How to vectorize conditional triple nested for loop - MATLAB

Question

I have two 3D arrays:

shape is a 240 x 121 x 10958 array
area is a 240 x 1 x 10958 array

The values of the arrays are of the type double. Both have NaN as fill values where there is no relevant data.

For each [240 x 121] page of the shape array, there are several elements filled by the same number. For example, there will be a block of 1s, a block of 2s, etc. For each corresponding page of the area array there is a single column of numeric values 240 rows long. What I need to do is progressively go through each page of the shape array (moving along the 3rd, 10958-long axis) and replace each numbered element in that page with the number that fills the row of the matching number in the area array.

For example, if I'm looking at shape(:,:,500), I want to replace all the 8s on that page with area(8,1,500). I need to do this for numbers 1 through 20, and I need to do it for all 10958 pages of the array.

If I extract a single page and only replace one number I can get it to work:

shapetest = shape(:,:,500);
shapetest(shapetest==8)=area(8,1,500);

This does exactly what I need for one page and for one number. Going through numbers 1-20 with a for loop doesn't seem like an issue, but I can't find a vectorized way to do this for all the pages of the original 3D array. In fact, I couldn't even get it work for a single page without extracting that page as its own matrix like I did above. I tried things like this to no avail:

shape(shape(:,:,500)==8)=area(8,1,500);

If I can't do it for one page, I'm even more lost as to how to do it for all at once. But I'm inexperienced in MATLAB, and I think I am just ignorant of the proper syntax.

Instead, I ended up using a cell array and the following very inefficient nested for loops:

MyCell=num2cell(shape,[2 1]);
shapetest3=reshape(MyCell,1,10958);

for w=1:numel(shapetest3)
     test_result{1,w}=zeros(121,240)*NaN;
end  
 
for k=1:10958
    for i=1:29040 % 121 x 240
        for n=1:20
            if shapetest3{1,k}(i)==n
                test_result{1,k}(i)=area(n,1,k);
            end
        end
    end
end

This gets the job done, and I can easily turn it back to an array, but it is very slow, and I am confident there is a much better vectorized way. I'd appreciate any help or tips. Thanks in advance.

I'm getting an "Array indices must be positive integers or logical values" error. The data type of the arrays is double. Could that be the issue? Also, both the shape and area arrays have fill values of NaN. I realized I should have mentioned that. — Archimedes
– Archimedes, Commented Jul 9, 2020 at 15:07

Cris Luengo · Accepted Answer · 2020-07-09 22:17:50Z

1

To vectorize the mapping operation, we can use shape as an index into area. But because the mapping is different for each plane, we need to loop over the planes to accomplish this. In short, it'll look like this:

test_result = zeros(size(shape)); % pre-allocate output
for k=1:size(area,3)              % loop over planes
   lut = area(:,1,k);
   test_result(:,:,k) = lut(shape(:,:,k));
end

The above only works if shape only contains integer values in the range [1,N], where N = size(area,1). That is, for other values in shape we'll be doing wrong indexing. We will need to fix shape to avoid this. The question here is, what do we want to do with those out-of-range values?

As an example, preparing shape to deal with NaN values:

code = size(area,1) + 1; % this is an unused code word
shape(isnan(shape)) = code;
area(code,1,:) = NaN;

This replaces all NaN values in shape with the value code, which is one larger than any code value we were mapping. Then, we extend area to have one more value, a value for the input code. The value we fill in here is the value that the output test_result will have where shape is NaN. In this case, we write NaN, such that NaN in the input maps to NaN in the output.

Something similar can be done with values below 0 and above 240 (shape(shape<1 | shape>240) = code), or with non-integer values (shape(mod(shape,1)~=0) = code).

edited Jul 9, 2020 at 22:17

answered Jul 9, 2020 at 15:29

Cris Luengo

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1 Comment

Archimedes Over a year ago

Fantastic! Thank you! Replaced an execution time of about 10 min with one of approximately 1.67 seconds.

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