Length of the structure array

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SS
SS el 9 de Jun. de 2020
Comentada: SS el 10 de Jun. de 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?
Thanks, in advance.

Respuesta aceptada

Walter Roberson
Walter Roberson el 10 de Jun. de 2020
count = nnz(arrayfun(@(S) all(10 < S.f1 & S.f1 <= 20), S))
  3 comentarios
Walter Roberson
Walter Roberson el 10 de Jun. de 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);
SS
SS el 10 de Jun. de 2020
Thank you.

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