# Length of the structure array

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SS on 9 Jun 2020
Commented: SS on 10 Jun 2020
Hi. I am working with a structure array S (1 X 20,000) with 3 fields. I want to count the number of S(i) that meet a condition on one of its fields. For example,
Here is the input,
S(1).f1=[11,17,3,18,15,13], S(1).f2=[100,20,50,60,70,140] and S(1).f3=[-10,20,-50,42,-70,140] ;
S(2).f1=[10,12,14,17,19], S(2).f2=[101,54,69,20,11] and S(2).f3=[17,-54,69,-20,37];
S(3).f1=S(1).f1=[19,17,13,14,15,10,11,16], S(3).f2=..... and S(3).f3=...........;
S(4).f1=[11,17,30,108,15,13,37,14], , S(4).f2=..... and S(4).f3=............;
.
.
S(i).f1=...., S(i).f2=.... and S(i).f3=............;
Let's say, I have a condition on f1: 10 < f1 <=20. Based, on this condition I want the count of S(i) whose f1 is strictly in the these limits. In this example, S(2).f1 and S(3).f1 has all the f1 in bewteen 10 and 20, the count is 2.
I want to implement this on S (1 X 20,000). Can someone help me with this?

Walter Roberson on 10 Jun 2020
count = nnz(arrayfun(@(S) all(10 < S.f1 & S.f1 <= 20), S))

SS on 10 Jun 2020
Thank you so much. How, can I find the number of elements in S(2).f1 and S(3).f1? I am also interested in calculating the mean and variance of the corresponding f2 elements meeting the above condition? Meaning, mean and variance of S(2).f2 and S(3).f2?
Any help is highly appreciated!
Walter Roberson on 10 Jun 2020
mask = arrayfun(@(s) all(10 < s.f1 & s.f1 <= 20), S);
selected_f2_means = cellfun(@mean, selected_f2);
selected_f2_variances = cellfun(@var, selected_f2);
SS on 10 Jun 2020
Thank you.