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Lorenne
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using find function and logical array

Lorenne
さんによって質問されました 2018 年 6 月 5 日
最新アクティビティ Lorenne
さんによって コメントされました 2018 年 6 月 5 日
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?

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3 件の回答

Birdman
回答者: Birdman
2018 年 6 月 5 日
編集済み: Birdman
2018 年 6 月 5 日
 採用された回答

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.

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Lorenne
2018 年 6 月 5 日
Thanks!

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回答者: 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

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Stephen Cobeldick
2018 年 6 月 5 日
+1 nice and clear explanation!
Lorenne
2018 年 6 月 5 日
Thanks!!

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Walter Roberson
回答者: 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)

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