首先，让我们创建一个虚拟函数，以便正确测试

crayonBreak = @()deal(rand,rand,rand);  %This is just a function that will generate three outputs.

原始代码在每次循环时覆盖Sl值，因此最终值是crayonBreal的最后结果加上数字10000。

for trials = 1:10000
    [S,M,L] = crayonBreak(); % function to run
    Sl = [S; trials]; % list for all S values
    Ml = [M; trials]; % list for all M values
    Ll = [L; trials]; % list for all L values
end

%>> Sl
%Sl =
%         0.477917969692359
%                     10000

这个代码的最小的调整是在每次循环中将最近的结果和之前的结果连接起来。看起来像这样。

Sl=[];  %Iniitalize the outputs to an emmpty array
Ml=[];
Ll=[];
for trials = 1:10000
    [S,M,L] = crayonBreak(); % function to run
    Sl = [Sl; S]; % At leash loop, concatenate the new value to the growing array
    Ml = [Ml; M]; % 
    Ll = [Ll; L]; % 
end

%>> Sl
%ans =
%         0.920537104151778
%          0.83184440865223
%         0.567461009088077
%         ...

一个更好的实现是首先预分配输出，然后在执行时插入结果。

nTrials = 10000;
Sl=nan(nTrials,1);  %Iniitalize the outputs to a properly sized array of NaNs
Ml=nan(nTrials,1);
Ll=nan(nTrials,1);
for trials = 1:10000
    [S,M,L] = crayonBreak(); % function to run
    Sl(trials) = S; % At leash loop, insert the value into the output
    Ml(trials) = M; % 
    Ll(trials) = L; % 
end

%>> Sl(1:5)
%ans =
%         0.779202300519089
%         0.510303051476698
%         0.415940683606639
%         ...