有没有更优雅的方式来编写这些循环？ [英] Is there a more elegant way to write these loops?

问题描述

我有一个脚本，需要少量的参数运行。我有兴趣探索结果作为参数改变，所以我定义了一些 scan 数组在顶部，包装整个代码在多个 loops 并将参数值设置为当前扫描值。

I have a script that requires a handful of parameters to run. I'm interested in exploring the results as the parameters change, so I define a few scan arrays at the top, wrap the whole code in multiple for loops and set the parameters values to the current scan values.

这是容易出错和不雅的。更改代码的过程是：1）重置扫描顶部的变量，2）注释例如 b = scan2（j2）和3）取消注释 b = b0 。

This is error prone and inelegant. The process for changing the code is: 1) reset scan variables at the top, 2) comment out eg b = scan2(j2) and 3) uncomment b=b0.

什么是更好的方法，允许变量设置为数组，然后运行所有这样的组合的代码？ ：

close 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

推荐答案

基于这个想法使用一个for循环来模拟多个循环，我试着适应你的情况。在实现良好的内存效率和可用性的同时，此解决方案比使用个别循环更慢。

Based on this idea to use one for loop to simulate multiple loops, I tried to adapt it to your case. While fulfilling a good memory efficiency and usability, this solution is slower than using individual for loops.

%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，它增加了 0.0002 每次迭代的计算开销， 0.0002。* prod（iterations） code>总计

For the scenarios I tested it adds an computation overhead below 0.0002s per iteration which is 0.0002.*prod(iterations)s in total.

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