obtain a matrix out of other matrix

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mehra
mehra 2019 年 10 月 24 日
コメント済み: mehra 2019 年 10 月 25 日
Hello
I have a 1*104 matrix which ı have to create a 8*13 matrix out of it.
For one column of the final 8*13 matrix I can do the following code but how can I write one single code to get the whole matrix
zs_d=zeros(8,1);
for r=1:8;
zs_d(8,1)=(zs(13*r-12));
end
This gives me the first coloumn and for the second column I can write :
for r=1:8;
zs_d(8,1)=(zs(13*r-11));
end
and it goes the same till end...
I tried to write it in one for loop like the following but it gave me an error:
zs_d=zeros(8,13);
for r=1:8,c=1:12;
zs_d(8,13)=(zs(13*r-13-c));
end
error : Subscript indices must either be real positive integers or logicals.
thanks for the help
  2 件のコメント
Fangjun Jiang
Fangjun Jiang 2019 年 10 月 24 日
Your code does not match your description. In any case, I think reshape() should help you.
mehra
mehra 2019 年 10 月 24 日
Maybe I can add that my 104*1 matrix is zs and I need to get zs(1), zs(14) , zs(27), zs(40), zs(53),zs(66),zs(79),zs(92) in the first column and the rest will also follow the trend to be in the 2nd to 13th columns. Please let me know if any other explanation is needed.

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採用された回答

Fangjun Jiang
Fangjun Jiang 2019 年 10 月 24 日
transpose(reshape((1:104)',13,[]))

その他の回答 (1 件)

Steven Lord
Steven Lord 2019 年 10 月 24 日
編集済み: Steven Lord 2019 年 10 月 24 日
reshape your vector to be 13-by-8 then take the transpose of it.
>> x = 1:104;
>> y = reshape(x, [13 8])
>> z = y.'
You can do this in one step, but I think it's important to see the intermediate result (y) so you can see how MATLAB arranges elements. This is useful knowledge for a technique called linear indexing.
  1 件のコメント
mehra
mehra 2019 年 10 月 25 日
thanks alot

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