Dear members
I have matrix with size M*N and vector with size 1*K
I want to create a null vector with length of N - length of K
I tried this :
V=zeros(1,length(N)-length(K));
but it doesn't work. Any solution please ?
Thank you.

3 件のコメント

Just Manuel
Just Manuel 2021 年 2 月 18 日
this would be easier if you told us what is not working. Do you get an error message? Or just an unexpected result?
As it stands, your code is valid.
Afluo Raoual
Afluo Raoual 2021 年 2 月 18 日
I get an unexpected result which is:
1*0 empty double row vector
Just Manuel
Just Manuel 2021 年 2 月 18 日
Yep, then go with Bjorn Gustavsson's answer.
Cheers
Manuel

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 採用された回答

Bjorn Gustavsson
Bjorn Gustavsson 2021 年 2 月 18 日

0 投票

First off if you've done something like:
N = 12;
M = 14;
K = 7;
Mtr = randn(M,N);
V = rand(1,K);
V=zeros(1,length(N)-length(K));
Then you only check the length of the 1-by-1 arrays N and K - and the difference of that is zero. If your N and K are your arrays you might have run into a situation where N < K, because this also happens when N < K. Perhaps you've mixed up the dimensions of your matrix and you meant to make something like this:
V=zeros(1,M - K);
HTH

4 件のコメント

Afluo Raoual
Afluo Raoual 2021 年 2 月 18 日
No, in my case N > K
I have N=7774 and K=5832
When I put zeros(1,7774-5832) it works, but with zeros(1, length(N)-length(K)) it doesn't work
Just Manuel
Just Manuel 2021 年 2 月 18 日
Yes, then omit the length call. Just do
zeros(1, N-K);
Cheers
Manuel
Afluo Raoual
Afluo Raoual 2021 年 2 月 18 日
It's done. Thank you :)
Just Manuel
Just Manuel 2021 年 2 月 18 日
You're welcome.

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