Reshaping (M,N)-matrix to (M,1)-matrix

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Jonathan
Jonathan 2013-11-8
评论： Jos (10584) 2013-11-8
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)

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2013-11-8
编辑：Andrei Bobrov 2013-11-8
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
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Jonathan
Jonathan 2013-11-8
Only one small problem with this:
It adds zeroes to the end of my numbers and I would like that not to happen since my numbers are timestamps. Do you know how I can stop Matlab from adding those extra zeroes?
Thanks in advance!

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更多回答（2 个）

Jos (10584)
Jos (10584) 2013-11-8
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 个评论
Jos (10584)
Jos (10584) 2013-11-8
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.

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Simon
Simon 2013-11-8
编辑：Simon 2013-11-8
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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