How do i remove duplicates from a word in 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!

Martin Thoma

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asked Oct 17, 2011 at 13:08

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.

answered Oct 17, 2011 at 13:40

MarkusMarkus

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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[]]]

answered Oct 17, 2011 at 13:12

spicavigospicavigo

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7

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[]]

answered Oct 17, 2011 at 13:21

NPENPE

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1

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]

answered Oct 17, 2011 at 13:22

3

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

answered Nov 9, 2018 at 8:33

3

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]

answered Oct 17, 2011 at 13:27

ekhumoroekhumoro

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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

answered Oct 17, 2011 at 13:10

Pablo Santa CruzPablo Santa Cruz

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2

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.]

answered Oct 17, 2011 at 22:13

Chris PhillipsChris Phillips

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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[]]]

answered Jun 8, 2020 at 12:04

Sulman MalikSulman Malik

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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]]

answered Apr 17, 2016 at 16:38

1

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

answered Jun 16, 2018 at 23:44

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!!!

answered Jun 25, 2018 at 7:22

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'] 

Taazar

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answered Mar 6, 2020 at 12:06

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] 

Peter Csala

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answered Oct 17, 2021 at 12:39

1

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]

answered Nov 9, 2018 at 8:28

3

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