Python 如果DataFrame多个子字符串匹配，也会将行的特定匹配子字符串放入新列中

我正试图从一份调查回复中提取一些记录。所有这些记录都需要至少包含一些关键字中的一个。例如： 现在我有了一个数据帧df：

svy_rspns_txt
I like it
I hate it
It's a scam
It's shaddy
Scam!
Good service
Very disappointed

现在如果我跑

kw="hate,scam,shaddy,disappoint"
sensitive_words=[unicode(x,'unicode-escape') for x in kw.lower().split(",")]
df=df[df["svy_rspns_txt"].astype('unicode').str.contains('|'.join(sensitive_words),case=False,na=False)]

我会得到这样的结果

svy_rspns_txt
I hate it
It's a scam
It's shaddy
Scam!
Very disappointed

现在，我如何添加一列“matched_word”来显示匹配的确切字符串，以便得到如下结果：

svy_rspns_txt            matched_word
I hate it                hate
It's a scam              scam
It's shaddy              shaddy
Scam!                    scam
Very disappointed        disappoint

---

将生成器表达式与

next

一起使用：

df = pd.DataFrame({'text': ["I like it", "I hate it", "It's a scam", "It's shaddy",
                            "Scam!", "Good service", "Very disappointed"]})

kw = "hate,scam,shaddy,disappoint"

words = set(kw.split(','))

df['match'] = df['text'].apply(lambda x: next((i for i in words if i in x.lower()), np.nan))

print(df)

                text       match
0          I like it         NaN
1          I hate it        hate
2        It's a scam        scam
3        It's shaddy      shaddy
4              Scam!        scam
5       Good service         NaN
6  Very disappointed  disappoint

您可以通过或注意

NaN！=南

：

res = df[df['match'].notnull()]
# or, res = df[df['match'].notna()]
# or, res = df[df['match'] == df['match']]

print(res)

                text       match
1          I hate it        hate
2        It's a scam        scam
3        It's shaddy      shaddy
4              Scam!        scam
6  Very disappointed  disappoint

---

如果多个单词匹配怎么办？

matched_word

列应该显示所有单词，还是只显示匹配的第一个单词？@TimJohns现在我只需要显示第一个单词。但是如果你能给我一些建议来显示所有匹配的单词，那就太好了。非常感谢。你能不能只做

res=df[df['match'].notna（）]

，让它更清楚？@TimJohns，是的，那也行，更清楚，谢谢。我通常使用

pd.Series.notnull

。