C - B
C[A==0] - 0
would be somewhat more efficient.
On Sun, 2 Mar 2008, jim holtman wrote:
Does this do what you want?
A - matrix(sample(0:2, 25, TRUE), ncol=5)
B - matrix(1:25, ncol=5)
C - ifelse(A == 0, 0, B)
A
[,1] [,2] [,3] [,4] [,5]
[1,]11121
[2,]10110
[3,]00102
[4,]01200
[5,]12122
B
[,1] [,2] [,3] [,4] [,5]
[1,]16 11 16 21
[2,]27 12 17 22
[3,]38 13 18 23
[4,]49 14 19 24
[5,]5 10 15 20 25
C
[,1] [,2] [,3] [,4] [,5]
[1,]16 11 16 21
[2,]20 12 170
[3,]00 130 23
[4,]09 1400
[5,]5 10 15 20 25
On Sun, Mar 2, 2008 at 7:11 AM, Diogo André Alagador
[EMAIL PROTECTED] wrote:
To all,
I am undergoing an analysis involving big matrices of about 3x200 which
I have to handle in a more efficient way. So I would like some advice to
build such efficient function to deliver the following result:
- starting with 2 matrices of the same dimension (eg. A and B)
0 0 3 5 6 0 0 5
A= 0 0 6 4 B= 0 4 3 5
0 0 5 0 1 0 0 9
- the function should deliver a C matrix (same dimension too),
where at each position C(i,j), compares A and B.
if A(i,j)=0, than C(i,j)=0,
if A(i,j)!=0, than C(i,j)=B(i,j)
6 0 0 5
C= 0 0 3 5
0 0 0 0
Although not an expert I could build a function with 2 cycles (reading
columns and rows) which is not quick. Maybe you can help me in this
challenge.
Much thanks in advance,
Diogo André Alagador
Biodiversity Global Change Lab, Museo Nacional de Ciencias Naturales,
CSIC, Madrid, España
Forest Research Centre, Instituto Superior de Agronomia, Universidade
Técnica de Lisboa, Lisboa, Portugal
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Jim Holtman
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+1 513 646 9390
What is the problem you are trying to solve?
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Brian D. Ripley, [EMAIL PROTECTED]
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University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
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