using find function and logical array
73 ビュー (過去 30 日間)
古いコメントを表示
If i have a matrix Z=[1 0 1;0 1 0;1 0 1];
>> find(Z==1)
ans =
1
3
5
7
9
>> find(Z(Z==1))
ans =
1
2
3
4
5
Why does the find(Z(Z==1)) produces the above output?
0 件のコメント
採用された回答
Birdman
2018 年 6 月 5 日
編集済み: Birdman
2018 年 6 月 5 日
Because
Z(Z==1)
will produce
1
1
1
1
1
which is a new column vector generated from the initial line. Then, you want to find the Z(Z==1) which is equivalent to Z(Z==1)~=0, therefore you obtain
1
2
3
4
5
which are the indices of all ones in your vector.
その他の回答 (2 件)
monika shivhare
2018 年 6 月 5 日
find(M) takes the matrix given in argument M, and returns matrix of indexes in M where value at that index in M is not zero. Z==1 gives a logical matrix of size(Z) showing 1 at index where value of Z is equal to 1.
>> Z==1
ans =
3×3 logical array
1 0 1
0 1 0
1 0 1
logical indexing of Z returns a array of elements of Z where logical value at that index is 1. So, Z(Z==1) gives array of elements of Z where (Z==1) has value 1
>> Z(Z==1)
ans =
1
1
1
1
1
Now, if we execute find(Z(Z==1)), it will return indexes of Z(Z==1) where value of Z(Z==1) is not zero. Value of Z(Z==1) is nonzero at index [1,2,3,4,5]. Hence,
>> find(Z(Z==1))
ans =
1
2
3
4
5
2 件のコメント
Walter Roberson
2018 年 6 月 5 日
What else could it produce? Z(Z==1) is logical indexing and returns the elements of Z where Z is 1 -- a vector of values. Because you asked for 1, all the values in that vector are 1. You then find() on that vector of 1's. None of the 1's are 0, and find() returns the locations of all non-zero values... but all of the values are non-zero, so that is all of the locations in that vector that was created by Z(Z==1)
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Resizing and Reshaping Matrices についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!