similar to: combining vectors into list

Hi R-users I can not change the name of one column only of my matrix. my_matrix <- matrix (1:12,ncol=3) colnames(my_matrix)[1] <- 'myname' Error in dimnames(x) <- dn : length of 'dimnames' [2] not equal to array extent thank you for your help Lorenzo [[alternative HTML version deleted]]

Dear All, I am experiencing some problems in resetting the values of some selected elements in a dataframe. Consider d<-seq(-1,1,length=16) dim(d)<-c(4,4) d<-as.data.frame(d) sel_pos<-which(d>0, arr.ind=TRUE) d[sel_pos]<- -9 which returns the error Error in `[<-.data.frame`(`*tmp*`, sel_pos, value = -9) : only logical matrix subscripts are allowed in replacement

It looks from str(SA) that Response IPS1 is a data.frame of class "anova", which probably cannot be coerced to vector. Maybe you can use unlist() instead of as.vector() Or something like SA[["Response IPS1"]]["as.factor(WSD)",] ## to select the first row only, even maybe with unlist() Without a better REPRODUCIBLE example, I cannot tell more (maybe some others

Hello everyone, I have versions 2.7.2 and 2.10.1 installed on a machine that has no access to internet. In 2.7.2 I can use ? to get help on functions, which in 2.10.1 that does not work, all I see is "starting httpd help server...done" and then nothing. Have I downloaded 2.10.1 incorrectly (=forgot to tick some box for local help file repository) or is the internet help now the

Dear All If I have the following dataset V1 V2 x y y x z b a c b j d l c o How do I use R command to get the total number of different letter in column "V1" column "V1" has 7 different letters. Thank you -- View this message in context: http://r.789695.n4.nabble.com/Summing-up-Non-numeric-column-tp3077710p3077710.html Sent from the R help mailing list archive

sorry I had a mistake sending my question without a subject. I do resend again. Please excuse me. > Hello > I have a data array with soil variables (caperf), in which the variable "clay" is factor (as I see entering str(caperf)) . I need to do a regression model, so I need to have arcilla (=clay) as a numeric variable. For that I have entered > >

Dear users, My problem concerns save() within a for loop. Here is my code: for (i in 1:4) { temp <- data.frame(a=(i+1):(i+10), b=LETTERS[(i+1):(i+10)]) filename <- paste("file", i, sep="") assign(filename, temp) save(filename, file=paste(filename, ".rda", sep="")) } As you can see, save() doesn't work as I would like: (1) the object

Hi, I have tried looking at the archives but havent found any answer that works till now (Sorry if i have missed anything) I am a newbie to R and i am trying to carry out hierarchical clustering using hclust -> as.dendrogram and then plotting the results as a dendrogram using the plot function plot(object). My question is : In the function "plot", can one decrease the leaf label

Hi. In a data set I have a variable that takes values from 1 to 14. For each subgroup of values of this variable, I would like to obtain some descriptive statistics of other variables present in the data set. I've been trying with a "for" loop but I couldn't get nothing. Could you please suggest me some lines? -- View this message in context:

Dear users, I have looked on different sources and found different functions to prompt the user to provide input. However, I couldn't find one that does exactly what I'm looking for. select.list() and menu() are nice because a graphic window appears to prompt the user. However, the user can only choose from a predefined list of choices. readline() and scan() are more free in the

Hello, How can I read a xlsx file using xlsx package? Thanks Albert [[alternative HTML version deleted]]

I have two vectors: both have possible values of 1,-1, or 0 trend1 <- c(1,1,1,1,1,1,-1,-1,-1,-1,-1,-1) trend2 <- c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,) i want to create a third vector that is conditional upon these two vectors: if (trend2 == 1 && trend1 == 1) {position <- 1} elseif (trend2 == -1 && trend1== -1) {position <- 1} else {position <- 0} based on this two

Dear All, Can you please suggest me a correct way to load the following R data frame from the internet and save it to the hard drive? The following is what I tried: > raceprofiling <- > read.table("http://rss.acs.unt.edu/Rdoc/library/twang/data/raceprofiling.RData") > save(raceprofiling, file = 'raceprofiling.RData') However, when I used dim() to check the dataset,

Hi, I'm running R 2.11 Does anyone know if it possible to transform one character vector to one character string ? Many thanks Benoit -- Benoit Wastine Laboratoire des Sciences du Climat et de l?Environnement (LSCE/IPSL) CEA-CNRS-UVSQ CE Saclay Orme des merisiers B?t 703 - Pte 13A 91191 Gif sur Yvette Cedex France Tel : 33 (0)1 69 08 21 97 Fax : 33 (0)1 69 08 77 16

Dear all, how can I perform a string operation like strsplit(x," ") on a column of a dataframe, and put the first or the second item of the split into a new dataframe column? (so that on each row it is consistent) Thanks Boris

Dear R helpers I seem to have one trivial problem but can't find solution to it. Suppose I have following input. A = c(1,  3,  0,  5,  8)                  # 3rd element is 0 B = c(100, 30,  0,  25,  40)          # 3rd element is 0 C = A/B > C [1] 0.01 0.10  NaN 0.20 0.20 Obviously, I can't divide 0/0 and hence NaN. My problem is how to replace this NaN say by 0. So that I can

Dear R experts, I've written some functions in a few source files (such as main.R, control.R ...). After loading them into R, how do I retrieve which functions are loaded from which source file? For example: main.R has functions: myPrint(), myScan() control.R has functions: setPrinter(), setData() After loading the sources (main.R and control.R) into R, I would want to know that

Dear R helpers Suppose val1 = c(10, 20, 35, 80, 12) val2 = c(3, 8, 11, 7) I want to select either val1 or val2 depending on value of third quantity val3. val3 assumes either of the values "Monthly" or "Yearly". If val3 = "Monthly", then val = val1 and if val3 = "Yearly", then val = val2. I tried the ifelse statement as ifelse(val3 =

Hello, I'd like to ask you something again. I have a database and it has a column which looks like this one here: small_factor <- factor(c("d_variable1","d_variable2","d_variable3")) small_factor Now the thing is that I would like to convert each element of this factorized column. Basically I want to rewrite the words in it without the first two characters:

Hi all, I generally use the Reduce() function to reduce the dimension of list object. However in my current session I have few objects which are actually list of list of list..(say n step). If I reduce their dimension then I have call Reduce() function many times. Therefore my question is, is there any 1-step way to reduce the dimension at the lowest level? Take this example: lis1 <-