迭代后过滤器对象变空? [英] filter object becomes empty after iteration?

问题描述

我正在学习如何使用 filter 函数.

这是我写的代码:

people = [{'name': 'Mary', 'height': 160},{'name': 'Isla', 'height': 80},{'姓名':'山姆'}]people2 = filter(lambda x: x 中的高度"，人)

如您所见，我正在尝试删除所有不包含 'height' 键的字典.

代码工作正常，事实上如果我这样做:

print(list(people2))

我明白了:

[{'name': 'Mary', 'height': 160}, {'name': 'Isla', 'height': 80}]

问题是如果我做两次:

print(list(people2))打印(列表(人 2))

第二次，我得到一个空列表.

你能解释一下为什么吗?

解决方案

这是一个经典的python3 doh！

过滤器是一种特殊的可迭代对象，您可以对其进行迭代.然而，就像一个生成器，你只能迭代它一次.因此，通过调用 list(people2)，您将迭代 filter 对象的每个元素以生成 list.在这一点上，您已经到达了可迭代对象的末尾，没有什么可以返回了.

因此，当您再次调用 list(people2) 时，您会得到一个空列表.

演示:

<预><代码>>>>l = 范围(10)>>>k = 过滤器(λ x: x > 5, l)>>>列表(k)[6, 7, 8, 9]>>>列表(k)[]

我应该提到，对于 python2，filter 返回一个列表，所以你不会遇到这个问题.当你把py3的惰性求值带入画面时，问题就出现了.

I'm learning how to use the filter function.

This is the code I've written:

people = [{'name': 'Mary', 'height': 160},
          {'name': 'Isla', 'height': 80},
          {'name': 'Sam'}]

people2 = filter(lambda x: "height" in x, people)

As you can see what I'm trying to do is to remove all the dictionaries that don't contain the 'height' key.

The code works properly, in fact if I do:

print(list(people2))

I get:

[{'name': 'Mary', 'height': 160}, {'name': 'Isla', 'height': 80}]

The problem is that if I do it twice:

print(list(people2))
print(list(people2))

the second time, I get an empty list.

Can you explain me why?

解决方案

This is a classic python3 doh!.

A filter is a special iterable object you can iterate over. However, much like a generator, you can iterate over it only once. So, by calling list(people2), you are iterating over each element of the filter object to generate the list. At this point, you've reached the end of the iterable and nothing more to return.

So, when you call list(people2) again, you get an empty list.

Demo:

>>> l = range(10)
>>> k = filter(lambda x: x > 5, l)
>>> list(k)
[6, 7, 8, 9]
>>> list(k)
[]

I should mention that with python2, filter returns a list, so you don't run into this issue. The problem arises when you bring py3's lazy evaluation into the picture.

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