检查列表的元素是否存在于另一个列表的元素中
检查列表的元素是否存在于另一个列表的元素中
提问于 2013-02-01 20:45:11
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)。
回答 3
Stack Overflow用户
回答已采纳
发布于 2013-02-01 20:51:21
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+|$)' for x in a))
c = [regex.sub("",x) for x in b ]
print (c) #['cat', 'dog', 'is stupid', 'good stuff', 'awesome', 'transferred dogcatcher']Stack Overflow用户
发布于 2013-02-01 21:03:19
好吧，我不知道这还算不算蟒蛇，因为reduce被流放到python3的functools，总得有人说一句：
a = ["Britney spears", "red dog", "\xa2xe3"]
b = ["cat","dog","red dog is stupid", "good stuff \xa2xe3", "awesome Britney spears"]
b = [reduce(lambda acc, n: acc.replace(n, ''), a, x).strip() for x in b]更快的是
[reduce(lambda acc, n: acc.replace(n, '') if n in acc else acc, a, x).strip() for x in b]但随着可读性的降低，我认为它变得越来越不像pythonic了。
这里有一个处理transferred dogcatcher的案例。我借用了mgilson的正则表达式，但我认为这是可以的，因为它非常琐碎:-)：
def reducer(acc, n):
if n in acc:
return re.sub('(?:\s+|^)' + re.escape(n) + '(?:\s+|$)', '', acc)
return acc
b = [reduce(reducer, a, x).strip() for x in b]为了提高可读性，我将lambda提取到一个命名函数中。
Stack Overflow用户
发布于 2013-02-01 20:48:51
嗯，最简单的方法是直接的列表理解，只要a很小，它甚至是一种非常有效的方法。
b = [i for i in b if i not in a]页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/14646606
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