Sum across columns with shift
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I have a matrix a, let say 20 rows and 10 columns. I want to obtain an array b where b(1)=a(1,1) b(2)=a(1,2)+a(2,1) b(3)=a(1,3)+a(2,2)+a(3,1) ... b(20+10-1)=a(20,10) In practice, every row of the a matrix is shifted right by one column with respect to the row above and then the elements of each column of the resulting (larger) matrix are summed. Is it possible to obtain this without loops and without building the big shifted matrix?
采纳的回答
May be some other more clever indexing, but the "deadahead" thing that comes to mind if I understand the desire
>> a=1:18;a=reshape(a,6,[]) % sample smaller dataset for illustration...
a =
1 7 13
2 8 14
3 9 15
4 10 16
5 11 17
6 12 18
The engine
[r,c]=size(a); % get the array dimensions
b=arrayfun(@(i) sum(diag(flipud(a),i)),-(r-1):c-1); % sum diagonals in desired sequence
Result
>> b
b =
1 9 24 27 30 33 29 18
>>
ADDENDUM:
Somewhat cleaner is to subtract earlier for the indexing cleanup...
[r,c]=size(a)-1; % array dimensions less one for 0-base count
b=arrayfun(@(i) sum(diag(flipud(a),i)),r:c); % sum diagonals in desired sequence
2 个评论
Andrea Console
2020-2-20
andrea console
2020-2-25
How hard do you think it could be to extend this answer to a three-dimensional case? I.e. sum of bidimensional matrices shifted across one of the axes
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