Removing duplicates in each item in a list of strings

Question

I have a list which contains the below values.

list1= ['AAB', 'CAA', 'ADA']

How should I remove duplicates in each item in a list such that i get the following output

list1= ['AB', 'CA', 'AD']

The order should be preserved here

Can you [edit] the question, showing what you have tried, that didn't work? — costaparas, Feb 04 '21 at 11:30
Does this answer your question? [Removing duplicate characters from a string](https://stackoverflow.com/questions/9841303/removing-duplicate-characters-from-a-string) — mhhabib, Feb 04 '21 at 11:32

score 1 · Accepted Answer · answered Feb 04 '21 at 11:32

You can use set combined with list comprehension if you don't want the letter ordering to be preserved:

list1= ['AAB', 'CAA', 'ADA']
list1 = [''.join(set(l)) for l in list1]
print(list1)

Or use OrderedDict if you want the ordering to be preserved:

from collections import OrderedDict 
list1= ['AAB', 'CAA', 'ADA']
list1 = [''.join(OrderedDict.fromkeys(l).keys()) for l in list1]
print(list1)

score 1 · Answer 2 · answered Feb 04 '21 at 11:41

1

You can create a dict using characters on the strings as keys to eliminate duplicates.

list1= ['AAB', 'CAA', 'ADA']
list1 = [''.join(dict.fromkeys(l).keys()) for l in list1]
print(list1)

answered Feb 04 '21 at 11:41

Seralpa

Reza Takhshid · Answer 3 · 2021-02-04T11:40:02.480

0

Loop over the list, cast to set, join

new_list = []
for word in list1:
    new_list.append(''.join(set(word)))

edited Feb 04 '21 at 11:40

answered Feb 04 '21 at 11:33

Reza Takhshid

score 0 · Answer 4 · answered Feb 04 '21 at 12:09

A regex solution for fun:

>>> import re
>>> [re.sub(r'(.)(?=.*\1)', '', s[::-1])[::-1] for s in list1]
['AB', 'CA', 'AD']

You can omit the reversals if you're ok with keeping the last occurrence of each:

>>> [re.sub(r'(.)(?=.*\1)', '', s) for s in list1]
['AB', 'CA', 'DA']

Proper solution would be:

>>> [''.join(dict.fromkeys(s)) for s in list1]
['AB', 'CA', 'AD']

(Not sure why the others use .keys(), that just unnecessarily costs extra code, time and space.)

4 Answers4