Multipling a multidimensinnal array with a vector

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John Hock
John Hock 2019-4-28
评论： Star Strider 2019-5-1
Hi Everyone
I am trying to multiply a multidimensional 13*2560 (13 rows and 2560 coloumns) with 1*2560 (1 row and 2560 coloumns).
But matlab is giving error that matrix dimensions is nit match
For example as for simplest example
A=[1 2 3 4 ; 5 6 7 8]; 2*4 array
>> B=[11 22 33 44 ];1*4 vector
I want first row of A get multiplied with B and give its mean as a output then same for other row.
weighted_mean = mean(B.*A)
Please help
Thanks in advance

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Star Strider
Star Strider 2019-4-28
编辑：Star Strider 2019-4-28
If you want the mean of each row, specify the dimension argument as 2:
weighted_mean = mean(B.*A, 2);
That should do what you want.
EDIT —
To calculate a mean of the rows with respect to the elements of the weighting vector (as described in Mathematical definition (link) of the Weighted aritmentic mean), this works:
WeightedMean = sum(bsxfun(@times, B, A),2) / sum(B);
equivalently (R2016b and later releases):
WeightedMean = sum(B.*A,2) / sum(B);
This of course assumes that ‘B’ is the weighting vector.
Experiment to get the result you want.
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John Hock
John Hock 2019-5-1
@Star Strider
Sorry to disturb you again but i again facing some problem
My matrix whose which I want to calculate weighted average z3(22*2562) and weighted vector
abc(1*2562)
Using the above mentioned command
WeightedMean = sum(bsxfun(@times, abc, z3),2) / sum(abc)
But I am expecting the output 1*2562 but it is giving output of just 22*1 values
Pictures are attached1111.PNG
Please help me in this regard
2222.PNG
Thanks in advance
3333.PNG
Star Strider
Star Strider 2019-5-1
No worries.
If you want a (1 x 2562) result, sum down the columns rather than across the columns:
WeightedMean = sum(bsxfun(@times, abc, z3),1) / sum(abc);
or using the default column sum:
WeightedMean = sum(bsxfun(@times, abc, z3)) / sum(abc)
My only concern is that may not be the same thing as the true weighted mean, since you are applying the weighting vector across the columns, although summing down the columns.

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