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numbering of matrices generation

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Ghulam Murtaza
Ghulam Murtaza 2016 年 6 月 23 日
編集済み: Walter Roberson 2016 年 6 月 23 日
hi if i have
for j=1:3;
B=[A(j,1); A(j,1)]
end;
then how i got the matrices B1 B2 B3 instead of only B .
  1 件のコメント
KSSV
KSSV 2016 年 6 月 23 日
How we would know how you got it without telling what is A and what you wanted?

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回答 (1 件)

Stephen23
Stephen23 2016 年 6 月 23 日
編集済み: Stephen23 2016 年 6 月 23 日
>> A = randi(9,3,1)
A =
6
5
9
>> B1 = A([1,1],1)
B1 =
6
6
>> B2 = A([2,2],1)
B2 =
5
5
>> B3 = A([3,3],1)
B3 =
9
9
But almost always creating numbered variables is a really bad idea. Just use indexing instead, because indexing is faster, more reliable, neater, and easier to read and understand than trying to create variable names dynamically. Read these to know why:
Or even better is to learn how to use MATLAB properly and avoid the loop altogether and use indexing:
>> A([1,1],1)
ans =
6
6
>> A([2,2],1)
ans =
5
5
Or if you really want to split A up into smaller matrices:
>> B = num2cell(A(:,[1,1]).',1);
>> B{1}
ans =
6
6
>> B{2}
ans =
5
5

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