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Confused by code?

6 messages · Peter Alspach, Bazman76, Rui Barradas

Bazman76
# 
x<-matrix(c(1,0,0,0,1,0,0,0,1),nrow=3)
The resultant matrix x is all zeros except for the last two diagonal cells
which are 1's.
While y is lower triangualr 0's with the remaining cells all ones.

I really don't understand how this deceptively simple looking piece of code
is giving that result can someone explain please.
I'm obviously missing something pretty basic so please keep your answer
suitably basic.



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Peter Alspach
# 
Tena koe

I think you probably meant:
x[as.logical(z)] <- y[as.logical(z)]

i.e., choosing those elements of ? and y where z is 1 (TRUE as logical).  Whereas what you have written:

?[z] <- y[z]

references the 0th (by default indexing starts at 1 so this is empty (see ?[0]) and the first element of ? and y (repeatedly).

Hope this helps ....

Peter Alspach

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Bazman76
Sent: Monday, 24 September 2012 8:53 a.m.
To: r-help at r-project.org
Subject: [R] Confused by code?

x<-matrix(c(1,0,0,0,1,0,0,0,1),nrow=3)
The resultant matrix x is all zeros except for the last two diagonal cells which are 1's.
While y is lower triangualr 0's with the remaining cells all ones.

I really don't understand how this deceptively simple looking piece of code is giving that result can someone explain please.
I'm obviously missing something pretty basic so please keep your answer suitably basic.



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Rui Barradas
# 
Hello,

It is pretty basic, and it is deceptively simple. The worst of all :)
When you index a matrix 'x' by another matrix 'z' the index can be a 
logical matrix of the same dimensions or recyclable to the dims of 'x', 
it can be a matrix with only two columns, a row numbers column and a 
column numbers one, or your case.

In your case, 'z' is coerced to vector, and the values in 'z' are taken 
to be indexes to 'x'. But since you only have two distinct values and 
one of them is zero, it will only return x[1] three times (there are 
three 1s in 'z'). The same goes for 'y'.

Correct:

# Create an index matrix
z.inx <- which(z == 1, arr.ind = TRUE)
z.inx

# Test
x1 <- x2 <- x3 <- x # Use copies to test
x1[z == 1] <- y[z == 1]
x2[z.inx] <- y[z.inx]
# 1 and 0 to T/F
x3[as.logical(z)] <- y[as.logical(z)]

x1
identical(x1, x2)
identical(x1, x3)


Hope this helps,

Rui Barradas

Em 23-09-2012 21:52, Bazman76 escreveu:
Bazman76
# 
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?

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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Rui Barradas
# 
Hello,

Inline.
Em 24-09-2012 15:31, Bazman76 escreveu:
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
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:
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.
Use this suggestion. It can't go wrong.

Rui Barradas