How to average 5 rows of a matrix iteratively?

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Samuel White
Samuel White 2018 年 9 月 5 日
コメント済み: Star Strider 2018 年 9 月 5 日
I have a large matrix. I wish to avaerage the data from 5 rows and write the values to a new matrix iteratively until the whole matrix has been averaged. How can I do this? Many thanks.
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Star Strider
Star Strider 2018 年 9 月 5 日
The matrix you posted has 44 rows and 10 columns.
What result do you want?
Samuel White
Samuel White 2018 年 9 月 5 日
Well, it actually has 5000 rows and 10 collumns. I want to average sets of 5 rows ideally :)

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回答 (2 件)

Joel Miller
Joel Miller 2018 年 9 月 5 日
I am unsure whether you want to average rows 1-5, 2-6, 3-7, etc., or rows 1-5, 6-10, 11-15, etc. I assume the number of rows in your array, y, is an integer multiple of 5. You can create a 3D array to get a solution for the former case:
for k=1:5
z(:,:,k)=circshift(y,1-k,1);
end
z=z(1:end-4,:,:);
m=mean(z,3);
Likewise, for the latter case, create a 3D array:
for k=1:size(y,2)/5
z(k,:,:)=permute(y((k-1)*5+1:(k-1)*5+5,:),[3 2 1]);
end
m=mean(z,3);
Star Strider
Star Strider 2018 年 9 月 5 日
This creates ‘blocks’ of 5 rows each, then takes the mean down the columns (across rows) of each ‘block’. I shortened the matrix you posted by 4 rows (to make the number of rows an integer multiple of 5), so I am re-posting it here as well as the code.
M = [0 0.1904 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0098 0 6.5535
0 0.1904 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0100 0 6.5535
0 0.1904 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0100 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0098 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0100 0.0002 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0098 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0100 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0098 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0100 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0098 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0100 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0098 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0100 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0100 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0100 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0098 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0100 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0100 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0098 0 6.5535
0 0.1903 -0.0235 0.0690 0.0106 -0.0001 -0.0179 0.0104 0 6.5535
0 0.1902 -0.0235 0.0691 0.0106 -0.0001 -0.0179 0.0095 0 6.5535
0 0.1901 -0.0235 0.0691 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1900 -0.0236 0.0692 0.0106 -0.0001 -0.0179 0.0099 0 6.5535
0 0.1899 -0.0236 0.0692 0.0106 -0.0001 -0.0179 0.0099 0 6.5535];
M3 = reshape(M', size(M,2), 5, []); % Reshape ‘M’
M3 = permute(M3, [2 1 3]); % Change Dimension Order
Mmean5 = squeeze(mean(M3,3)) % Take Row Means Of 5-Row ‘Blocks’
Experiment to get the result you want.
  2 件のコメント
Samuel White
Samuel White 2018 年 9 月 5 日
Thank you for your response. Unfortunately I am quite new to matlab and so I'm not sure I quite understand. I tried the code but it keeps returning an error:
Error using reshape Product of known dimensions, 45, not divisible into total number of elements, 24957.
Error in filefix (line 3) M3 = reshape(M', size(M,2), 5, []); % Reshape M
Many thanks
Star Strider
Star Strider 2018 年 9 月 5 日
My pleasure.
My code returns the mean values of ‘blocks’ of 5 rows, and I shortened your matrix to 40 rows (an integer multiple of 5) so that it would be compatible with 5-row blocks.
The matrix you are working with has 2 incompatible rows:
ExtraRows = rem(24957, 5)
ExtraRows =
2
Probably the easiest way to deal with that is:
M3 = reshape(M(1:24955,:)', size(M,2), 5, []); % Reshape ‘M’
M3 = permute(M3, [2 1 3]); % Change Dimension Order
Mmean5 = squeeze(mean(M3,3)) % Take Row Means Of 5-Row ‘Blocks’
then:
Last2 = mean(M(end-1:end,:))
You can then append (concatenate) ‘Last2’ to the end of ‘Mmean5’ as:
Mmean5 = [Mmean5; Last2];
or just keep them separate.
This last is UNTESTED CODE, since I do not have your matrix. It should work.

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