similar to: Same sum, different sets of integers

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

I came up with this, using recursion. Short and should work for n greater than 9 :) Peter sumsToN = function(n) { if (n==1) return(1); out = lapply(1:(n-1), function(i) { s1 = sumsToN(n-i); lapply(s1, c, i) }) c(n, unlist(out, recursive = FALSE)); } > sumsToN(4) [[1]] [1] 4 [[2]] [1] 3 1 [[3]] [1] 2 1 1 [[4]] [1] 1 1 1 1 [[5]] [1] 1 2 1 [[6]] [1] 2 2 [[7]] [1] 1 1 2

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/

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 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

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

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

... 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

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

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, How would you calculate the "mean colour" of several colours, for example c("#FF7C00","#00BF40","#FFFF00")? Yours, Atte Tenkanen

Hi, Is there a function for raising a matrix to a power? For example if you like to compute A%*%A%*%A, is there an abbreviation similar to A^3? Atte Tenkanen > A=rbind(c(1,1),c(-1,-2)) > A [,1] [,2] [1,] 1 1 [2,] -1 -2 > A^3 [,1] [,2] [1,] 1 1 [2,] -1 -8 But: > A%*%A%*%A [,1] [,2] [1,] 1 2 [2,] -2 -5

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"

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

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

?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

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

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, When plotting the density estimation with DrawDensity3D-function (in package VecStatGraphs3D) it often happens that the end product ie. layers are not intact , see figure in the link: https://www.dropbox.com/s/vzyaiu0vso8hjw2/20623868_10213333234554853_1371245353_n.png?dl=0 Can we somehow effect on that an get intact layers? -- Atte Tenkanen