python3中map和list迭代器之间的区别 [英] Difference between map and list iterators in python3

问题描述

在python3中使用map和list迭代器时遇到了意外的行为。在这个MWE中，我首先生成一张地图图。然后，我希望每个地图的第一个元素在一个列表中，其余部分在原始地图中：

I ran into unexpected behaviour when working with map and list iterators in python3. In this MWE I first generate a map of maps. Then, I want the first element of each map in one list, and the remaining parts in the original map:

# s will be be a map of maps
s=[[1,2,3],[4,5,6]]
s=map(lambda l: map(lambda t:t,l),s)

# uncomment to obtain desired output
# s = list(s) # s is now a list of maps

s1 = map(next,s)

print(list(s1))
print(list(map(list,s)))

在python 3.4.2中运行MWE会产生s1的预期输出：

Running the MWE as is in python 3.4.2 yields the expected output for s1:

s1 = ([1,4]),

但s的空列表[]。取消注释标记的行会产生正确的输出，如上所述s1，但也有s的预期输出：

but the empty list [] for s. Uncommenting the marked line yields the correct output, s1 as above, but with the expected output for s as well:

s=[[2,3],[5,6]].

文档说地图需要迭代。直到今天，我看到map和list迭代器之间没有区别。有人可以解释这种行为吗？

The docs say that map expects an iterable. To this day, I saw no difference between map and list iterators. Could someone explain this behaviour?

PS：奇怪的是，如果我取消注释第一个print语句，则会打印s的初始状态。所以也可能是这种行为与地图的一种懒惰（？）评估有关？

PS: Curiously enough, if I uncomment the first print statement, the initial state of s is printed. So it could also be that this behaviour has something to do with a kind of lazy(?) evaluation of maps?

推荐答案

A map（）是一个迭代器;你只能迭代一次。例如，您可以使用 next（）获取单个元素，但是一旦用完了项目，就无法获得更多值。

A map() is an iterator; you can only iterate over it once. You could get individual elements with next() for example, but once you run out of items you cannot get any more values.

我给你的对象一些更容易记住的名字：

I've given your objects a few easier-to-remember names:

>>> s = [[1, 2, 3], [4, 5, 6]]
>>> map_of_maps = map(lambda l: map(lambda t: t, l), s)
>>> first_elements = map(next, map_of_maps)

迭代 first_elements 这里将依次迭代 map_of_maps 。你只能一次，所以一旦我们用完了元素，任何进一步的迭代都会失败：

Iterating over first_elements here will in turn iterate over map_of_maps. You can only do so once, so once we run out of elements any further iteration will fail:

>>> next(first_elements)
1
>>> next(first_elements)
4
>>> next(first_elements)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration

list（）做同样的事情;它需要一个可迭代的参数，并将迭代该对象以从结果中创建一个新的列表对象。但是如果你给它一个已经用尽的 map（） ，就没有任何东西可以复制到新的列表了。因此，您会得到一个空结果：

list() does exactly the same thing; it takes an iterable argument, and will iterate over that object to create a new list object from the results. But if you give it a map() that is already exhausted, there is nothing to copy into the new list anymore. As such, you get an empty result:

>>> list(first_elements)
[]

您需要重新创建 map（）从头开始：

You need to recreate the map() from scratch:

>>> map_of_maps = map(lambda l: map(lambda t: t, l), s)
>>> first_elements = map(next, map_of_maps)
>>> list(first_elements)
[1, 4]
>>> list(first_elements)
[]

请注意秒 list（）再次调用 map（）对象产生一个空列表对象。

Note that a second list() call on the map() object resulted in an empty list object, once again.

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