如果某個單詞不在搜索單詞之前，則添加到列表python

Question

我希望該程序檢測某個單詞是否在搜索單詞之前，以及是否不將其添加到列表中。

這是我自己想到的：

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)

預期的輸出為[“ take”]，但與[“ take，” take]相同

Answer 1

看着它應該如何制定： 收集所有take未使用前逐字逐句出現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)

輸出：

['take']

Answer 2

首先將您的句子轉換為單詞列表可能會更簡單。

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']