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

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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.

Accepted Answer

Matt J
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);
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More Answers (1)

the cyclist
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.
  3 Comments
SS
SS on 31 Aug 2019
Edited: SS on 31 Aug 2019
Thanks for the reply. I have tried using the above lines but, unfortunately it creates another structure of size 1 X N in which, some entries are numeric and others are NaN. I have used mean(new created structure) to see whether, it will give one single answer but, it gives NaN as the answer.
How can I get one single (explicit) mean and standard deviation?
the cyclist
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.

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