Sort a list by the number of occurrences of the elements in the list [duplicate]

Question

I want to sort a list by the number of occurrences of the elements in the list.
When I use this form:

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

the result is not what I want: [2, 1, 3, 4, 2, 2, 3].
But, when I write like it using sorted:

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

the result is right: [1, 4, 3, 3, 2, 2, 2].
what's the reason for this behavior?

Answer 1

This is by design and intentional. CPython temporarily "disallows" access to the list while the list is being sorted in place, the behavior is documented here:

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

You can inspect that by printing A inside the key function - you'll get an empty list:

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) 

But, if you do that for sorted():

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] 

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It is also documented inside the sort() implementation:

/* 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).  */.