[R] Confused by code?

[Rui Barradas]

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 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,

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 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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>> PLEASE do read the posting guide 
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> R-help at 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.