Array indexing question for vectorization

I have three arrays A, B, and IND, with A containing some values that I want to update, B containing the values that I want to use to update A, and IND containing the indeces to update in A. IND may reference the same index in A several times. For example, I may have:
A = ones(1, 10);
B = 1:10;
IND = [1 1 2 5 5 5 5 6 8 9];
and the update I want to do is multiplication. I can achieve this by doing
for ii = 1:10
A(IND(ii)) = A(IND(ii)) * B(ii);
end
which gives me my desired answer of
A =
2 3 1 1 840 8 1 9 10 1
Is there a way I can do this in a vectorized operation, avoiding the for loop (I'm potentially doing this kind of operation thousands of times on large arrays)? Doing
A(IND) = A(IND) .* B
results in
A =
2 3 1 1 7 8 1 9 10 1
Any tips greatly appreciated!

2 Kommentare

Are A and B always like that? Could they be more like:
A = randi(N,1,M);
B = randi(K,1,J);
or not?
Richard
Richard am 13 Nov. 2012
Hi Matt,
My initial problem specification was indeed lacking - see my reply to Jan below for a more accurate representation of the problem...

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

Honglei Chen
Honglei Chen am 13 Nov. 2012
Bearbeitet: Honglei Chen am 13 Nov. 2012

0 Stimmen

You can try the following trick
c = unique(IND)
d = accumarray(IND',B',[],@prod)
A(c) = A(c).*d(c)'

6 Kommentare

Richard
Richard am 13 Nov. 2012
Fantastic, thanks! I didn't know about accumarray(), looks very useful.
José-Luis
José-Luis am 13 Nov. 2012
Bearbeitet: José-Luis am 13 Nov. 2012
This is not a good idea though. A for loop will be much faster in this case.
unique() is expensive
accumarray() is expensive
  1. Loop: Elapsed time is 0.000443 seconds.
  2. Accumarray: Elapsed time is 0.021888 seconds.
Matt J
Matt J am 13 Nov. 2012
Bearbeitet: Matt J am 13 Nov. 2012
There's no real need for UNIQUE, though, as far as I can see
z=accumarray(IND(:),B,[10,1],@prod,1).';
A=A.*z;
Jan
Jan am 13 Nov. 2012
@Jose-Luis: Please post this as an answer such that I can vote for it. @Richard: Please post the typical size of the data. While the loop will have a nearly linear scaling, unique is less efficient dur to the internal sorting. If your data are sorted already, there are smarter versions than unique for the first part. accumarray suffers from calling Matlab from a Mex function.
José-Luis
José-Luis am 13 Nov. 2012
True, but it still is slower than a for loop:
Elapsed time is 0.000003 seconds. %for
Elapsed time is 0.000828 seconds. %unique+accumarray
Elapsed time is 0.000515 seconds. %accumarray
Richard
Richard am 13 Nov. 2012
Interesting, thanks. I'm still in the mindset of "for-loop bad", and aiming to replace them without necessarily thinking of the built-in function overheads...

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Jan
Jan am 13 Nov. 2012
Bearbeitet: Jan am 13 Nov. 2012

0 Stimmen

Is your A pre-allocated? If IND is sorted, running the loop backwards wil be faster:
for ii = 10:-1:1
A(IND(ii)) = A(IND(ii)) * B(ii);
end
If IND is not sorted, pre-allocate explicitly:
A = zeros(1, max(IND));
For more speed create a C-Mex function. If you are interested and have a compiler installed, I can post the few required lines of code.

2 Kommentare

Hi Jan,
Yes, A is preallocated. Actually the problem is a bit more complicated than I posted (note to self: make sure your example code is suitable!): in my actual application, A is a 2D matrix, and B is a 1D array with fewer elements than A, so the solution I've accepted doesn't quite work... e.g.
% A more accurate example...
A = zeros(20000, 10);
indR = randi(20000, 1, 500); % row indeces
indC = randi(10, 1, 500); % column indeces
IND = sub2ind(size(A1), indR, indC);
B = rand(1, 500);
tic
for ii = 1:length(IND)
A(IND(ii)) = A(IND(ii)) * B(ii);
end
toc
(in this example with random indeces I think the instances of duplicate values in IND are quite low compared with the actual program). Thinking about it, my IND values could be sorted if I reorganised my data a bit prior to this (which may be worth the reorganisation overhead as this bit of the program is running in a simulation loop over many time steps)...
Thanks for the useful discussion everyone.
Jan
Jan am 13 Nov. 2012
@Richard: A = zeros(...); ... A(k) = A(k) * ... leads to zeros. As in your original question your need ones().

Melden Sie sich an, um zu kommentieren.

Kategorien

Produkte

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by