Length of the structure array
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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=............;
.
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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?
Thanks, in advance.
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Accepted Answer
Walter Roberson
on 10 Jun 2020
count = nnz(arrayfun(@(S) all(10 < S.f1 & S.f1 <= 20), S))
3 Comments
Walter Roberson
on 10 Jun 2020
mask = arrayfun(@(s) all(10 < s.f1 & s.f1 <= 20), S);
count = nnz(mask);
selected_f1_element_counts = cellfun(@numel, {S(mask).f1});
selected_f2 = {S(mask).f2};
selected_f2_means = cellfun(@mean, selected_f2);
selected_f2_variances = cellfun(@var, selected_f2);
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