Как найти повторяющиеся слова в строке питон

Jane Doe2Jane Doe2

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You just have to compare the first part with the second, you can do this with slicing like this:

def doubleWord(word):
    return word[len(word) // 2:] == word[:len(word) // 2]

answered Nov 5, 2016 at 21:45

Martin Thoma

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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

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

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

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

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

Chris PhillipsChris Phillips

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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

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

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) 

OkroshiashviliOkroshiashvili

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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

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