フィルターのクリア
フィルターのクリア

matrix comparison and substitute value

1 回表示 (過去 30 日間)
Nafila Farheen
Nafila Farheen 2019 年 11 月 6 日
回答済み: Walter Roberson 2019 年 11 月 6 日
Hi, I have one matrix
A=[4 2 3
9 5 6]
I have another vector b=[1 0]. I want to compare the first column of A with b, and if the position in A correspondes to a 1 in b, then I would like to replace that value (4 multiply 0.99) and if 0, it will keep as it is . I was wondering is there any way to do that without executing nested for loops?TIA

サインインしてこの質問に回答する。

回答 (1 件)

Walter Roberson
Walter Roberson 2019 年 11 月 6 日
A = A * (1 - 0.01 * (repmat(b.', 1, size(A,2)) == 1))

カテゴリ

Help Center および File Exchange で Data Types 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