How to check if all elements of 1 list are in the *same quantity* and in any order, in the list2?

Question

I know its a very common question at first, but I haven't found one that specific. (If you do, please tell me.) And all ways I found didnt work for me. I need to check if all elements of list 1 appears in the same amount in the list2.

Ex :

#If list1 = [2,2,2,6]    
# and list2 =[2,6,2,5,2,4]     
#then all list1 are in list2.
#If list2 = [2,6] then all list1 are not in list2.

i'm trying this way :

list1 = [6,2]

import itertools

for i in itertools.product((2,4,5,1), repeat=3) :
    asd = i[0] + i[1]
    asd2= i[1] + i[2]

    list2 = [asd, asd2]
    if all(elem in list2  for elem in list1):
        print (i,list2)

It works when the elements are not repeated in the list1, like [1,2]. But when they are repeated, all repeated elements is beeing counted as only 1 : [2,2,2] its beeing understanded as [2]. Or so i think.

Answer 1

Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.

from collections import Counter

def a_all_in_b(a, b):
    """True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""
    return Counter(a).items() <= Counter(b).items()

Note that Counter works on hashable elements only, because it's a subclass of dict.