在列表中查找元素，以使元素不应具有相似的数字

Question

Consider a list = [23,52,44,32,78] 23,52,32 all these elements have at least one digit common so the set i want to filter is [44,78] as they don't have any common numbers.

Another example: [52,12,255,211,223,123,64,87,999] would be filtered as [64,87,999]

My current idea is to convert all the numbers into list like [2,3],[5,2]... and take a intersection of those but i couldn't understand how to compare all these sub-lists and filter out the desired numbers.

def convert_into_sublist(i):
    sublist = [int(x) for x in str(i)] 

def intersection(l1, l2): 
    l3 = [value for value in l1 if value in l2]
    if(len(l3)==0):
        return 1

Answer 1

handle the numbers as a list of characters and use set conversion and test if disjoint. Maybe not the most performant one-liner but works:

lst = [23,52,44,32,78]
# optional: remove duplicates:
lst = set(lst)

unique = [l for l in lst if all(set(str(x)).isdisjoint(str(l)) for x in lst if x != l)]

result:

>>> unique
[44, 78]

maybe slightly faster: convert to string once, process strings, convert back to integers in the end:

lst = [str(x) for x in lst]
unique = [int(l) for l in lst if all(set(x).isdisjoint(l) for x in lst if x != l)]

Answer 2

You can use Counter to find non-common digits:

from collections import Counter
from itertools import chain
from operator import itemgetter

lst = [23, 52, 44, 32, 78]

sets = [set(str(i)) for i in lst]
# [{'3', '2'}, {'5', '2'}, {'4'}, {'3', '2'}, {'7', '8'}]

c = Counter(chain.from_iterable(sets))
# Counter({'2': 3, '3': 2, '5': 1, '4': 1, '7': 1, '8': 1})

new_lst = []
for num, set_ in zip(lst, sets):
    counts = itemgetter(*set_)(c)
    if counts == 1 or set(counts) == {1}:
        new_lst.append(num)

print(new_lst)
# [44, 78]