria como convertir estas frases con expresiones regulares? 1.3ptd -> 1.3 ptd 1.3ptdm -> 1.3 ptdm 4.4ptdm23j -> 4.4 ptdm 23j 7.716s -> 7.7 16s 1.4hola -> 1.4 hola 1.4hola.hola -> 1.4 hola.hola 5.5v6 -> 5.5 v6 5.5v6sdp -> 5.5 v6 sdp 5.5v10sdp -> 5.5 v10 sdp de forma que esta frase "hola 1.3ptd 1.3ptdm 4.4ptdm23j 7.716s 1.4hola pepe

Hola esta es una solución library(data.table) library(stringr) dt <- data.table( V1a = sample(c("1","0"), 10, TRUE) , V1b = sample(c("1","0"), 10, TRUE) , V2a = sample(c("1","0"), 10, TRUE) , V2b = sample(c("1","0"), 10, TRUE) , V3a =