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using find function and logical array

Asked by Lorenne on 5 Jun 2018
Latest activity Commented on by Lorenne on 5 Jun 2018
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 Answers

Answer by Birdman
on 5 Jun 2018
Edited by Birdman
on 5 Jun 2018
 Accepted Answer

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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Thanks!

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Answer by monika shivhare on 5 Jun 2018

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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+1 nice and clear explanation!
Thanks!!

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Answer by Walter Roberson
on 5 Jun 2018

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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