How to remove numbers from a list which contain repeated digits?

user_11877543 0
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I am trying to solve the following problem:

I have a pretty long list of integer numbers in a given range, most of them contain numbers with repeated digits like in the example below.

[123456, 889756, 854123, 997886, 634178]

My goal is to remove the ones with repeated digits or get a new list with numbers having only distinct digits:

[123456, 854123, 634178]

Is there a nice way how to do this? Thank you very much in advance!

list algorithm python
5 Answers

You can use set on the string representation of the number to see how many digits make it into the set. If that is the same number of digits as the original number has, it passes the test:

lst = [123456, 889756, 854123, 997886, 634178]
result = [n for n in lst if len(set(str(n))) == len(str(n))]

print(result)

As commented below, benchmarks confirm that it is advantageous to perform an inline assignment to a temporary variable:

result = [n for n in lst if len(set(s := str(n))) == len(s)]
user_54598390

Another solution, with re:

import re

r = re.compile(r"(\d).*\1")
lst = [123456, 889756, 854123, 997886, 634178]

lst = [i for i in lst if not r.search(str(i))]
print(lst)

Prints:

[123456, 854123, 634178]

EDIT: Small benchmark:

from timeit import timeit


lst = [123456, 889756, 854123, 997886, 634178] * 10000


def re_method(lst):
    r = re.compile(r"(\d).*\1")
    return [i for i in lst if not r.search(str(i))]


def trincot1(lst):
    return [n for n in lst if len(set(str(n))) == len(str(n))]


def trincot2(lst):
    return [n for n in lst if len(set(s := str(n))) == len(s)]


def afaalgo(lst):
    answer = []
    for value in lst:
        value = str(value)
        new_list = []
        for nums in value:
            new_list.append(nums)
        if sorted(list(set(new_list))) == sorted(new_list):
            answer.append(int(value))
    return answer


t1 = timeit(lambda: re_method(lst), number=10)
t2 = timeit(lambda: trincot1(lst), number=10)
t3 = timeit(lambda: trincot2(lst), number=10)
t4 = timeit(lambda: afaalgo(lst), number=10)

print(t1)
print(t2)
print(t3)
print(t4)

Prints on my machine (3700x/Python 3.8.5):

0.2806989410019014
0.33745980000821874
0.263871792005375
0.8039937680005096

So version with set() and := is fastest in this case.

user_100359850

Here is some pseudocode (not in Python):

newList <- []
for each number in oldList
  if hasNoRepeatDigits(number)
    newList.add(number)
  endif
endfor
user_8201270 
lst = [123456, 889756, 854123, 997886, 634178]

answer = []
for value in lst:
    value = str(value)
    new_list = []
    for nums in value:
        new_list.append(nums)
    if sorted(list(set(new_list))) == sorted(new_list):
        answer.append(int(value))
print(answer)
user_58580060

When using PyPy, this seems about 2 to 7 times faster than the other ones:

def has_duplicates(n):
  s = 0
  while n:
    d = n % 10
    if (1<<d) & s:
      return True
    s |= 1<<d
    n = n // 10
  return False
 
def f(lst):
  return [i for i in lst if not has_duplicates(i)]
user_20347870
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