How can I calculate a weighted mean in Matlab?

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I have a 180-by-360 matrix of (surface temperature) values and I want to calculate a weighted average of all values given in this matrix. However, I need to weight these values with respect to latitude. Is there a way to calculate a weighted mean in Matlab? Please help me.
Thanks in advance for any help!
- Jan-Erik
  1 Comment
shelley
shelley on 20 Oct 2015
Normally taking dim 1 mean of some, say 3x5 matrix A, can be seen as w=[1/3 1/3 1/3], and w*A returns you the answer.
Now if we want to take dim 1 weighted mean of A, we can set our weight in w = [1/6 2/3 1/6], then take w*A, which should return you a 1x5 row vector.

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Accepted Answer

Matt J
Matt J on 18 Feb 2013
Edited: Matt J on 18 Feb 2013
If A is your 180x360 matrix and W is a 180x1 vector of weights for the latitudes, do
weighted_mean = mean(W.'*A,2);
  6 Comments
Jason Garcia
Jason Garcia on 28 Jan 2014
weighted_mean = mean(bsxfun(@times,A,w'),2)
Matt J
Matt J on 28 Jan 2014
weighted_mean = mean(bsxfun(@times,A,w'),2)
If that's what the OP intended, it would be more efficient to do
weighted_mean=w'.*mean(A,2);

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More Answers (2)

John Hock
John Hock on 28 Apr 2019
Hi Everyone
I just want to ask one thing that if i have 2d array 2*4 and want to get weighted average with vector of 1*4
i want first row of A get multiplied with B and give weighted average as an output similarly for next row.
Please help
  2 Comments
Image Analyst
Image Analyst on 28 Apr 2019
Try
B = repmat(B, [2, 1]) % Make B 2*4
weightedMeans = sum(A.*B, 1); % Get weighted means within a row going across columns.
stelios loizidis
stelios loizidis on 27 May 2019
Hello,
I have the following questuin: I have a matrix e.g
[5 3 3 1
3 4 5 2
5 0 0 0
3 4 5 2]
For this matrix I have to make a weighted average for each row. That is [5+3+(4/5)*(3+1)] /4=16/5. Also, if there is only one number and the zero (row 3) the weighted average should be [5+0+0+0] / 1 =5 and the end we choose the largest weight average. How does this work?

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Micke Malmström
Micke Malmström on 3 Oct 2019

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