如果某个单词不在搜索单词之前，则添加到列表python

Question

我希望该程序检测某个单词是否在搜索单词之前，以及是否不将其添加到列表中。

这是我自己想到的：

sentence = "today i will take my dog for a walk, tomorrow i will not take my dog for a walk"

all = ["take", "take"] 

all2= [w for w in all if not(re.search(r'not' + w + r'\b', sentence))]
print(all2)

预期的输出为[“ take”]，但与[“ take，” take]相同

Answer 1

看着它应该如何制定： 收集所有take未使用前逐字逐句出现not ：

import re

sentence = "today i will take my dog for a walk, tomorrow i will not take my dog for a walk"
search_word = 'take'
all_takes_without_not = re.findall(fr'(?<!\bnot)\s+({search_word})\b', sentence)
print(all_takes_without_not)

输出：

['take']

Answer 2

首先将您的句子转换为单词列表可能会更简单。

from itertools import chain

# Get individual words from the string
words = sentence.split()

# Create an iterator which yields the previous word at each position
previous = chain([None], words)

output = [word for prev, word in zip(previous, words) if word=='take' and prev != 'not']