CODEWITHSUNDEEP
pythonlistalgorithm
Lullaby
Lullaby

Reputation: 133

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

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!

Upvotes: 2

Views: 1571

Answers (5)

גלעד ברקן
גלעד ברקן

Reputation: 23955

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)]

Upvotes: 1

trincot
trincot

Reputation: 350147

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)]

Upvotes: 5

Andrej Kesely
Andrej Kesely

Reputation: 195418

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.

Upvotes: 2

Anmol Parida
Anmol Parida

Reputation: 688

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)

Upvotes: 1

rossum
rossum

Reputation: 15685

Here is some pseudocode (not in Python):

newList <- []
for each number in oldList
  if hasNoRepeatDigits(number)
    newList.add(number)
  endif
endfor

Upvotes: 1

Related Questions