use of find command for 3d array

How to use find command for a 3d array

eg. A = rand(5,5,3)

I want to get i,j,k location of the array A for value [0.5 ,0.2,0.3]

psuedo code

[i,j,k] = find(A(:,:,1) == .5 & A(:,:,2)== 0.2 & A(:,:,3) ==0.3)

but this is not working any help is highly appreciated

Answer 1

[Guillaume]

For arrays with more than two dimensions you have to get the linear index (1 output version of find) and then transform that into ND indices using ind2sub:

 idx = find(yourlogicalarray);
 [row, col, page] = ind2sub(size(yourlogicalarray), idx);

However, in your example the array you're passing to find is a 2D array, not 3D. So, in your case:

[row, col] = find(A(:,:,1) == .5 & A(:,:,2)== 0.2 & A(:,:,3) ==0.3);

is all that is required. And to get the 3 values that match these coordinates:

A(row, col, :)

[Jan]

+1. A marginal note: Some years ago I asked the support if expressions like ".5" and "5." are guaranteed to work and if this is documented. The answer was: "We don't know. Everything, which can be created by sprintf, is supported. Stay at this." All interpreters I know accept ".5" so I think this is just a question of taste - but not documented. :-)

[Guillaume]

To be honest, I just copy/pasted the original code for that bit. Didn't even notice the .5 (and the inconsistent use of blank spaces).

That the answer is "We don't know" is alarming! I would expect the grammar of the parser to be well defined, not left to random chance. As far as I know, both are perfectly fine.

Answer 2

[Adam]

doc ind2sub

Use the standard form of find that returns linear indices and convert them using ind2sub to 3d.