Binning matrix data (negative and positive decimals): mean of each 100 rows and create a new matrix
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William Norbey Sanchez Luna
on 26 Aug 2024
Commented: William Norbey Sanchez Luna
on 29 Aug 2024
I have a A= 7382300X12 matrix. I want to reduce it by averaging every 1000 rows in all columns. So at the end I will have a B= 73823X12 matrix.
The matrix A have negative and positive decimals, so blockproc() didn't work for me.
How can I do it?
2 Comments
Walter Roberson
on 26 Aug 2024
You have the small issue that 7382300 is not evenly divisible by 1000. Should the final 300 entries be treated as a complete group?
Accepted Answer
Sahas
on 27 Aug 2024
As per my understanding, you would like to reduce the size of an input matrix by taking the mean of every 1000 rows and all the columns.
The following code snippet achieves the desired functionality. You can modify the input matrix to suit your preferred data types.
inputMatrix = randi([-100, 100], 3300, 12);
numRows = size(inputMatrix, 1);
numCols = size(inputMatrix, 2);
blockSize = 1000;
% Calculate the number of complete blocks
numBlocks = floor(numRows / blockSize);
% Reshape inputMatrix to group every 'blockSize' rows together for each column
reshapedA = reshape(inputMatrix(1:blockSize*numBlocks, :), blockSize, numBlocks, numCols);
% Calculate the mean along the first dimension
semiFinalMatrix = squeeze(mean(reshapedA, 1))
% Assuming treating the leftover rows as one group and getting their mean as well
remainingRows = mod(numRows, blockSize);
if remainingRows > 0
% Calculate the mean of the remaining rows
remainingMean = mean(inputMatrix(end-remainingRows+1:end, :), 1);
% Append the mean of the remaining rows to semiFinalMatrix
finalMatrix = [semiFinalMatrix; remainingMean]
end
Refer to the following MathWorks documentation links for more information on “reshape” and “squeeze” functions.
- https://www.mathworks.com/help/matlab/ref/reshape.html
- https://www.mathworks.com/help/matlab/ref/squeeze.html
Hope this is beneficial!
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