How can I remove duplicate characters from a string using Python? For example, let's say I have a string:
foo = 'mppmt'
How can I make the string:
foo = 'mpt'
NOTE: Order is not important
If order does not matter, you can use
"".join(set(foo))
set() will create a set of unique letters in the string, and "".join() will join the letters back to a string in arbitrary order.
If order does matter, you can use a dict instead of a set, which since Python 3.7 preserves the insertion order of the keys. (In the CPython implementation, this is already supported in Python 3.6 as an implementation detail.)
foo = "mppmt"
result = "".join(dict.fromkeys(foo))
resulting in the string "mpt". In earlier versions of Python, you can use collections.OrderedDict, which has been available starting from Python 2.7.
Create a list in Python and also a set which doesn't allow any duplicates. Solution1 :
def fix(string):
s = set()
list = []
for ch in string:
if ch not in s:
s.add(ch)
list.append(ch)
return ''.join(list)
string = "Protiijaayiiii"
print(fix(string))
Method 2 :
s = "Protijayi"
aa = [ ch for i, ch in enumerate(s) if ch not in s[:i]]
print(''.join(aa))
As was mentioned "".join(set(foo)) and collections.OrderedDict will do. A added foo = foo.lower() in case the string has upper and lower case characters and you need to remove ALL duplicates no matter if they're upper or lower characters.
from collections import OrderedDict
foo = "EugeneEhGhsnaWW"
foo = foo.lower()
print "".join(OrderedDict.fromkeys(foo))
prints eugnhsaw
d = {}
s="YOUR_DESIRED_STRING"
res=[]
for c in s:
if c not in d:
res.append(c)
d[c]=1
print ("".join(res))
variable 'c' traverses through String 's' in the for loop and is checked if c is in a set d (which initially has no element) and if c is not in d, c is appended to the character array 'res' then the index c of set d is changed to 1. after the loop is exited i.e c finishes traversing through the string to store unique elements in set d, the resultant res which has all unique characters is printed.
close as dupeand supply that link. Thank you - Martin Beckett