python - 按列表中元素的出现次数对列表进行排序

Question

这个问题在这里已经有了答案:

Accessing the list while being sorted (2 个回答)

关闭 3年前。

我想按列表中元素出现的次数对列表进行排序。
当我使用这种形式时:

A=[2,1,3,4,2,2,3]
A.sort(key=lambda x:A.count(x))  
print(A)

结果不是我想要的:[2, 1, 3, 4, 2, 2, 3].
但是，当我使用 sorted 写出这样的内容时:

B=sorted(A,key=lambda x:A.count(x))
print(B)

结果是对的:[1, 4, 3, 3, 2, 2, 2].
这种行为的原因是什么？

Answer 1

这是设计和故意的。 CPython 暂时“禁止”访问列表，而列表正在就地排序，行为是 documented here :

CPython implementation detail: While a list is being sorted, theeffect of attempting to mutate, or even inspect, the list isundefined. The C implementation of Python makes the list appear emptyfor the duration, and raises ValueError if it can detect that the listhas been mutated during a sort.

您可以通过在 key 函数中打印 A 来检查 - 您将得到一个 空列表:

In [2]: def key_function(x):
    ...:     print(A, x)
    ...:     return A.count(x)
    ...: 

In [3]: A.sort(key=key_function)  
([], 2)
([], 1)
([], 3)
([], 4)
([], 2)
([], 2)
([], 3)

但是，如果你为 sorted() 这样做:

In [4]: sorted(A, key=key_function)
([2, 1, 3, 4, 2, 2, 3], 2)
([2, 1, 3, 4, 2, 2, 3], 1)
([2, 1, 3, 4, 2, 2, 3], 3)
([2, 1, 3, 4, 2, 2, 3], 4)
([2, 1, 3, 4, 2, 2, 3], 2)
([2, 1, 3, 4, 2, 2, 3], 2)
([2, 1, 3, 4, 2, 2, 3], 3)
Out[4]: [1, 4, 3, 3, 2, 2, 2]

---

它也记录在 sort() implementation 中。 :

/* The list is temporarily made empty, so that mutations performed
 * by comparison functions can't affect the slice of memory we're
 * sorting (allowing mutations during sorting is a core-dump
 * factory, since ob_item may change).
 */.