争论次数可变的Python itertools.product [英] Python itertools.product with variable number of arguements

问题描述

我正在尝试编写一个模块，以使用itertools.product组合可变数量的列表.

I am trying to write a module to combine a variable number of lists using itertools.product.

我能得到的最接近的是:

The closest I can get is:

import itertools
lists = [["item1","item2"],["A","b","C"], ["etc..."]]
searchterms = list(itertools.product(lists))
print searchterms

这不起作用，因为列表是单个列表，所以它只返回原始序列.但是我不知道如何将列表变量的每个元素传递给itertools.

This doesn't work, because lists is a single list, so it just returns the original sequence. But I can't figure out how to pass each element of the lists variable to itertools.

谢谢您的建议.

推荐答案

您需要使用*将单个列表分成其组成列表:

You need to use * to separate the single list into its constituent lists:

searchterms = list(itertools.product(*lists))

请参见解压缩参数列表上的Python教程部分.

See the Python Tutorial section on Unpacking Argument Lists.

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