如何使用Python删除字符串中的重复单词？

Question

以下示例:

string1 = "calvin klein design dress calvin klein"

我怎样才能删除第二个一式两份"calvin",并"klein"？

结果应该是这样的

string2 = "calvin klein design dress"

只应删除第二个重复项,并且不应更改单词的顺序!

Answer 1

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

这会根据原始单词列表中的单词索引对字符串中所有(唯一)单词的集合进行排序.

Answer 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()))

Answer 3

在Python 2.7+中,您可以使用collections.OrderedDict:

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

Answer 4

从itertools食谱中剪切和粘贴

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

我真的希望他们能够继续前进,尽快从这些食谱中制作一个模块.我非常希望能够在from itertools_recipes import unique_everseen每次需要时都能使用剪切和粘贴.

使用这样:

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)

Answer 5

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

解释：

.split()- 它是一种将字符串拆分为列表的方法（没有参数，它按空格拆分）
set()- 它是排除重复项的无序集合类型
'separator'.join(list)- 意味着您希望使用元素之间的“分隔符”将列表从参数连接到字符串

Answer 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)