Python: Best Way to remove duplicate character from string

Question

How can I remove duplicate characters from a string using Python? For example, let's say I have a string:

foo = "SSYYNNOOPPSSIISS"

How can I make the string:

foo = SYNOPSIS

I'm new to python and What I have tired and it's working. I knew there is smart and best way to do this.. and only experience can show this..

def RemoveDupliChar(Word):
        NewWord = " "
        index = 0
        for char in Word:
                if char != NewWord[index]:
                        NewWord += char
                        index += 1
        print(NewWord.strip())

NOTE: Order is important and this question is not similar to this one.

falsetru · Accepted Answer · 2013-09-14 06:37:48Z

20

Using itertools.groupby:

>>> foo = "SSYYNNOOPPSSIISS"
>>> import itertools
>>> ''.join(ch for ch, _ in itertools.groupby(foo))
'SYNOPSIS'

edited Sep 14, 2013 at 6:37

answered Sep 14, 2013 at 6:32

falsetru

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11 Comments

roman Over a year ago

Is it possible to change grp to _?

falsetru Over a year ago

@RahulPatil, _ (grp before the answer modification) is iterable which yield individual items (characters here) that is grouped together.

Rahul Patil Over a year ago

I spent some time to create that function, I don't aware of itertools.groupby how you found that ?

TerryA Over a year ago

@RahulPatil Is it commonly used in loops as a placeholder name. You never use it, but it is put there because you need to put something. itertools.groupby is part of the itertools module in the standard library. There is a link in falsetru's answer

falsetru Over a year ago

@RahulPatil, I see Python Module Index to find useful modules in standard library.

|

G M · Accepted Answer · 2013-09-14 08:07:30Z

4

This is a solution without importing itertools:

foo = "SSYYNNOOPPSSIISS"
''.join([foo[i] for i in range(len(foo)-1) if foo[i+1]!= foo[i]]+[foo[-1]])

Out[1]: 'SYNOPSIS'

But it is slower than the others method!

edited Sep 14, 2013 at 8:07

answered Sep 14, 2013 at 7:25

G M

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Comments

Elliott · Accepted Answer · 2015-04-03 00:24:49Z

3

How about this:

oldstring = 'SSSYYYNNNOOOOOPPPSSSIIISSS'
newstring = oldstring[0]
for char in oldstring[1:]:
    if char != newstring[-1]:
        newstring += char

answered Apr 3, 2015 at 0:24

Elliott

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Comments

Espoir Murhabazi · Accepted Answer · 2017-07-15 08:48:01Z

1

def remove_duplicates(astring):
  if isinstance(astring,str) :
    #the first approach will be to use set so we will convert string to set and then convert back set to string and compare the lenght of the 2
    newstring = astring[0]
    for char in astring[1:]:
        if char not in newstring:
            newstring += char    
    return newstring,len(astring)-len(newstring)
  else:
raise TypeError("only deal with alpha  strings")

I've discover that solution with itertools and with list comprehesion even the solution when we compare the char to the last element of the list doesn't works

answered Jul 15, 2017 at 8:48

Espoir Murhabazi

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Comments

rrao · Accepted Answer · 2016-11-19 12:30:02Z

0

def removeDuplicate(s):  
    if (len(s)) < 2:
        return s

    result = []
    for i in s:
        if i not in result:
            result.append(i)

    return ''.join(result)

edited Nov 19, 2016 at 12:30

rrao

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answered Oct 22, 2014 at 9:58

jtu

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Comments

rrao · Accepted Answer · 2016-11-19 13:41:51Z

0

How about

foo = "SSYYNNOOPPSSIISS"


def rm_dup(input_str):
    newstring = foo[0]
    for i in xrange(len(input_str)):
        if newstring[(len(newstring) - 1 )] != input_str[i]:
            newstring += input_str[i]
        else:
            pass
    return newstring

print rm_dup(foo)

edited Nov 19, 2016 at 13:41

answered Nov 15, 2016 at 12:03

rrao

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Dino · Accepted Answer · 2019-10-20 17:04:30Z

-1

You can try this:

string1 = "example1122334455"
string2 = "hello there"

def duplicate(string):
    temp = ''

    for i in string:
        if i not in temp: 
            temp += i

    return temp;

print(duplicate(string1))
print(duplicate(string2))

edited Oct 20, 2019 at 17:04

Dino

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answered Oct 20, 2019 at 10:50

mohit Kumar

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