Length of the structure array

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SS
SS 2020 年 6 月 9 日
コメント済み: SS 2020 年 6 月 10 日
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.

採用された回答

Walter Roberson
Walter Roberson 2020 年 6 月 10 日
count = nnz(arrayfun(@(S) all(10 < S.f1 & S.f1 <= 20), S))
  3 件のコメント
Walter Roberson
Walter Roberson 2020 年 6 月 10 日
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 2020 年 6 月 10 日
Thank you.

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