Length of the structure array

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SS
SS il 9 Giu 2020
Commentato: SS il 10 Giu 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.

Risposta accettata

Walter Roberson
Walter Roberson il 10 Giu 2020
count = nnz(arrayfun(@(S) all(10 < S.f1 & S.f1 <= 20), S))
  3 Commenti
Walter Roberson
Walter Roberson il 10 Giu 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 il 10 Giu 2020
Thank you.

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