检查列表元素是否存在于另一个列表的元素中

Question

我在列表中遇到了一些问题。所以，基本上，我有一个列表：

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

，我有另一个列表，看起来像：

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

我想要做的是检查是否在a元素中某些元素的一部分是什么b - 如果是这样，请将其从b的元素中移除。所以，我想b的样子：

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

什么是最Python的（在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，但我不确定这是否非常有效 - 我有一个大小约为50k的列表（b）。

Answer 1

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，正则表达式引擎可以优化测试替代品的清单。

这里有一些替代方法来显示不同的正则表达式如何表现。

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

那么，最简​​单的将是一个直接的列表理解，只要a是小的，它甚至是一个非常有效的方法。

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

Answer 3

好了，我不知道，知道，如果这样继续下去了算作Python的因为reduce得到了在python3流放到functools，有人把一个班轮在桌子上：

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] 

更快的将是

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

但可读性下降，它越来越少Python的我想。

下面是一个处理transferred dogcatcher的情况。我借mgilson的正则表达式，但我认为这是可以的，因为它是相当琐碎:-)：

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] 

我提取lambda为可读性命名函数。