Question

我在列表方面遇到了一些麻烦。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"]

我想要检查a中的元素是否属于b中某个元素的一部分 - 如果是，请将其从b的元素中删除。所以，我希望b看起来像：

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

实现此目的的最pythonic（以2.7.x）方式是什么？

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替代列表的测试。

以下是一系列显示不同正则表达式行为的替代方案。

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 ，因为reduce被放逐到python3中的functools，有人必须放一个单行表：

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]

甚至更快

[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，我认为它越来越少 pythonic 。

这是处理transferred dogcatcher案件的人。我借用了mgilson的正则表达式，但我认为这很好，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提取到命名函数以便于阅读。

Answer 3

嗯，最简单的是直接列表理解，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]

检查列表的元素是否存在于另一个列表的元素中

3 个答案: