Run a function inside a for loop 10,000 times and record the 3 separate outputs for each trial

조회 수: 2 (최근 30일)
I'm very new to MATLAB and I'm trying to import a function and run it 10,000 times inside a for loop.
The function produces 3 separate outputs and I need to record each output from each trial in an array so that I can then pull the mean value from each array. However I cannot get it to work. Here is for loop I have currently.
Please help.
for trials = 1:10000
[S,M,L] = crayonBreak();
Sl = [Sl; S];
Ml = [Ml; M];
Ll = [Ll; L];
end

답변 (2개)

VBBV
VBBV 2022년 11월 28일
Sl(trials) = S;
Ml(trials) = M;
Ll(trials) = L;
  댓글 수: 3
Kate
Kate 2022년 11월 28일
When I try that it still only shows one value for Sl, Ml, and Ll. I need it to store all 10,000 for each.
VBBV
VBBV 2022년 11월 28일
I need to record each output from each trial in an array so that I can then pull the mean value from each array
As mentioned in your question, mean will give only value for any length of array. Do you look at Sl, Ml, Ll arrays below ? they are 1 x 10000 it implies all 10000 values are stored in those arrays as shown below
for trials = 1:10000
[S,M,L] = crayonBreak();
Sl(trials) = S;
Ml(trials) = M;
Ll(trials) = L;
end
Sl, Ml, Ll,
Sl = 1×10000
0.6659 0.6360 0.1235 0.2869 0.8678 0.8752 0.2682 0.3422 0.1486 0.7197 0.4954 0.7328 0.3132 0.1744 0.7686 0.2773 0.2526 0.6079 0.7764 0.9991 0.8646 0.9189 0.0676 0.4408 0.2586 0.1732 0.3026 0.3628 0.3977 0.1412
Ml = 1×10000
0.6169 0.9453 0.6292 0.7253 0.2442 0.9436 0.7622 0.3416 0.7905 0.5936 0.7303 0.8699 0.3346 0.3015 0.8931 0.6656 0.1587 0.0749 0.9722 0.6737 0.6742 0.2671 0.5881 0.4916 0.3278 0.3150 0.0546 0.2817 0.9364 0.3957
Ll = 1×10000
0.9494 0.9364 0.2546 0.4788 0.3378 0.0563 0.8846 0.2533 0.4457 0.5506 0.6268 0.2566 0.4805 0.3051 0.2895 0.0662 0.4002 0.2488 0.2092 0.6633 0.6558 0.5478 0.6782 0.9287 0.8313 0.2034 0.0531 0.5057 0.3346 0.1618
mean(Sl)
ans = 0.5024
mean(Ml)
ans = 0.5034
mean(Ll)
ans = 0.5003
function [S,M,L] = crayonBreak()
S = rand(1);
M = rand(1);
L = rand(1);
end

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Stephen23
Stephen23 2022년 11월 29일
편집: Stephen23 2022년 11월 29일
Assuming that each output is a scalar double:
N = 1e4;
S = nan(1,N);
M = nan(1,N);
L = nan(1,N);
for k = 1:N
[S(k),M(k),L(k)] = crayonBreak();
end
Now lets calculate the mean of one of those vectors:
display(S)
S = 1×10000
0.7019 0.0534 0.8125 0.5301 0.4659 0.2806 0.3872 0.4478 0.5410 0.4365 0.1543 0.6310 0.8740 0.9585 0.0317 0.5657 0.8402 0.4991 0.9272 0.1651 0.6194 0.1985 0.2538 0.7495 0.7372 0.7060 0.3727 0.0894 0.7250 0.6181
mean(S)
ans = 0.4969
function [S,M,L] = crayonBreak()
S = rand(1);
M = rand(1);
L = rand(1);
end

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