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&lt;-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&lt;- 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?



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