Reshaping (M,N)-matrix to (M,1)-matrix
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Hello everyone,
I have a matrix, for example:
A = [1, 2, 3;
4, 5, 6;
7, 8, 9]
and now I want to create a matrix:
A = [123;
456;
789]
Does anybody know how I can do this efficiently?
(I need to do this 62.000 times for my matrix)
0 件のコメント
採用された回答
Andrei Bobrov
2013 年 11 月 8 日
編集済み: Andrei Bobrov
2013 年 11 月 8 日
z = floor(log10(A)+1);
z(isinf(z)) = 1;
ex = fliplr(cumsum([zeros(size(A,1),1), fliplr(z(:,2:end))],2));
out = sum(A.*10.^ex,2);
eg:
> A =[14 10 3 16
12 0 10 18
3 11 14 7
9 13 0 20];
> z = floor(log10(A)+1);
> z(isinf(z)) = 1;
> ex = fliplr(cumsum([zeros(size(A,1),1), fliplr(z(:,2:end))],2));
> out = sum(A.*10.^ex,2)
out =
1410316
1201018
311147
913020
3 件のコメント
その他の回答 (2 件)
Jos (10584)
2013 年 11 月 8 日
Another trick using strings:
A = [1 2 3 ; 10 11 12 ; 90 0 99]
B = str2num(sprintf([repmat('%d',1,size(A,2)) ' '],A.')).'
2 件のコメント
Jos (10584)
2013 年 11 月 8 日
Note that you can play around with the "%d". For instance, you can use "%02d" so that a single digit number will be have a leading zero.
Simon
2013 年 11 月 8 日
編集済み: Simon
2013 年 11 月 8 日
Hi!
So your resulting matrix is a vector. right?
Try this
mat10 = ones(size(A, 1), 1) * 10.^((size(A, 2)-1):-1:0);
sum(A .* mat10, 2)
This is applicable for matrices of any size.
In string form
str2num(char(regexprep(cellstr(num2str(A)), ' ', '')))
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