Why does it become four number in a line?

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zhenyu zeng
zhenyu zeng 2019 年 3 月 1 日
コメント済み: Steven Lord 2019 年 3 月 1 日
>>A(0 0 0;0 2 2)
>>A(1:2)=[]
A=
0 2 0 2
Why A becomes 0 2 0 2 not becomes (0 2;0 2)?

採用された回答

Stephan
Stephan 2019 年 3 月 1 日
Hi,
you do linear indexing which lets Matlab reshape A. To avoid this use:
A(:,1)=[]
Best regards
Stephan
  3 件のコメント
Stephan
Stephan 2019 年 3 月 1 日
There are 2 ways of indexing in Matlab. Depending on what you want to do the one maybe better or the other. Knowing both and how they behave lets you make the right decision. I think having 2 possibilities is the advantage.
Steven Lord
Steven Lord 2019 年 3 月 1 日
If you delete elements using linear indices, if MATLAB didn't reshape the resulting elements into a vector you could end up with an array with different numbers of elements in each row or column. That's not allowed. Arrays in MATLAB must be "rectangular".
thisWillNotWork = [1 2; 3]
In your particular case the elements remaining after the deletion would still be arranged in a rectangle. But in general they wouldn't. Ideally you shouldn't need to check after performing that deletion operation what shape the result is.
In fact, even if you don't delete any elements at all we reshape the result. You tried to delete elements, so we put your data in the shape of a vector.
A = magic(4);
A([]) = [];
size(A) % [1 16]

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その他の回答 (1 件)

TADA
TADA 2019 年 3 月 1 日
you accessed a 2d matrix with linear indexing,
If you want to retain the dimentions of your matrix use subscripts instead:
A(:,1) = []
A =
0 0
2 2

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