How can I remove duplicate words in a string with Python?

Following example:

string1 = "calvin klein design dress calvin klein"

How can I remove the second two duplicates "calvin" and "klein"?

The result should look like

string2 = "calvin klein design dress"

only the second duplicates should be removed and the sequence of the words should not be changed!

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

string1 = "calvin klein design dress calvin klein"
words = string1.split()
print (" ".join(sorted(set(words), key=words.index)))

This sorts the set of all the (unique) words in your string by the word’s index in the original list of words.

Method 2

def unique_list(l):
    ulist = []
    [ulist.append(x) for x in l if x not in ulist]
    return ulist

a="calvin klein design dress calvin klein"
a=' '.join(unique_list(a.split()))

Method 3

In Python 2.7+, you could use collections.OrderedDict for this:

from collections import OrderedDict
s = "calvin klein design dress calvin klein"
print ' '.join(OrderedDict((w,w) for w in s.split()).keys())

Method 4

Cut and paste from the itertools recipes

from itertools import ifilterfalse

def unique_everseen(iterable, key=None):
    "List unique elements, preserving order. Remember all elements ever seen."
    # unique_everseen('AAAABBBCCDAABBB') --> A B C D
    # unique_everseen('ABBCcAD', str.lower) --> A B C D
    seen = set()
    seen_add = seen.add
    if key is None:
        for element in ifilterfalse(seen.__contains__, iterable):
            seen_add(element)
            yield element
    else:
        for element in iterable:
            k = key(element)
            if k not in seen:
                seen_add(k)
                yield element

I really wish they could go ahead and make a module out of those recipes soon. I’d very much like to be able to do from itertools_recipes import unique_everseen instead of using cut-and-paste every time I need something.

Use like this:

def unique_words(string, ignore_case=False):
    key = None
    if ignore_case:
        key = str.lower
    return " ".join(unique_everseen(string.split(), key=key))

string2 = unique_words(string1)

Method 5

string2 = ' '.join(set(string1.split()))

Explanation:

.split() – it is a method to split string to list (without params it split by spaces)
set() – it is type of unordered collections that exclude dublicates
'separator'.join(list) – mean that you want to join list from params to string with ‘separator’ between elements

Method 6

string = 'calvin klein design dress calvin klein'

def uniquify(string):
    output = []
    seen = set()
    for word in string.split():
        if word not in seen:
            output.append(word)
            seen.add(word)
    return ' '.join(output)

print uniquify(string)

Method 7

You can use a set to keep track of already processed words.

words = set()
result = ''
for word in string1.split():
    if word not in words:
        result = result + word + ' '
        words.add(word)
print result

Method 8

Several answers are pretty close to this but haven’t quite ended up where I did:

def uniques( your_string ):    
    seen = set()
    return ' '.join( seen.add(i) or i for i in your_string.split() if i not in seen )

Of course, if you want it a tiny bit cleaner or faster, we can refactor a bit:

def uniques( your_string ):    
    words = your_string.split()

    seen = set()
    seen_add = seen.add

    def add(x):
        seen_add(x)  
        return x

    return ' '.join( add(i) for i in words if i not in seen )

I think the second version is about as performant as you can get in a small amount of code. (More code could be used to do all the work in a single scan across the input string but for most workloads, this should be sufficient.)

Method 9

Use numpy function
make an import its better to have an alias for the import (as np)

import numpy as np

and then you can bing it like this
for removing duplicates from array you can use it this way

no_duplicates_array = np.unique(your_array)

for your case if you want result in string you can use

no_duplicates_string = ' '.join(np.unique(your_string.split()))

Method 10

11 and 2 work perfectly:

    s="the sky is blue very blue"
    s=s.lower()
    slist = s.split()
    print " ".join(sorted(set(slist), key=slist.index))

and 2

    s="the sky is blue very blue"
    s=s.lower()
    slist = s.split()
    print " ".join(sorted(set(slist), key=slist.index))

Method 11

Question: Remove the duplicates in a string

 from _collections import OrderedDict

    a = "Gina Gini Gini Protijayi"

    aa = OrderedDict().fromkeys(a.split())
    print(' '.join(aa))
   # output => Gina Gini Protijayi

Method 12

You can remove duplicate or repeated words from a text file or string using following codes –

from collections import Counter
for lines in all_words:

    line=''.join(lines.lower())
    new_data1=' '.join(lemmatize_sentence(line))
    new_data2 = word_tokenize(new_data1)
    new_data3=nltk.pos_tag(new_data2)

    # below code is for removal of repeated words

    for i in range(0, len(new_data3)):
        new_data3[i] = "".join(new_data3[i])
    UniqW = Counter(new_data3)
    new_data5 = " ".join(UniqW.keys())
    print (new_data5)


    new_data.append(new_data5)


print (new_data)

P.S. -Do identations as per required.
Hope this helps!!!

Method 13

Without using the split function (will help in interviews)

def unique_words2(a):
    words = []
    spaces = ' '
    length = len(a)
    i = 0
    while i < length:
        if a[i] not in spaces:
            word_start = i
            while i < length and a[i] not in spaces:
                i += 1
            words.append(a[word_start:i])
        i += 1
    words_stack = []
    for val in words:  #
        if val not in words_stack:  # We can replace these three lines with this one -> [words_stack.append(val) for val in words if val not in words_stack]
            words_stack.append(val)  #
    print(' '.join(words_stack))  # or return, your choice


unique_words2('calvin klein design dress calvin klein')

Method 14

initializing list

listA = [ 'xy-xy', 'pq-qr', 'xp-xp-xp', 'dd-ee']

print("Given list : ",listA)

using set() and split()

res = [set(sub.split('-')) for sub in listA]

Result

print("List after duplicate removal :", res)

Method 15

You can do that simply by getting the set associated to the string, which is a mathematical object containing no repeated elements by definition. It suffices to join the words in the set back into a string:

def remove_duplicate_words(string):
        x = string.split()
        x = sorted(set(x), key = x.index)
        return ' '.join(x)

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0