similar to: prcomp() and the lenght of PC:s

Dear R-users, I would like to test, whether a sample distribution differs significantly from a population distribution. They are not normally distributed. How should I proceed? Using somehow glm-models? How? The population and the sample data are here. They can be loaded using the load-command. http://users.utu.fi/attenka/D_Pop http://users.utu.fi/attenka/D_Samp Best regards, Atte Tenkanen

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. -------------------------------------------------------------------------------

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 all, I have a distribution, and take a sample of it. Then I compare that sample with the mean of the population like here in "Wilcoxon signed rank test with continuity correction": > wilcox.test(Sample,mu=mean(All), alt="two.sided") Wilcoxon signed rank test with continuity correction data: AlphaNoteOnsetDists V = 63855, p-value = 0.0002093 alternative hypothesis:

Hi, How can I test, whether all the values that is.element() -function returns are TRUE's Eg. > (is.element(c(1,4,2),c(1,2,3))) [1] TRUE FALSE TRUE This doesn't work: > (is.element(c(1,4,2),c(1,2,3)))==TRUE [1] TRUE FALSE TRUE Best, Atte Tenkanen, FT, MuM http://users.utu.fi/attenka/

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/

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

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

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

Hi, Does somebody know if there is some trick to betoken the starting points ("on the floor") of the vertical lines in scatterplot3d-plots. I'd prefer something like that one in the example here http://users.utu.fi/attenka/proposal3d.jpg If this is not possible, could this option be added? Sometimes it is not easy to distinguish objects behind the others and for instance in this

... and if you need to convert back: ?as.hexmode -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Sat, Apr 16, 2016 at 8:20 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > ?strtoi > > You'll have to remove

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

Dear R-users, How do you save a big table or matrix as an independent object and attach it to your Rdata-project when needed? Atte Tenkanen University of Turku, Finland Department of Musicology +023335278

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"

Since these are color strings, you can use functions in the grDevices package (other others) to manipulate them. E.g., you can convert them to various color spaces and perhaps use the mean in one of those spaces as your 'average color'. > myColors <- c(One="#FF7C00",Two="#00BF40",Three="#FFFF00") > col2rgb(myColors) One Two Three red

?strtoi You'll have to remove the "#" first, e.g. via substring() -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Sat, Apr 16, 2016 at 5:47 AM, Atte Tenkanen <attenka at utu.fi> wrote: > Hi, > > How

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

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 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

This is not the most efficient, but gets the idea across. This is the largest sum I can compute on my laptop with 16GB of memory. If I try to set N to 9, I run out of memory due to the size of the expand.grid. > N <- 8 # value to add up to > # create expand.grid for all combinations and convert to matrix > x <- as.matrix(expand.grid(rep(list(0:(N - 1)), N))) > > # generate