Python中的参数化嵌套循环 [英] Parametric nested loops in Python

问题描述

模型中的一个参数控制数组的尺寸；它可以从1到任何正整数之间变化:就我的目的而言，最大为20.
程序的流程经过多个循环，具体取决于此维度.

One parameter in my model controls the dimension of arrays; it can vary from 1 to any positive integer: for my purposes this can be up to 20.
The flow of the program goes through a number of loops that depend on this dimension.

例如，如果参数的值为1，我将拥有:

For example, if the value of the parameter is one I would have:

for i1 in range(0,100,1):
   do stuff

如果参数的值为2，那么我将得到类似的信息:

If the value of the parameter is two, then I would have something like:

for i1 in range (0,100,1):
    for i2 in range (0,100,1):
        do stuff 

或者，如果参数的值为3，我将拥有:

Or, if the value of the parameter is three I would have:

for i1 in range (0,100,1):
    for i2 in range (0,100,1):
        for i3 in range (0,100,1):
            do stuff

维度可以更改，因此无法预先指定需要多少个嵌套循环；这必须以某种参数化方式来编写.

The dimension can change, so it's not possible to specify in advance how many nested loops will be needed; this has to written in some parametric way.

推荐答案

由于您没有在中间循环中列出任何处理-仅在最内部的循环中，我认为您真正需要的是序列的迭代器索引:

Since you've listed no processing in the intermediate loops -- only in the innermost loop, I feel that what you really need is an iterator for your sequence of indices:

让max_dim为空间的维数，即维数.

Let max_dim be the dimensionality of your space, the quantity of dimensions.

max_val = 100
one_dim = list(range(max_val))
all_dim = [one_dim] * max_val

all_dim现在是一个列表列表，每个维度一个.每个列表包含值0-99，这是嵌套循环使用的值.现在从itertools开始

all_dim is now a list of lists, one for each dimension. Each list contains the values 0-99, the very values your nested loops are using. Now for the magic steps from itertools:

from itertools import product
for index_list in product(*all_dim):
    # do your stuff; the index_list is [i1, i2, i3, ...]

这将迭代您想要的尺寸.举一个小例子，这是product序列在只有两个值和三个维度的情况下的样子:

This will iterate through your desired dimensions. For a small example, here's how the product sequence looks with only two values and three dimensions:

>>> all_dim = [[0,1]] * 3
>>> all_dim
[[0, 1], [0, 1], [0, 1]]
>>> list(product(*all_dim))
[(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)]

这样可以很好地解决您的问题吗?

Does that handle your problem well enough?

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