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replacing loop with cell of index values

2 vues (au cours des 30 derniers jours)
Catherine
Catherine le 16 Jan 2021
Commenté : Catherine le 16 Jan 2021
I don't know how to explain with words what I'm trying to do, so here is an example
%I have an array of values
s = [1:100];
% and I have a cell array of different index values
c{1} = 1;
c{2} = 1:5;
c{3} = 2:2:10;
% I would like a cell array t, where
t{1} = s(c{1});
t{2} = s(c{2});
% etc.
% I can do this with a loop, but I'm wondering if there is a way to do this without a loop as my set of indices are large.

Réponse acceptée

Stephen23
Stephen23 le 16 Jan 2021
Modifié(e) : Stephen23 le 16 Jan 2021
s = 1:100;
c{1} = 1;
c{2} = 1:5;
c{3} = 2:2:10;
out = cellfun(@(x)s(x),c,'uni',0);
out{:}
ans = 1
ans = 1×5
1 2 3 4 5
ans = 1×5
2 4 6 8 10
A well-written (i.e. correctly preallocated, etc.) loop will be faster.
  1 commentaire
Catherine
Catherine le 16 Jan 2021
Thank you. Good to know that the loop might be faster.

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