Confused by code?
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 2
which 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 between
0 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 of
mP[1] and all other elements will remain unaffected?
Is this correct?
Yes and no, 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) 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 Barradas
If so I am still confused as this is not what I thought was supposed to by happening 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.html Sent from the R help mailing list archive at Nabble.com.
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