group sum with matrix
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With below matrix, I would like to have a sum of the third column and a sum of the fourth column by groups that are the combination of the first and second column.
1 1 1 0.5
1 1 2 0.5
1 1 3 0.5
1 1 4 0.5
1 1 5 0.5
1 2 6 0.5
1 2 7 0.5
1 2 8 0.5
1 2 8 0.5
1 2 9 0.5
2 1 0.1 2
2 1 0.2 2
2 1 0.3 2
2 1 0.4 2
2 1 0.5 2
2 2 0.6 2
2 2 0.7 2
2 2 0.8 2
2 2 0.8 2
2 2 0.9 2
So, here is what I am desiring for
1 1 15 2.5
1 2 38 2.5
2 1 1.5 10
2 2 3.8 10
1 个评论
Guillaume
2018-5-1
This is very similar to the question you asked yesterday. The same method can be used for both, so please read and learn from the answers you're given as otherwise people will start ignoring your questions.
回答（1 个）
Ameer Hamza
2018-5-1
The following code will summarise the table as described in question
groups = findgroups(A(:, 1), A(:, 2));
[~, uniqueGroupsIndex] = unique(groups);
uniqueGroups = A(uniqueGroupsIndex, 1:2);
s1 = splitapply(@(x) sum(x), A(:, 3:end), findgroups(A(:, 1), A(:, 2)) );
final = [uniqueGroups s1];
4 个评论
Guillaume
2018-5-1
You can simplify the whole
groups = findgroups(A(:, 1), A(:, 2));
[~, uniqueGroupsIndex] = unique(groups);
uniqueGroups = A(uniqueGroupsIndex, 1:2);
with
[uniqeGroups, ~, groups] = unique(A(:, [1 2]), 'rows')
and there's no need to call findgroups in the splitapply call. You've already calculated the result as group, so:
s1 = splitapply(@(x) sum(x), A(:, 3:end), groups);
Note: In my answer to the near identical question, I use accumarray instead of splitapply. The two are more or less equivalent, with splitapply being newer and slightly more versatile (needed here but not in the other question) but generally a lot slower than accumarray.
Ameer Hamza
2018-5-1
@Guillaume Thanks for further elaborating. Properly using unique() can really make code simple sometimes.
Boram Lim
2018-5-1
Boram Lim
2018-5-1
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