Sum two cells and write to a third cell

43 ビュー (過去 30 日間)
User05
User05 2015 年 5 月 11 日
コメント済み: User05 2015 年 5 月 12 日
I have three cells A{10}, B{10} and C{10} of size 5x5. On each cell I want to do: A(:,1) = B(:,1) + C(:,1) I might also want to do: A(2:4,5) = B(2:4,5) + C(2:4,5)
Currently if I just write something like above, I will get error as Index exceeds matrix dimensions.
Any ideas on how this can be done (as fast as possible)

採用された回答

Alfonso Nieto-Castanon
Alfonso Nieto-Castanon 2015 年 5 月 11 日
編集済み: Alfonso Nieto-Castanon 2015 年 5 月 11 日
Probaby a loop is going to be fastest, e.g.:
for n=1:numel(A)
A{n}(:,1)=B{n}(:,1)+C{n}(:,1);
end
but if you really dislike for loops you could always do something like:
A = cellfun(@(a,b,c)[b(:,1)+c(:,1) a(:,2:end)], A, B, C, 'uni',0);
  3 件のコメント
Alfonso Nieto-Castanon
Alfonso Nieto-Castanon 2015 年 5 月 11 日
編集済み: Alfonso Nieto-Castanon 2015 年 5 月 11 日
not sure what you mean by:
in1{1} = tin1(:,1) + tin2(:,1);
that will change the variable in1 instead of the variable out, it will make the first element of that cell array a 10 by 1 vector instead of a 10 by 10 matrix, and it will only modify that first element of the in1 cell array and none of the other elements. Is that what you really want?
If, instead, what you want is to change all five elements of out, but only the first column of each, and keep the rest of those 10x10 matrices the same as they were, then use the original code again (but first initialize out to something that shows you unambiguously what happened, e.g.
out = repmat({randn(10)}, 1,5);
User05
User05 2015 年 5 月 12 日
Sorry that in1 was typo. Initializing out to something in beginning works. Below code achieves what I want to do:
in1 = cell(1,5);
in1(:) = {int32(zeros(10,10))};
in2 = cell(1,5);
in2(:) = {int32(ones(10,10))};
out = cell(1,5);
out(:) = {int32(zeros(10,10))};
for n=1:5
out{n}(:,1)=in1{n}(:,1)+in2{n}(:,1);
end

サインインしてコメントする。

その他の回答 (1 件)

Walter Roberson
Walter Roberson 2015 年 5 月 11 日
編集済み: Walter Roberson 2015 年 5 月 12 日
cellfun()
However, the method to do it "as fast as possible" is likely to involve creating some mex code written in C and optimizing the heck out of it. It would probably take you a few months to learn how to do properly -- I know the resident expert James Tursa is still learning about using mex after what must be at least 8 years of working with it.
  3 件のコメント
Walter Roberson
Walter Roberson 2015 年 5 月 11 日
Use an anonymous function.
A = cellfun(@(a, b, c) [a(:,1:end-1), b(:,1) + c(:,1)], A, B, C, 'Uniform', false);
User05
User05 2015 年 5 月 12 日
I tried this anonymous function suggestion, but I dont quite get whats it exactly doing. My code:
in1 = cell(1,5);
in1(:) = {int32(zeros(10,10))};
in2 = cell(1,5);
in2(:) = {int32(ones(10,10))};
out = cell(1,5);
out(:) = {int32(zeros(10,10))};
out = cellfun(@(a, b, c) [a(:,1:end-1), b(:,1) + c(:,1)], out, in1, in2, 'Uniform', false);
After execution I see that 10th col of each cell in out is getting assigned value 1. What I wanted and expected was the 1st col should be all 1. So isnt that what is suppose to happen when I write b(:,1)+c(:,1)? Also what is a(:,1:end-1) doing?

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeLogical についてさらに検索

タグ

Community Treasure Hunt

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

Start Hunting!

Translated by