"Find" function with 3d matrix doesn't work

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Wojtek
Wojtek 2012-8-21
Hello, I have a 3d matrix named A3D (size 200x200x200). Values in this matrix are from 0 to 1. If I type for example:
A3D(50,92,100)
I get an answer:
ans =
1.0000
So I'm sure there are values 1 (actually I know it also from "image" of a slice of matrix - there I can see a lot of values "1"). But when I try to use "find":
[val1,val2,val3] = ind2sub(size(A3D),find(A3D == 1));
then vectors val1 val2 and val3 are empty. These vectors are filled when I type:
[val1,val2,val3] = ind2sub(size(A3D),find(A3D < 1));
It that case Matlab says that all of the values in my matrix are <1. But that is not true! Why matlab doesn't want to find values == 1? Thank you very much for your help!

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采纳的回答

Titus Edelhofer
Titus Edelhofer 2012-8-21
Hi,
the answer
ans =
1.0000
instead of
ans =
1
tells you the problem: the value is 1, but only rounded to 5 digits. It's not exactly one. Change your code to
[val1,val2,val3] = ind2sub(size(A3D),find(abs(A3D-1)<1e-10));
where 1e-10 is the tolerance. Choose the tolerance in such a way that you only catch those entries that are 1 (within the tolerance).
Titus
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Matt Fig
Matt Fig 2012-8-21
编辑：Matt Fig 2012-8-21
If that is all you are doing, why even bother with FIND and IND2SUB?
A3D(abs(A3D-1)<1e-10) = 0;
In fact, what you are trying to do above will not even work. Look at a simpler example so you can see for yourself.
A = rand(3,3,3)
B = A; % A copy, to compare.
[I,J,K] = ind2sub(size(A),find(abs(A-1)<.2)) % Find all on [.8 1];
A(I,J,K) = nan % Does this look right to you?? NO!
B(abs(B-1)<.2) = nan % That looks better!!
Wojtek
Wojtek 2012-8-21
编辑：Wojtek 2012-8-21
I get it now. I wanted to change values in this matrix (for example values == 1) with values from another matrix but with the same coordinates. That's why I wanted to save coordinates from the 1st matrix to apply them to the 2nd one. But of course now I can do:
A3D(abs(A3D-1) < 1e-4) = B(abs(A3D-1) < 1e-4);
Well, thanks a lot! I would wait for a long time, because matlab didn't show any error - it just went "busy". Thanks once again!

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更多回答（1 个）

Azzi Abdelmalek
Azzi Abdelmalek 2012-8-21
use
[val1,val2,val3] = ind2sub(size(A3D),find(A3D <= 1 & A3D>1-eps));
maybe 1 is 0.9999999999999999999 for numeric considerations
  2 个评论
Jan
Jan 2012-8-21
编辑：Jan 2012-8-21
There are not much numbers, which can be represented by a IEEE754 double value between 1-eps and 1... Think of the definition of EPS.

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