Loop through hundreds of matrices to change values larger than 1 to 0

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Zoe 2017-10-19

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评论： Zoe 2017-10-23

I have more than 400 matrices, currently I am using "find" to change the values larger than 1 in the matrix to 0 one by one manually with this code:

val = find(yf > 1);
yf(val) = 0;

Is there a way to do it with loop? Please help, thank you.

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James Tursa 2017-10-19

"... 400 matrices ..."

How are these matrices stored? In individual variables? As part of a cell or struct array? Or ...?

Zoe 2017-10-19

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Hi Walter, those matrices are actually image data, the value I WANT is actually between 0 and 1 (definitely smaller than 1), and values larger than 1 are noise (background data, etc.)

Hi James, in a 1*452 cell array , so I use

images{1, #}

to read them one by one right now.

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Answer 1

Image Analyst 2017-10-19

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If you want a loop and your images are in the cell array, you can do it like this:

for k = 1 : length(images)
  % Extract his one image.
  thisImage = images{1, k};
  % Reduce/Clip values 1.000001 and larger to 1.
  thisImage(thisImage > 1) = 1;
  % Stick back into cell array
  images{1, k} = thisImage;
end

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Image Analyst 2017-10-21

You're welcome, but you probably never should have stored all your images in a cell array in the first place. It takes up a lot of memory and there was probably never a need to have them all in memory simultaneously. You probably could have read your images one at a time and processed them immediately in the loop without storing them all. Anyway, thanks for Accepting the answer.

Zoe 2017-10-23

Thank you for letting me know this:)

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Answer 2

Star Strider 2017-10-19

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Instead of find, I would simply use ‘logical indexing’.

Example —

yf = 0.5 + rand(4,5)
yf(yf > 1) = 0
yf =
      0.64115       1.2321       1.0209       1.3162       1.1876
       1.0121       1.2498      0.71908       1.2939       1.4869
       1.2213      0.90732       1.3424      0.96911       1.2699
       1.4288      0.73949       1.1629      0.80952       1.3296
yf =
      0.64115            0            0            0            0
            0            0      0.71908            0            0
            0      0.90732            0      0.96911            0
            0      0.73949            0      0.80952            0

This should be more efficient, although you will still have to loop through every matrix.

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Zoe 2017-10-19

Thank you for replying! But my question is how are you gonna do the loop?? Sorry I am relatively new to matlab, and I do know lots of functions but I dont know how to loop through my matrices.