min function on two arrays

51 ビュー (過去 30 日間)
Robert Demyanovich
Robert Demyanovich 2021 年 6 月 8 日
コメント済み: Walter Roberson 2021 年 6 月 8 日
I've read the documentation on the min function, but still don't understand how it works on two arrays. I have the following:
dS=zeros(1,N)
dS=min(cA(i+1,:),cB(i+1,:))
where cA and cB are equally sized arrays. Doesn't the min function just take the value at equivalent locations in cA or cB that is the lowest. So if position 10,50 in cA is 5 and the same position in cB is 3, min returns 3 in the resulting row vector, dS, at column 50. Is that correct?
  3 件のコメント
Robert Demyanovich
Robert Demyanovich 2021 年 6 月 8 日
No, I'm just a MatLab beginner trying to understand some complex (for me, anyway) code.
What if dS had been defined as an array? I guess the result is the same because in both cA and cB, the i+1 row has been specified.
James Tursa
James Tursa 2021 年 6 月 8 日
編集済み: James Tursa 2021 年 6 月 8 日
This line creates dS as a vector:
dS=zeros(1,N)
Then this line completely overwrites the dS you just created and instead assigns dS the result of the min( ) function call:
dS=min(cA(i+1,:),cB(i+1,:))
I.e., the first line is completely useless and accomplishes nothing because it gets overwritten in the second line.
And yes, cA(i:1,:) and cB(i+1,:) will be a row vectors if i is a scalar.

サインインしてコメントする。

採用された回答

SALAH ALRABEEI
SALAH ALRABEEI 2021 年 6 月 8 日
%
dS=min([cA(i+1,:),cB(i+1,:)])
  3 件のコメント
SALAH ALRABEEI
SALAH ALRABEEI 2021 年 6 月 8 日
Oops, yes it must be semicol; Thanks.
% correction
dS=min([cA(i+1,:);cB(i+1,:)])
Walter Roberson
Walter Roberson 2021 年 6 月 8 日
That would work in the case that cA and cB have the same number of columns.
The original code
dS=min(cA(i+1,:),cB(i+1,:))
would also work if one of cA or cB had a single column and the other one did not.

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeMatrices and Arrays についてさらに検索

タグ

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by