44

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!

17 Answers 17

59
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.

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Comments

27
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()))

7 Comments

Unfortunately it's O(N²) – the in goes through the whole ulist each time. Don't use it for long lists.
Thanks Pablo. I found that list comprehension part about 2 years ago on SO itself. Have been using it ever since.
@Petr. Thats true. I provided it here under the assumption that the list is not going to be too long.
I find your use of append in a list comprehension disturbing.
A list comprehension is inappropriate and should not be used unless you're using the output. Use a proper for x in l: if x not in ulist: ulist.append(x).
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13

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())

1 Comment

' '.join(OrderedDict.fromkeys(s.split())).
8

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)

3 Comments

I timed a few of these… this one is very fast, even for long lists.
@lazyr: As for your wish, it turns out you can do exactly that. Just install the package from PyPI.
@Petr This news does not suprise me in the slightest. I'd be amazed if there weren't a PyPI package for just that. What I meant was that it should be part of the included batteries in python, since they are used so frequently. I'm rather puzzled as to why they're not.
7
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

3 Comments

While this might answer the authors question, it lacks some explaining words and/or links to documentation. Raw code snippets are not very helpful without some phrases around them. You may also find how to write a good answer very helpful. Please edit your answer.
This potentially changes the order of the words in the string.
This will not remove duplicates if you want to split on other element than space. fe: "cisco, cisco systems, cisco".join(set(a.split())) will output: 'cisco, systems, cisco'
5
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)

Comments

2

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

2 Comments

Note that set is a built-in type. No need to import it (unless you use an ancient version of Python).
You should make result a list, append the words to it, and then return " ".join(result) in the end. This is much more efficient.
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.)

Comments

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

1 Comment

Starting from Python 3.7, insertion order is guaranteed in dicts. So no need for OrderedDict.
1

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()))

Comments

1

To remove duplicate words from sentence and preserve the order of the words you can use dict.fromkeys method.

string1 = "calvin klein design dress calvin klein"

words = string1.split()

result = " ".join(list(dict.fromkeys(words)))

print(result)

Comments

0

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))

1 Comment

How is this key argument work? I couldn't find it in the documentation.
0

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!!!

Comments

0

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') 

Comments

0

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) 

1 Comment

Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center.
0
import re

# Calea către fișierul tău
file_path = "g:\Pyton+ChatGPT\dictionar_no_duplicates.txt"

# Citește conținutul fișierului
with open(file_path, "r", encoding="utf-8") as file:
    text = file.read()

# Elimină cuvintele duplicate
result = re.sub(r'\b(\w+)\b(?=.*\b\1\b)', '', text)

# Elimină spații suplimentare sau virgule consecutive
result = re.sub(r'\s+', ' ', result).strip().replace(" ,", ",")

# Rescrie fișierul cu conținutul fără duplicate
with open(file_path, "w", encoding="utf-8") as file:
    file.write(result)

OR THIS

def remove_duplicates(words):
    words_stack = []
    for val in words:
        if val not in words_stack:
            words_stack.append(val)
    return words_stack

input_file = r'g:\Pyton+ChatGPT\dictionar.txt'
output_file = r'g:\Pyton+ChatGPT\dictionar_no_duplicates.txt'

with open(input_file, 'r', encoding='utf-8') as f:
    words = f.read().splitlines()

unique_words = remove_duplicates(words)

with open(output_file, 'w', encoding='utf-8') as f:
    for word in unique_words:
        f.write(word + '\n')

print("Duplicate removal completed.")

OR THIS

import re

# Calea către fișierul tău
file_path = "g:\Pyton+ChatGPT\dictionar_no_duplicates.txt"

# Citește conținutul fișierului
with open(file_path, "r", encoding="utf-8") as file:
    text = file.read()

# Crează o listă pentru cuvintele eliminate
removed_words = []

# Funcție callback pentru a adăuga cuvintele duplicate în listă
def replace_and_collect(match):
    word = match.group(1)
    if word not in removed_words:
        removed_words.append(word)
    return ''

# Elimină cuvintele duplicate și virgula asociată folosind funcția callback
result = re.sub(r'\b(\w+)\b,?(?=.*\b\1\b)', replace_and_collect, text)

# Elimină spații suplimentare sau virgule consecutive
result = re.sub(r'\s+', ' ', result).strip().replace(" ,", ",").strip(", ")

# Rescrie fișierul cu conținutul fără duplicate
with open(file_path, "w", encoding="utf-8") as file:
    file.write(result)

# Afișează informații despre cuvintele eliminate
print(f"Numărul de cuvinte duplicate eliminate: {len(removed_words)}")
print(f"Cuvintele eliminate: {', '.join(removed_words)}")

Comments

-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)

3 Comments

While this might answer the authors question, it lacks some explaining words and/or links to documentation. Raw code snippets are not very helpful without some phrases around them. You may also find how to write a good answer very helpful. Please edit your answer.
This potentially changes the order of the words in the string.
Thanks @parvus I have modified my answer

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