How to remove duplicate digits from integers in list

A list is mutable; in Python mutable containers are not hashable. set(names) needs to hash the elements of names to sort them but your names list has list as it's elements (["cat", 9112, "dog123", 5625] and ["luck", 1232, "bad23"]) and hence, it can't be converted to a set.

Try this:

names = [ ["cat", 9112, "dog123", 5625], ["luck", 1232, "bad23"] ]

li = [[x for x in y if isinstance(x, int)] for y in names]
final = [["".join(sorted(set(str(x)), key=str(x).index)) for x in y] for y in li]
print(li)
print(final)

It gives the following output:

[[9112, 5625], [1232]] 
[['912', '562'], ['123']] 

EDIT:

This solution will give the desired result. It may not be best and optimal solution and OP hasn't mentioned anything related to performance.

names = [ ["cat", 9112, "dog123", 5625], ["luck", 1232, "bad23"],["123"] ]
updated_name=[]
for n_list in names:
    undated_n_list=[]
    for n in n_list:
        if type(n)==int:
            new_str = []
            for digit in str(n):
                if digit not in new_str:
                    new_str.append(digit)
            undated_n_list.append(int("".join(map(str, new_str))))
    if undated_n_list:
        updated_name.append(undated_n_list)
print(updated_name)

Output:

[[912, 562], [123]]

It is bit lengthy but hopefully it works for you.

Here's a function to turn integers into ones with unique digits:

def to_uniq_digit_int(n):
      seen = set() # A set that collects seen digits
      result = 0
      for i in str(n): # A lazy way to iterate over digits in an integer
          if i not in seen:
              seen.add(i)
              # Since we are iterating from the most significant to the least significant, we can multiply the result by ten each time to move the integer one digit left
              result = result * 10 + int(i)
      return result

Using a helper function may help with readability of your code.