Looping with indices that are not equally spaced
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John F
am 23 Jun. 2011
Kommentiert: Walter Roberson
am 22 Dez. 2019
I'm trying to run a loop on a group of indices I obtained using "find". The indices will not always be consecutive. So, running a for loop like:
won't work. I'm trying to avoid doing something like:
for i = 1:length(VECTOR)
Any ideas?
2 Kommentare
Oleg Komarov
am 23 Jun. 2011
not clear why it won't work. Depends how you structure the operations inside the loop. Post more code.
Daniel Shub
am 23 Jun. 2011
What do you mean it doesn't work? What would you expect to get with:
indices = [1,2,3,5,7,13,11];
for i = indices, i, end
Akzeptierte Antwort
Laura Proctor
am 23 Jun. 2011
Actually, it will work.
for idx = [ 1 -2 10 12.5 0 ]
disp(idx)
end
Isn't MATLAB cool?
0 Kommentare
Weitere Antworten (2)
John F
am 23 Jun. 2011
2 Kommentare
Laura Proctor
am 23 Jun. 2011
Bearbeitet: Walter Roberson
am 22 Dez. 2019
You are correct - check out Loren's Blog, it explains this behavior much better than I can:
Daniel Shub
am 23 Jun. 2011
Bearbeitet: Walter Roberson
am 22 Dez. 2019
Yeah, but check out what it does do with a column. You should have a read of:
Frederick Abangba Akendola
am 22 Dez. 2019
Please, how do I write a “For” loop with irregular interval? For example; 2,4,8,16,32
1 Kommentar
Walter Roberson
am 22 Dez. 2019
for K = 2.^(1:5)
result = whatever involving K
end
However, most of the time you want to create one output per input. The general way to do that is
K_vals = 2.^(1:5);
numK = numel(K_vals);
results = zeros(size(K_vals));
for K_idx = 1 : numK
K = K_vals(K_idx);
results(K_id) = whatever involving K
end
plot(K_vals, results)
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