python 列表推导式；压缩列表列表? [英] python list comprehensions; compressing a list of lists?

问题描述

伙计们.我正在尝试为问题找到最优雅的解决方案，并想知道 python 是否为我正在尝试做的事情内置了任何东西.

guys. I'm trying to find the most elegant solution to a problem and wondered if python has anything built-in for what I'm trying to do.

我正在做的是这个.我有一个列表，A，还有一个函数 f，它接受一个项目并返回一个列表.我可以使用列表理解来转换 A 中的所有内容；

What I'm doing is this. I have a list, A, and I have a function f which takes an item and returns a list. I can use a list comprehension to convert everything in A like so;

[f(a) for a in A]

但这会返回一个列表列表；

But this return a list of lists;

[a1,a2,a3] => [[b11,b12],[b21,b22],[b31,b32]]

我真正想要的是得到扁平化的列表；

What I really want is to get the flattened list;

[b11,b12,b21,b22,b31,b32]

现在，其他语言都有了；它在函数式编程语言中传统上称为 flatmap，而 .Net 将其称为 SelectMany.python有类似的吗?有没有一种巧妙的方法可以将函数映射到列表上并将结果展平?

Now, other languages have it; it's traditionally called flatmap in functional programming languages, and .Net calls it SelectMany. Does python have anything similar? Is there a neat way to map a function over a list and flatten the result?

我试图解决的实际问题是这样的；从目录列表开始，找到所有子目录.所以;

import os
dirs = ["c:\usr", "c:\temp"]
subs = [os.listdir(d) for d in dirs]
print subs

currentliy 给了我一个列表列表，但我真的想要一个列表.

推荐答案

您可以在单个列表推导式中嵌套迭代:

You can have nested iterations in a single list comprehension:

[filename for path in dirs for filename in os.listdir(path)]

相当于(至少在功能上):

which is equivalent (at least functionally) to:

filenames = []
for path in dirs:
    for filename in os.listdir(path):
        filenames.append(filename)

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