Indexing a 3D array with a vector

% My question is best explained with an example;

% I would like to change each of the elements in column three of each page of exampleData to -100, 
% if they are greater than 5. I have done this with a for loop, but would like to know if there is 
% a more elegant way, using assignments.

% Creating example data;

x = [[1,2,3,4];[5,6,7,8];[2,8,9,7]];

y = [[1,3,5,3];[7,9,0,2];[2,4,5,4]];

z = [[2,4,6,6];[3,5,7,1];[8,9,3,3]];

exampleData = cat(3,x,y,z);

% For loop method - this gets the desired result, but I am interested to see if it can be done

% via an assignement;

b = exampleData;  % using "b" just to make it more readable.

for pageInd = 1:length(exampleData(1,1,:));

      cInd = b(:,3,pageInd)>5;
      b(cInd,3,pageInd) = -100;
  end

% Assignment method - this does not produce the desired result.

c = exampleData;  % using "c" just to make it more readable.

cInd = c(:,3,:)>5;

c(cInd) = -100;

% Thanks very much for any help you give.

% John

>> b (desired result)

b(:,:,1) =

     1     2     3     4
     5     6  -100     8
     2     8  -100     7

b(:,:,2) =

     1     3     5     3
     7     9     0     2
     2     4     5     4

b(:,:,3) =

     2     4  -100     6
     3     5  -100     1
     8     9     3     3

>> c (undesired result)

c(:,:,1) =

     1     2  -100     4
  -100     6  -100     8
  -100     8     9     7

c(:,:,2) =

     1     3     5     3
     7     9     0     2
     2     4     5     4

c(:,:,3) =

     2     4     6     6
     3     5     7     1
     8     9     3     3

Answer 1

[David Young]

c = b(:,3,:);
c(c > 5) = -100;
b(:,3,:) = c;

[John]

David,

Great answer thanks for that!

I’ve been looking through it trying to follow it.

I think at one stage we have the same index, mine was “b2(:,3,:)>5” and yours was “c = b(:,3,:)” followed by “(c > 5) = -100”. They then both obtain the same index values;

>> ind

ind(:,:,1) =

     0
     1
     1

ind(:,:,2) =

     0
     0
     0

ind(:,:,3) =

     1
     1
     0

With your answer you then used this index on the third column only “c” and then applied that to exampleData. Whereas I tried to apply the index directly into exampleData.

I really like your solution, thanks very much for your help.

Answer 2

[pietro]

try this:

x = [[1,2,3,4];[5,6,7,8];[2,8,9,7]];

y = [[1,3,5,3];[7,9,0,2];[2,4,5,4]];

z = [[2,4,6,6];[3,5,7,1];[8,9,3,3]];

exampleData = cat(3,x,y,z);
[I,J,K] = ind2sub(size(exampleData),find(exampleData>5));
b=exampleData;
for i=1:length(I)
    b(I(i),J(i),K(i))=-100;
end

[John]

Hi pietro.

I tried your answer, but didn’t get the desired result. Also I was trying to avoid a for loop simply as I felt it should be possible by assignment.

Please note I’ve edited the original question as I got the example answers round the wrong way. Thanks very much for your help.

Your proposal gave; >> b

b(:,:,1) =

     1     2     3     4
     5  -100  -100  -100
     2  -100  -100  -100

b(:,:,2) =

     1     3     5     3
  -100  -100     0     2
     2     4     5     4

b(:,:,3) =

     2     4  -100  -100
     3     5  -100     1
  -100  -100     3     3