best way to compare contents of 2 lists?

[Terry Reedy]

Steven D'Aprano wrote:
> On Thu, 23 Apr 2009 21:51:42 -0400, Esmail wrote:
> 
>>> set(a) == set(b)    # test if a and b have the same elements
>>>
>>> # check that each list has the same number of each element # i.e.   
>>> [1,2,1,2] == [1,1,2,2], but [1,2,2,2] != [1,1,1,2] for elem in set(a):
>>>   a.count(elem) == b.count(elem)
>> Ah .. this part would take care of different number of duplicates in the
>> lists. Cool.
> 
> At significant cost of extra work.
> 
> Counting the number of times a single element occurs in the list is O(N). 
> Counting the number of times every element occurs in the list is O(N**2). 

A frequency dict should be O(n) also, and hence faster than sorting.

> Sorting is O(N*log N), so for large lists, sorting will probably be much 
> cheaper.
> 
>