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0

sum of matrix omitting one dimension

Rafael Schwarzenegger さんによって質問されました 2017 年 11 月 2 日
最新アクティビティ Stephen Cobeldick
さんによって コメントされました 2017 年 11 月 2 日
Hi, I would like to ask how I could sum over a n dimensional matrix except one dimension, so the output is a vector. The ndims are not known in forehand. The summation is giving always a vector. (In my case a marginal pdf in statistics). Something like sum(K(:) except i-th dimension)
The best in a cyclus ( ndims not known).
For example having matrix K, the sums would be [6 22], [12 16], [10 18]
K(:,:,1) =
0 1
2 3
K(:,:,2) =
4 5
6 7

  3 件のコメント

KL
2017 年 11 月 2 日
You have to rephrase your question, what is your input and what do you expect as output?
Cedric Wannaz
2017 年 11 月 2 日
There may be a mistake in your example. Isn't the first 12 supposed to be 22?
Yeah 22, thank you.
I have as an input a m^n matrix and as and output n vectors of the length m.

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2 件の回答

回答者: Stephen Cobeldick
2017 年 11 月 2 日
編集済み: Stephen Cobeldick
2017 年 11 月 2 日
 採用された回答

Here is one very simple way:
>> d = 3; % dimension
>> v = 1:ndims(K);
>> v([1,d]) = v([d,1]);
>> s = sum(reshape(permute(K,v),size(K,d),[]),2)
s =
6
22
It works by swapping the first and the desired dimension, then reshaping so that all trailing dimensions are reduced to one. The sum is then trivially along each row.
Note that the method I show here also has the significant advantage that it does not create any variables with copies of the data from the input array: this will be important for scaling to larger input arrays. Also note that slow and complex arrayfun or cellfun are not required for this task!

  4 件のコメント

Stephen Cobeldick
2017 年 11 月 2 日
Just change the dimension d. It is on the very first line of code, I even labeled it for you.
" It gives [6 22]. "
No, it gives all of the answers that you showed in your question. How do I know this? Because I tested it. You just need to change the dimension value d. It would be trivial to do this in a loop, if you need to.
Oh yeah, that is correct. (I thought first, it is the total dimension.) Thank you very much!
Stephen Cobeldick
2017 年 11 月 2 日
@Rafael Schwarzenegger: the dimensions of the input array are automatically measured using ndims(K).

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回答者: Cedric Wannaz
2017 年 11 月 2 日

Here is one way, but there is probably a simpler approach:
buffer = arrayfun(@(k) permute(KL, circshift(1:ndims(KL), k-1)), 1:ndims(KL), 'UniformOutput', false) ;
s = cellfun(@(M) sum(reshape(M, [], size(M,2))), buffer, 'UniformOutput', false) ;
where s is a cell array of sum vectors.
EDIT : I won't have time, but maybe you can see how dimensions work with SHIFTDIM and understand how to have the outputs ordered as needed:
buffer = arrayfun(@(k) shiftdim(KL, k-1), 1:ndims(KL), 'UniformOutput', false) ;
s = cellfun(@(M) sum(reshape(M, [], size(M,2))), buffer, 'UniformOutput', false) ;
PS : you'll have to test whether it really does what you need. The approach is based on the fact that MATLAB reads memory column first

  4 件のコメント

Stephen Cobeldick
2017 年 11 月 2 日
@Rafael Schwarzenegger: See my answer for a simpler, neater, and much more efficient solution.
Cedric Wannaz
2017 年 11 月 2 日
As mentioned by Stephen, arrayfun and cellfun are not necessary. They are hidden loops and I used them for conciseness. You can build a FOR loop if you prefer, and again, as Stephen mentions it would have the advantage not to require storing copies of your data (if you are dealing with large arrays).
@Cedric Wannaz: Thank you anyway.

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