# Mean and standard deviation of a field - condition on another field

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SS on 30 Aug 2019
Commented: SS on 4 Oct 2019
Hi. I am working with a structure array S (1 X 50,000) with 10 fields. I want to calculate the mean and standard deviation of field f1 based on a condition on field f2.
For example,
Here is the input,
S(1).f1=[10,20,30 40,50,60] and S(1).f2=[100,20,50,60,70,140];
S(2).f1=[56,98,74,87,99] and S(2).f2=[101,54,69,20,11]
S(3).f1=...... and S(3).f2=.....
S(4).f1=.... and S(4).f2=.....
.
.
S(i).f1=.... and S(i).f2=....
I want to calculate the mean of f1 elements whose corresponding f2 field values are 50 < f2 <=100
Meaning, S(1).f1=[10,40,50] and S(2).f1=[98,74] should be considered in calculation of mean.
I want to calculate several mean values of f1 based on different conditions on f2.

Matt J on 31 Aug 2019
Edited: Matt J on 31 Aug 2019
How can I get one single (explicit) mean and standard deviation?
F1=[S.f1];
F2=[S.f2];
F=F1(F2>50 & F2<=100);
MEAN=mean(F);
STD=std(F);
SS on 4 Oct 2019
Thanks.

### More Answers (1)

the cyclist on 30 Aug 2019
Edited: the cyclist on 30 Aug 2019
The mean and standard deviation of f1, for the elements where 50 < f2 <= 100, can be calculated as follows:
arrayfun(@(x)mean(x.f1(x.f2>50 & x.f2<=100)),S);
arrayfun(@(x) std(x.f1(x.f2>50 & x.f2<=100)),S);
arrayfun will apply a function to each element of a structure array, so it is a vectorized way of doing the operation.
the cyclist on 31 Aug 2019
For a length-L structure, I thought you wanted the L means and standard deviations, one for each element in the structure array. The reason there were some NaNs is presumably that sometime f2 didn't have any elements that fit the condition.
Looks like Matt J got you what you wanted.