Matlab 有没有更优雅的方法来编写这些循环？
我有一个脚本，需要几个参数才能运行。我感兴趣的是随着参数的变化探索结果，因此我在顶部定义了几个Matlab 有没有更优雅的方法来编写这些循环？,matlab,coding-style,Matlab,Coding Style,我有一个脚本，需要几个参数才能运行。我感兴趣的是随着参数的变化探索结果，因此我在顶部定义了几个scan数组，将整个代码包装成多个for循环，并将参数值设置为当前扫描值 这很容易出错，而且不雅观。更改代码的过程是：1）重置顶部的scan变量，2）注释掉例如b=scan2（j2）和3）取消注释b=b0 有什么更好的方法允许将变量设置为数组，然后为所有这些组合运行代码？现在我的代码示例： close all clear all %scan1 = linspace(1,4,10); scan1 = 0
scanfor循环scanb=scan2（j2）b=b0close all
clear all
%scan1 = linspace(1,4,10);
scan1 = 0;
scan2 = linspace(0,1,10);
scan3 = linspace(-1,0,10);
for j3 = 1:length(scan3)
for j2 = 1:length(scan2)
for j1 = 1:length(scan1)
a = a0;
%b = scan2(j2);
b = b0;
%c = c0;
c = scan3(j3);
d = scan2(j2);
%(CODE BLOCK THAT DEPENDS ON variables a,b,c,d...)
end
end
end
一种方法是使用ndgrid生成包含所有参数组合的单个向量。对于足够大的参数扫描，这可能会成为内存问题，但在其他方面至少要干净得多，只需要一个循环，以后在代码中不需要重新分配：
a0vec = 1;
b0vec = linspace(1,4,4);
c0vec = linspace(11,15,5);
d0vec = linspace(101,104,4);
e0vec = 5;
[a0s,b0s,c0s,d0s,e0s] = ndgrid(a0vec,b0vec,c0vec,d0vec,e0vec);
N = numel(a0s);
for j = 1:N
a0 = a0s(j);
b0 = b0s(j);
c0 = c0s(j);
d0 = d0s(j);
e0 = e0s(j);
%__ (CODE THAT DEPENDS ON a0 - e0 here) __
end
还是想看看你的建议 一种方法是使用ndgrid生成包含所有参数组合的单个向量。对于足够大的参数扫描，这可能会成为内存问题，但在其他方面至少要干净得多，只需要一个循环，以后在代码中不需要重新分配：
a0vec = 1;
b0vec = linspace(1,4,4);
c0vec = linspace(11,15,5);
d0vec = linspace(101,104,4);
e0vec = 5;
[a0s,b0s,c0s,d0s,e0s] = ndgrid(a0vec,b0vec,c0vec,d0vec,e0vec);
N = numel(a0s);
for j = 1:N
a0 = a0s(j);
b0 = b0s(j);
c0 = c0s(j);
d0 = d0s(j);
e0 = e0s(j);
%__ (CODE THAT DEPENDS ON a0 - e0 here) __
end
%define your parameters
p.a = 1;
p.b = linspace(1,4,4);
p.c = linspace(11,15,5);
p.d = linspace(101,104,4);
p.e = 5;
iterations=structfun(@numel,p);
iterator=cell(1,numel(iterations));
for jx = 1:prod(iterations)
[iterator{:}]=ind2sub(iterations(:).',jx);%.'
%This line uses itertor to extract the corresponding elemets of p and creates a struct which only contains scalars.
q=cell2struct(cellfun(@(a,b)(a(b)),struct2cell(p),iterator(:),'uniform',false),fieldnames(p));
%__ (CODE THAT DEPENDS ON q.a to q.e here) __
end
0.00020.0002.*prod（iterations）%define your parameters
p.a = 1;
p.b = linspace(1,4,4);
p.c = linspace(11,15,5);
p.d = linspace(101,104,4);
p.e = 5;
iterations=structfun(@numel,p);
iterator=cell(1,numel(iterations));
for jx = 1:prod(iterations)
[iterator{:}]=ind2sub(iterations(:).',jx);%.'
%This line uses itertor to extract the corresponding elemets of p and creates a struct which only contains scalars.
q=cell2struct(cellfun(@(a,b)(a(b)),struct2cell(p),iterator(:),'uniform',false),fieldnames(p));
%__ (CODE THAT DEPENDS ON q.a to q.e here) __
end
对于我测试的场景，每次迭代增加的计算开销低于
0.00020.0002.*prod（迭代）scanscanajajbscanscanajajbpythonpython结构ind2sub（迭代（：）。，j）ind2sub（迭代次数（：）.”，3）结构