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将迭代器转换为列表的最快方法

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问题：将迭代器转换为列表的最快方法
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回答 1

问题：将迭代器转换为列表的最快方法

有一个iterator对象，是否有比列表理解更快，更好或更正确的方法来获取迭代器返回的对象的列表？

user_list = [user for user in user_iterator]
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Having an iterator object, is there something faster, better or more correct than a list comprehension to get a list of the objects returned by the iterator?

user_list = [user for user in user_iterator]

回答 0

list(your_iterator)
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list(your_iterator)

回答 1

从python 3.5开始， 您可以使用*可迭代的拆包运算符：

user_list = [*your_iterator]

但 pythonic的方法是：

user_list  = list(your_iterator)
点击查看英文原文

since python 3.5 you can use * iterable unpacking operator:

user_list = [*your_iterator]

but the pythonic way to do it is:

user_list  = list(your_iterator)