检查列表元素是否存在于另一个列表的元素中

Question

我在使用列表时遇到了麻烦。 因此，基本上，我有一个列表：

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

我还有另一个列表，看起来像：

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，则从b的元素中删除它们。 所以，我希望b看起来像：

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

什么是最pythonic（2.7.x）的方式来实现这一目标？

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

Answer 1

我想我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，正则表达式引擎可以针对替代项列表优化测试。

这是一堆替代方法，以显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']

Answer 2

好吧，我不知道这是否算作pythonic了，因为reduce被放到python3的functools中，所以有人不得不在表上放一个衬里：

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]

会更快

[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，我认为它的pythonic越来越少。

这是一个处理transferred dogcatcher案件的人。 我借用了mgilson的正则表达式，但我认为它没关系，因为它非常简单:-)：

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]

我将lambda提取到一个命名函数以提高可读性。

Answer 3

好吧，最简单的方法就是直接理解列表，只要a很小，它甚至是相当有效的方法。

b = [i for i in b if i not in a]