Reshape matrix with averages of 4 elements in each row

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Erin
Erin 2022 年 9 月 12 日
コメント済み: Erin 2022 年 9 月 13 日
Hello,
I want to take the average of every 4 elements in each row in A and put that into a smaller matrix B as shown below.
A = [ 0 2 2 4 0 6 4 6
1 1 2 8 0 0 4 0
9 9 2 8 0 0 0 8...]
B = [ 2 4
3 1
7 2...]
I am pretty new to MatLab, any ideas how I can go about this?
Any help is much appreciated, thanks in advance!
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Erin
Erin 2022 年 9 月 12 日
yes

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採用された回答

Walter Roberson
Walter Roberson 2022 年 9 月 12 日
A = [ 0 2 2 4 0 6 4 6
1 1 2 8 0 0 4 0
9 9 2 8 0 0 0 8]
A = 3×8
0 2 2 4 0 6 4 6 1 1 2 8 0 0 4 0 9 9 2 8 0 0 0 8
B = reshape(mean(reshape(A.', 4, []), 1), [], size(A,1)).'
B = 3×2
2 4 3 1 7 2
This is not the most obvious of methods. You first transpose rows and columns so that what were originally the row values become consecutive in memory. You can then reshape into groups of 4 and take the mean along the rows, and then reshape and transpose back.
The above process has the advantage of having the 4 be adjustable to any length that divides the number of columns.
In the specific case of 4, you can instead use
temp = double(A);
B = (temp(:,1:4:end) + temp(:,2:4:end) + temp(:,3:4:end) + temp(:,4:4:end))/4
B = 3×2
2 4 3 1 7 2
The double(A) step is there because A is not necessarily an integer data type. If it were, for example, uint8 (such as an image) and you were to add the values, then you would likely "saturate" the uint8 representation. So you need to do the addition as double (or at least something that is certain to be able to handle the entire possible range of sums) in case it is not already. If you know for sure that A is already double, you can skip the step,
B = (A(:,1:4:end) + A(:,2:4:end) + A(:,3:4:end) + A(:,4:4:end))/4
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Erin
Erin 2022 年 9 月 13 日
Thank you so much!

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その他の回答 (1 件)

the cyclist
the cyclist 2022 年 9 月 13 日
Here is one way:
n = 4;
A = [ 0 2 2 4 0 6 4 6
1 1 2 8 0 0 4 0
9 9 2 8 0 0 0 8];
tmp = movmean(A,[0 n-1],2);
B = tmp(:,1:n:end)
B = 3×2
2 4 3 1 7 2
  1 件のコメント
Erin
Erin 2022 年 9 月 13 日
Thank you for this answer as well!

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