Problem with find and logical array

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Hi,
I have a matrix with the values : A = [260,343; 344,433; 434,530; 531,631; 632,723];
where A(:,1) is the lower range and A(:,2) = upper range for each row. I am trying to find the index of A where a number may exist within the upper and lower range.
For ex. I tried B = (443 > A(:,1) & A(:,2) > 443) to check in which row the number 443 would lie, but got the ans as 0, even though when B comes as = [0,0,1,0,0]. I performed index = find(B),
I am probably doing something stupid. What am I doing wrong?
Thanks, Urvashi
  3 个评论
Urvashi
Urvashi 2014-6-6
I know. I will download 'find' again. May be it's a problem with my function
Geoff Hayes
Geoff Hayes 2014-6-6
What do you mean by "download 'find' again"? Are you using a different version of find from the built-in one?

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采纳的回答

Sara
Sara 2014-6-6
编辑:Sara 2014-6-6
You could do:
B = find(443 > A(:,1) & A(:,2) > 443);
B is the row or rows where the condition is satisfied.
  2 个评论
Urvashi
Urvashi 2014-6-6
I did that too, and still got the answer as B = 0. There might be something wrong with the function 'find' in my version. I should probably download it again.
Sara
Sara 2014-6-6
Yeah, could be, because everybody here is getting the right result.

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更多回答(2 个)

Image Analyst
Image Analyst 2014-6-6
I don't see any problem:
A = [260,343; 344,433; 434,530; 531,631; 632,723]
targetValue = 443
logicalRowsInRange = (targetValue > A(:,1) & A(:,2) > targetValue)
rowInRange = find(logicalRowsInRange)
A =
260 343
344 433
434 530
531 631
632 723
targetValue =
443
logicalRowsInRange =
0
0
1
0
0
rowInRange =
3

Urvashi
Urvashi 2014-6-6
编辑:Urvashi 2014-6-6
Yeah, thanks a lot. It worked. I had missed a ) in one of my previous statements of find, so find ended up being an empty matrix. And I didn't realize that the variable find was generated as there was a huge list of variables being generated during the program run.
My bad. Really, my bad.

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