I've just reread my answer and it's not very clear. Not at all. Inline. Em 24-09-2012 18:34, Rui Barradas escreveu:
Hello,Inline. Em 24-09-2012 15:31, Bazman76 escreveu:Thanks Rui Barrudas and Peter Alspach, I understand better now: x<-matrix(c(1,0,0,0,2,0,0,0,2),nrow=3) y<-matrix(c(7,8,9,1,5,10,1,1,0),nrow=3) z<-matrix(c(0,1,0,0,0,0,6,0,0),nrow=3) x[z]<-y[z] viewData(x) produces an x matrix 7 0 0 0 2 0 0 10 2which makes sense the first element of y 7 is inserted into z in slot x[1]and the and 6th element of y 10 is slotted into the x[6]. However the original code runs like this: mI<- mRU(de.d, de.nP)>de.CR mPV[mI]<mP[mI] where mPv and MP are both (de.d, de.nP) matrices. and mRU<-function(m,n){ return(array(runif(m*n), dim=c(m,n))) }i.e. it returns an array of m*n random numbers uniformly distributed between0 and 1. de.CR is a fixed value say 0.8. So mI<- mRU(de.d, de.NP)>de.CR returns a de.d*de.nP array where each element is 1 is its more than 0.8 and zero otherwise.So in this case element mPv[1] will be repeatedly filled with the value ofmP[1] and all other elements will remain unaffected? Is this correct?Yes and no,
Yes, it is absolutely correct. As is, the matrix mI is coerced to vector first and then, since it only has values 0 and 1 the element mPv[1] will be repeatedly filled with thesame value of mP[1].
The rest of my answer is right, though. But 'it', the very first word in my post after this comment, refers to what? To the condition that creates the index matrix ml but this is not at all as clear as it should.
it should return a logical matrix, not a numeric one. Since it seems to be returning numbers 0/1, you can use as.logical like I've shown in my first post, or, maybe better,mI<- which(mRU(de.d, de.nP) > de.CR, arr.ind = TRUE)
Use this suggestion. It can't go wrong. Rui Barradas
Like this you'll have an index matrix, whose purpose is precisely what its names says, to index. Matrices. (I'm also a bit confused as to why the logical condition is returning numbers, are you sure of that?)Anyway, the right way would be to index 'mPV' using a logical or an index matrix.Hope this helps, Rui BarradasIf so I am still confused as this is not what I thought was supposed to byhappening but I know that the code overall does its job correctly? --View this message in context: http://r.789695.n4.nabble.com/Confused-by-code-tp4643946p4644010.htmlSent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
