how to index 3D array with 2D array ?

i have a colored image 100x100x3 and a logical matrix 100x100. how do i keep the values of those at which the logical matrix is 1 and set the those at 0 to 0 ?

I would imagine it something like this but it doesnt work.

image_a;
image_b;
map;
image_b = image_a(map);

Answer 1

[Stephen23]

Method one create a new image:

 image_b = bsxfun(@times,double(map),image_a)

or on newer MATLAB versions with implicit array expansion:

 image_b = double(map) .* image_a;

The most generalized method is probably:

 bsxfun(@times, image_a, cast(map,class(image_a)));

Method two change original image:

 image_a(~map(:,:,[1,1,1])) = 0

[RuiQi]

I tried

 image_b = bsxfun(@times,map,image_a)

and I got an error "Mixed integer class inputs are not supported."

   image_b = map .* image_a;

gives me this error "Error using .* Integers can only be combined with integers of the same class, or scalar doubles"

My image_a is a 3-channel uint8. and map is a logical

[Stephen23]

You should convert all of the inputs to the same numeric class, e.g. if the image is of type double:

 image_b = bsxfun(@times,double(map),image_a)

[Jan]

Applying a mask to an RGB image is a common task. I'm still astonished, that you need to create a large temporary array for this:

idx = cat(3, map, map, map);

"image_a(map, 1:3)" would contain enough information to be interpreted correctly.

[Stephen23]

@Jan Simon: on MATLAB R2012b I get something quite different:

>> A = randi(9,2,3,2)
A(:,:,1) =
     7     5     9
     3     7     9
A(:,:,2) =
     5     2     8
     2     3     3
>> X = [true,false,false;false,true,true]
X =
     1     0     0
     0     1     1
>> A(X,1:2) = 0
A(:,:,1) =
     0     0     9
     3     7     9
     0     0     0
     0     0     0
     0     0     0
     0     0     0
A(:,:,2) =
     5     2     8
     2     3     3
     0     0     0
     0     0     0
     0     0     0
     0     0     0

But this works (I know, there is still a large intermediate variable):

>> A = randi(9,2,3,2)
A(:,:,1) =
     7     5     3
     7     1     9
A(:,:,2) =
     2     5     1
     8     9     4
>> X = [true,false,false;false,true,true]
X =
     1     0     0
     0     1     1
>> A(X(:,:,[1,1])) = 0
A(:,:,1) =
     0     5     3
     7     0     0
A(:,:,2) =
     0     5     1
     8     0     0