`sorted(list)` 与 `list.sort()` 有什么区别?
问题描述
list.sort() 对列表进行排序并替换原始列表，而 sorted(list) 返回列表的排序副本，而不更改原始列表.
list.sort() sorts the list and replaces the original list, whereas sorted(list) returns a sorted copy of the list, without changing the original list.
- 什么时候比另一个更受青睐?
- 哪个更有效率?多少?
list.sort()执行后列表能否恢复为未排序状态?
- When is one preferred over the other?
- Which is more efficient? By how much?
- Can a list be reverted to the unsorted state after
list.sort()has been performed?
解决方案
sorted() 返回一个 new 排序列表，不影响原始列表.list.sort() 对列表进行 就地 排序，改变列表索引，并返回 None(与所有就地操作一样).
sorted() returns a new sorted list, leaving the original list unaffected. list.sort() sorts the list in-place, mutating the list indices, and returns None (like all in-place operations).
sorted() 适用于任何可迭代对象，而不仅仅是列表.字符串、元组、字典(你会得到键)、生成器等，返回一个包含所有元素的列表，排序.
sorted() works on any iterable, not just lists. Strings, tuples, dictionaries (you'll get the keys), generators, etc., returning a list containing all elements, sorted.
当你想要改变列表时使用
list.sort()，当你想要一个新的排序对象返回时使用sorted().sorted() 当你想对一个可迭代的东西进行排序时，使用 还.
Use
list.sort()when you want to mutate the list,sorted()when you want a new sorted object back. Usesorted()when you want to sort something that is an iterable, not a list yet.
对于列表，list.sort() 比 sorted() 快，因为它不必创建副本.对于任何其他可迭代对象，您别无选择.
For lists, list.sort() is faster than sorted() because it doesn't have to create a copy. For any other iterable, you have no choice.
不，您无法检索原始位置.一旦你调用了 list.sort()，原来的顺序就消失了.
No, you cannot retrieve the original positions. Once you called list.sort() the original order is gone.
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