Python 2 与 Python 3 - 具有三个参数的地图行为差异? [英] Python 2 vs Python 3 - Difference in map behavior with three arguments?

问题描述

以下代码在 Python 2 和 Python 3 中的行为不同:

The following code behaves differently in Python 2 vs Python 3:

all(map(lambda x,y: x, [1, 2], [1, 2, 3]))

Python 2 给出 False 而 Python 3 给出 True.Python 2 的 文档 说它将提供 None 如果较短的列表已用尽但 Python 3 没有 这样做.

Python 2 gives False whereas Python 3 gives True. The documentation for Python 2 says that it will supply None if the shorter list is exhausted but Python 3 doesn't do that.

我正在编写一个出于某种原因确实需要保持长度的代码.获得旧行为的最干净的方法是什么?我知道我可以使用 from past.builtin import map as old_map，但是有没有更优雅的解决方案可以同时在两个版本中使用?

I am working on a code that really needs the length to be maintained for some reason. What is the cleanest way to get the old behavior? I know I can use from past.builtin import map as old_map, but is there a more elegant solution that would work in both versions?

推荐答案

本质上，map 为参数设置多个可迭代对象将 zip 可迭代对象，然后调用函数将 zip 中的元组作为 var-args.因此，您可以使用 itertools.starmap<获得相同的行为/code> 和 zip:

Essentially, map with multiple iterables for the arguments will zip the iterables, and then call the function with the tuples from the zip as var-args. So, you can get the same behaviour using itertools.starmap and zip:

>>> a = [10, 20]
>>> b = [1, 2, 3]
>>> f = lambda x, y: x
>>> list(map(f, a, b))
[10, 20]
>>> from itertools import starmap
>>> list(starmap(f, zip(a, b)))
[10, 20]

然后可以通过将 zip 替换为 itertools.zip_longest:

Then the behaviour you want can be achieved by replacing zip with itertools.zip_longest:

>>> from itertools import starmap, zip_longest
>>> list(starmap(f, zip_longest(a, b)))
[10, 20, None]

itertools 中的两个函数也存在于 Python 2 中，除了第二个函数名为 izip_longest 代替.你可以只 import ... as ... 来解决这个问题.

Both functions from itertools also exist in Python 2, except the second one is named izip_longest instead. You can just import ... as ... to get around that.

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