问检查列表的元素是否存在于另一个列表的元素中
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Stack Overflow用户

提问于 2013-02-01 12:45:11

回答 3查看 832关注 0票数 2

我在列表上遇到了一些麻烦。所以，基本上，我有一个清单：

a=["Britney spears", "red dog", "\xa2xe3"]

我还有另一个列表，看起来像这样：

b = ["cat","dog","red dog is stupid", "good stuff \xa2xe3", "awesome Britney spears"]

我想要做的是检查a中的元素是否是b中某个元素的一部分--如果是的话，将它们从b的元素中删除。所以，我希望b看起来像这样：

b = ["cat","dog","is stupid","good stuff","awesome"]

在2.7.x中，实现这一点的最具pythonic风格的方法是什么？

我假设我可以循环检查每个元素，但我不确定这是否非常有效-我有一个大小约为50k的列表(b)。

python

python-2.7

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回答 3

Stack Overflow用户

回答已采纳

发布于 2013-02-01 12:51:21

我想我会在这里使用正则表达式：

import re

a=["Britney spears", "red dog", "\xa2xe3"]

regex = re.compile('|'.join(re.escape(x) for x in a))

b=["cat","dog","red dog is stupid", "good stuff \xa2xe3", "awesome Britney spears"]

b = [regex.sub("",x) for x in b ]
print (b)  #['cat', 'dog', ' is stupid', 'good stuff ', 'awesome ']

这样，正则表达式引擎就可以针对备选方案列表优化测试。

这里有一堆替代方案来展示不同的正则表达式的行为。

import re

a = ["Britney spears", "red dog", "\xa2xe3"]
b = ["cat","dog",
     "red dog is stupid", 
     "good stuff \xa2xe3", 
     "awesome Britney spears",
     "transferred dogcatcher"]

#This version leaves whitespace and will match between words.
regex = re.compile('|'.join(re.escape(x) for x in a))
c = [regex.sub("",x) for x in b ]
print (c) #['cat', 'dog', ' is stupid', 'good stuff ', 'awesome ', 'transfercatcher']

#This version strips whitespace from either end
# of the returned string
regex = re.compile('|'.join(r'\s*{}\s*'.format(re.escape(x)) for x in a))
c = [regex.sub("",x) for x in b ]
print (c) #['cat', 'dog', 'is stupid', 'good stuff', 'awesome', 'transfercatcher']

#This version will only match at word boundaries,
# but you lose the match with \xa2xe3 since it isn't a word
regex = re.compile('|'.join(r'\s*\b{}\b\s*'.format(re.escape(x)) for x in a))
c = [regex.sub("",x) for x in b ]
print (c) #['cat', 'dog', 'is stupid', 'good stuff \xa2xe3', 'awesome', 'transferred dogcatcher']


#This version finally seems to get it right.  It matches whitespace (or the start
# of the string) and then the "word" and then more whitespace (or the end of the 
# string).  It then replaces that match with nothing -- i.e. it removes the match 
# from the string.
regex = re.compile('|'.join(r'(?:\s+|^)'+re.escape(x)+r'(?:\s+|$)' for x in a))
c = [regex.sub("",x) for x in b ]
print (c) #['cat', 'dog', 'is stupid', 'good stuff', 'awesome', 'transferred dogcatcher']

票数 4

Stack Overflow用户

发布于 2013-02-01 13:03:19

好吧，我不知道这还算不算蟒蛇，因为reduce被流放到python3的functools，总得有人说一句：

a = ["Britney spears", "red dog", "\xa2xe3"]
b = ["cat","dog","red dog is stupid", "good stuff \xa2xe3", "awesome Britney spears"]

b = [reduce(lambda acc, n: acc.replace(n, ''), a, x).strip() for x in b]

更快的是

[reduce(lambda acc, n: acc.replace(n, '') if n in acc else acc, a, x).strip() for x in b]

但随着可读性的降低，我认为它变得越来越不像pythonic了。

这里有一个处理transferred dogcatcher的案例。我借用了mgilson的正则表达式，但我认为这是可以的，因为它非常琐碎:-)：

def reducer(acc, n):
    if n in acc:
        return re.sub('(?:\s+|^)' + re.escape(n) + '(?:\s+|$)', '', acc)
    return acc

b = [reduce(reducer, a, x).strip() for x in b]