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Large binned matrix reshape

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Avik Mahata
Avik Mahata 2021-10-8
评论： Image Analyst 2021-10-9
I was trying to find a similar solution to my problem, but I couldn't find any exact solution. I have a large matrix of 2000x100, similar data is binned in to 10 different matrices. Each of the bins are 2000x10.
That matrix is right now, bin1(2000x10) bin2 (2000x10) ......... bin10(2000x10).
I am trying to reshape the such a way that it will be ,
bin1 (2000x10)
bin2 (2000x10)
.
.
.
bin10. (2000x10)
So I should be ending up with a dataset of 20000x10. The bins need to be converted in the same sequence.

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采纳的回答

Matt J
Matt J 2021-10-8
编辑：Matt J 2021-10-8
tmp=num2cell(reshape(yourMatrix,200,10,10),[1,2]);
result=vertcat(tmp{:});
  2 个评论
Avik Mahata
Avik Mahata 2021-10-9
Thanks Matt for taking time to reply. This trick worked. Thanks a lot. Wish you a happy long weekend.
Image Analyst
Image Analyst 2021-10-9
Not sure why going to a cell array, tmp, was necessary. Since you said your "data is binned in to 10 different matrices" of size 2000 x 100 (which is not large by the way) then why can't you simply combine your 10 different matrices like this:
% Vertically stack the "10 different matrices" that the user has:
allBins = vertcat(bin1, bin2, bin3, bin4, bin5, bin6, bin7, bin8, bin9, bin10);
as I suggested in my answer? Why convert each matrix into a cell array?

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更多回答（1 个）

Image Analyst
Image Analyst 2021-10-8
Wouldn't it just be
% Vertically stack the "10 different matrices" that the user has:
allBins = vertcat(bin1, bin2, bin3, bin4, bin5, bin6, bin7, bin8, bin9, bin10);
??? Or is that not the ordering you wanted?

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