Searching and summing array

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Fischer Zheng
Fischer Zheng on 30 Oct 2015
Answered: Fischer Zheng on 2 Nov 2015
Hi All, I have three arrays. One small one called A and two huge ones called M1 and M2. M1 and M2 are of the same size.
A = ['K1','K2','K3'....];
M1 = ['K9','K45','K2'..];
M2 = [123,34.2,1654,...];
I am trying to sum all the values in M2 base on M1 into the buckets in A. As the current example, 'K2' bucket would have 1654 plus anything else that is 'K2' in M1.
Thanks in advance, Fischer
  1 Comment
Jan
Jan on 30 Oct 2015
Note, that ['K1','K2','K3'] is the string 'K1K2K3'. Is this really the representation in your code? Or is A a cell string?

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Accepted Answer

Fischer Zheng
Fischer Zheng on 2 Nov 2015
This is what did on my first try and it takes 30mins to run through the code. However, when I did two for loops it is much faster. Taking only 15 seconds.
I guess the reason is that there are usually only two maybe three elements per key. running using vectorization is much slower because it go through the big array too many times

More Answers (1)

Geoff Hayes
Geoff Hayes on 30 Oct 2015
Fischer - think about how you would do this by hand. You would start with the first element of A, 'K1', and then look at each element of M1. Every match you find you would then add to the sum corresponding to 'K1'. Try doing something like the following where we assume that A and M1 are cell arrays and we use find to determine the indices of those elements of M1 that match the element of A that we are interested in.
% create an array that will manage the sums for each element of A
sumsOfA = zeros(size(A));
% iterate over each element of A
for k=1:length(A)
% get the indices of M1 that match A{k}
idcs = find(strcmp(A{k},M1))
% sum the elements of M1
sumsOfA(k) = sum(M2(idcs));
end
  2 Comments
Jan
Jan on 30 Oct 2015
This works faster without the find command.
Geoff Hayes
Geoff Hayes on 31 Oct 2015
For sure! The result of
strcmp(A{k},M1)
will return an array of zeros and ones which can the be used to access the elements of M2.
Thanks, Jan!

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