Summing every nth column

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Adnan Habib
Adnan Habib on 29 Nov 2021
Commented: Adnan Habib on 29 Nov 2021
Hi,
I have a 1336 by 89700 matrix. I want to sum every 300th colum starting from the first column so that I end up with a 1336 by 300 matrix. So the first colum in the new matrix will be the sum of colum 1, 300, 599....etc. of the previous matrix. And the second column of the new matrix will be the sum of column 2, 301, 600...etc. of the previous matrix. I hope I was able to clearly explain what I want. Any help will be appreciated.

Accepted Answer

Image Analyst
Image Analyst on 29 Nov 2021
Edited: Image Analyst on 29 Nov 2021
Try this and see if it does what you want:
data = rand(1336, 89700); % Create sample data.
sums = zeros(1336, 299); % Preallocate.
for col = 1 : size(data, 2) / 299
% Extract every 300th column, starting with column "col", into a new matrix.
extractedMatrix = data(:, col : 300 : end); % A new matrix 1/300th as wide as the original
% Sum all the values in that matrix, and put that sum into means.
% Get the sums for just this one set of columns.
theseSums = sum(extractedMatrix, 2);
% Store these sums in the column "col" of our master array.
sums(:, col) = theseSums; % Get a single sum from the entire 2-D matrix.
end
whos sums
Name Size Bytes Class Attributes sums 1336x300 3206400 double
  1 Comment
Adnan Habib
Adnan Habib on 29 Nov 2021
This worked! Thank you so much

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More Answers (2)

Yongjian Feng
Yongjian Feng on 29 Nov 2021
Your logic is not consistent. from 1 to 300, there are 298 rows in between. But from 300 to 600, there are 299 rows in between.
You might want 1, 301, 601, etc?
You can use reshape and sum to do this. Here a simpler example of summing 1, 4th, 7th.
ma = [1 2 3; 4 5 6; 7 8 9; 10 11 12; 13 14 15; 16 17 18;19 20 21; 22 23 24; 25 26 27]
mb = reshape(ma', 9, 3)
mc = sum(mb, 2)
me = reshape(mc, 3, 3)'
ma =
1 2 3
4 5 6
7 8 9
10 11 12
13 14 15
16 17 18
19 20 21
22 23 24
25 26 27
mb =
1 10 19
2 11 20
3 12 21
4 13 22
5 14 23
6 15 24
7 16 25
8 17 26
9 18 27
mc =
30
33
36
39
42
45
48
51
54
me =
30 33 36
39 42 45
48 51 54
  1 Comment
Adnan Habib
Adnan Habib on 29 Nov 2021
Thanks for this. Yes I made a mistake. It should be 1+300+599....i.e. 298 columns in between.

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Image Analyst
Image Analyst on 29 Nov 2021
I think this does what you said, though I'm not sure it's what you want:
data = rand(1336, 89700); % Create sample data.
means = zeros(1, 299); % Preallocate.
for col = 1 : 299
% Extract every 300th column, starting with column "col", into a new matrix.
extractedMatrix = data(:, col : 300 : end); % A new matrix 1/300th as wide as the original
% Sum all the values in that matrix, and put that sum into means.
means(col) = sum(extractedMatrix(:)); % Get a single sum from the entire 2-D matrix.
end
  2 Comments
Image Analyst
Image Analyst on 29 Nov 2021
OK, let's examine this. So your original matrix is 1336 rows tall by 89700 columns wide. So if you take every 299'th column (1, 300, 599, etc.) you get an extracted matrix of 1336 rows tall by 299 columns wide.
So now you say "So the first colum in the new matrix will be the sum of colum 1, 300, 599....etc. of the previous matrix." so we're to sum the extracted matrix row-by-row over all the columns so we'd end up with a new matrix of 1336 by 1 for the sum over the columns. So we'd do sums = sum(extractedMatrix, 2).
See my other answer elsewhere on the page for a full solution.

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