similar to: Data input performance

me again, new to R, haven't programmed in a long time at all sorry to ask what is probably basic a while ago i asked how to remove an element from a list apparently what i am using is not a list but a vector, didn't realize there was a list data type could also use a matrix for what i am doing, is there a way to remove single elements from either of those? thanks for any help, if i was

I've sometimes thought it would be useful to have another mail list called "no I haven't read the documentation but someone can save me a lot of time," or perhaps just "newbie" for short. This can be very useful for people just getting started in R and with school starting soon I expect the help list may become overwhelmed. Of course, I think that people should read all

Hi, How do you get the most common row from a matrix? If I have a matrix like this array(1:3,dim=c(4,5)) [,1] [,2] [,3] [,4] [,5] [1,] 1 2 3 1 2 [2,] 2 3 1 2 3 [3,] 3 1 2 3 1 [4,] 1 2 3 1 2 in which rows 1 and 4 are similar, I want to find that vector c (1,2,3,1,2). Atte Tenkanen University of Turku, Finland

I'd like to pick every imbricated five character long subsets from a vector. I guess there is some efficient way to do this without loops... Here is a for-loop-version and a model for output: VECTOR=c(1,4,2,6,5,0,11,10,4,3,6,8,6); ADDRESSES=c(); for(i in 1:(length(VECTOR)-4)){ ADDRESSES[i]=paste(VECTOR[i:(i+4)],collapse="") } > ADDRESSES [1] "14265"

I have a similar situation and have assigned all users to varying primary groups. It works OK but could be better. I am including an e-mail from the samba list that improves on my login/out scripts. I have not had a chance to try the improvements yet. Good luck. Forwarded From: "Naccarato, Robert" <naccarar@bis.adp.com> > > > I use the following script in the root

Hi, Can somebody tell me, which is the fastest way to make comparisons between all rows in a matrix (here A) and put the results to the new symmetric matrix? I have here used cosine distance as an example, but the comparison function can be any other, euclidean dist etc. A=rbind(c(2,3),c(4,5),c(-1,2),c(5,6)) M=matrix(nrow=length(A[,1]),ncol=length(A[,1])) for(i in 1:length(A[,1])) { for(j in

Dear all, I did a non-linear least square model fit y ~ a * x^b (a) > nls(y ~ a * x^b, start=list(a=1,b=1)) to obtain the coefficients a & b. I did the same with the linearized formula, including a linear model log(y) ~ log(a) + b * log(x) (b) > nls(log10(y) ~ log10(a) + b*log10(x), start=list(a=1,b=1)) (c) > lm(log10(y) ~ log10(x)) I expected coefficient b to be identical

Could somebody program this kind of plot type to R, if none exists, based on mds or correlation tables or some more suitable method? What do you think about idea? Does it work? None similar or better exists? http://weightedassociationmap.blogspot.com/ Atte Tenkanen University of Turku, Finland

Is there some (much) more efficient way to do this? VECTOR=c(3,2,4,5,5,3,3,5,1,6,6); NEWVECTOR=VECTOR[1]; for(i in 1:(length(VECTOR)-1)) { if((identical(VECTOR[i], VECTOR[i+1]))==FALSE){ NEWVECTOR=c(NEWVECTOR,VECTOR[i+1])} } > VECTOR [1] 3 2 4 5 5 3 3 5 1 6 6 > NEWVECTOR [1] 3 2 4 5 3 5 1 6 _______________________________ Atte Tenkanen University of Turku, Finland

Is there some way to change the line widths of plot borders? I couldn't find any parameters for that purpose. Atte Tenkanen University of Turku, Finland

Hi, What's wrong here?: > v=c(`-`,`+`,1,`^`,`^`,NA,NA,"X",9,"X",2) > i2=16 > v[i2] [[1]] NULL > is.null(v[i2]) [1] FALSE Is it a bug or have I misunderstood something? Atte Tenkanen University of Turku, Finland

Is there any possibility to divide too long text in a plot to two or more lines, when using labels-parameter in the text()-command? Here is an example picture: http://users.utu.fi/attenka/253.jpeg My example script is something like this: text(1,0.7,labels=Chordnames[fnid(pcs%%12)]) # according to Larry Solomon's table http://solomonsmusic.net/pcsets.htm Chordnames is a long vector with

Hi, Is it possible to break using if-condition during the recursive function? Here is a function which almost works. It is for inorder-tree-walk. iotw<-function(v,i,Stack,Indexes) # input: a vector and the first index (1), Stack=c(), Indexes=c(). { print(Indexes) # if (sum(i)==0) break # Doesn't work... if (is.na(v[i])==FALSE & is.null(unlist(v[i]))==FALSE) {Stack=c(i,Stack);

Hi there, I've installed R somewhen in German but I'd like to have it in English. The installation of a new version does not change the language setting though I think I've chosen "English" in the installation process (this choice was available, right?). Can anybody give me a hint? (Error-Messages in German are not easy to get a solution for...) Antje

I'm a beginner in R and, therefore, I don't know how serious my trouble is. After running a script: ** *t**<-c(14598417794,649693)* *data**=data.frame(read.spss("C:\\Ginters\\Kalibracija\\cal_data.sav"))* *Xs=**as.matrix(data[,1:2]) * *koef**=data.frame(read.spss("C:\\Ginters\\Kalibracija\\f.sav")) * *piks=**as.matrix(koef[,1**])*

Hi, I have data that looks like this: ASA1 ASA2 C1_C2 C M 9.0225 S S 2.4315 M C 3.4894 M S 4.5282 C M 1.3183 C S 1.3735 S C 1.0488 S M 7.948 M C 4.5827 I need to plot Boxplots for a given ASA1 (either C,S, or M) with respect to C1_C2. However, instead of one boxplot I want to plot

Dear R-users, I want to check if certain values are from random distribution, that includes values between 0-1. So, it is not really normal even though shapiro.test says it is highly normal... Can I do something like this and think that the values given are right. z.test is from package TeachingDemos. -------------------------------------------------------------------------------

We're almost at the stage of rolling out our new RH6.1/Samba server - one last item. We have some 60 users and 7 departments. Can I use/how do I use a logon script to log groups onto the server. I currently have 1 logon script per user, but a small change in 1 share means updating every script. Can I log them on by group ID but still map their personal shares with the %U flag so I then

Hello, Before I was able to interrupt R-calculations by pushing ESC, but now with R 2.6.2 (OSX Intel) it doesn't work. Is it bug or not? Regards, Atte Tenkanen University of Turku, Finland Department of Musicology +023335278

Hello, I have tried to change the margin widths so that mtext (here "sd of consecutive pc intervals", look at the picture) and plot(...,xlab="bar") fits to the picture. Here is an example: http://users.utu.fi/attenka/margins.png This doesn't help: par(mar=c(5.1, 7.1, 4.1, 2.1)) And here are the commands: png(filename="/Users/kone/Vaitostutkimus/Pictures/