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

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!

Answer 1

[Titus Edelhofer]

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

[Jan]

You can confirm this by checking the value of:

A3D(50,92,100) - 1

[Wojtek]

Thanks for the answer! You were right. I found almost 150 000 of values close to 1. Now I'm trying to change these values: so I'm doing:

A3D(val1,val2,val3) = 0 ;

But then matlab is "busy" forever. Yesterday I waited for 20 hours (and restarted matlab). Is there something wrong with my code, or is it normal that it takes so much time to change values in the matrix 200x200x200 in 150 000 cells?

[Matt Fig]

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]

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!

Answer 2

[Azzi Abdelmalek]

use

 [val1,val2,val3] = ind2sub(size(A3D),find(A3D <= 1 & A3D>1-eps));

maybe 1 is 0.9999999999999999999 for numeric considerations

[Jan]

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.

[Azzi Abdelmalek]

you are right; maby 1-10^(-10) can work