python - 精确的单词匹配并显示在列中

Question

我有以下数据框(df)

   Comments                       ID
0        10         Looking for help
1        11  Look at him but be nice
2        12                  Be calm
3        13               Being good
4        14              Him and Her
5        15                  Himself

以及列表中的一些单词，我需要搜索完全匹配

word_list = ['look','be','him']

这是我想要的输出

   Comments                       ID Word_01 Word_02 Word_03
0        10         Looking for help                        
1        11  Look at him but be nice    look     be      him
2        12                  Be calm    be                
3        13               Being good                        
4        14              Him and Her    him                
5        15                  Himself  

我尝试过一些东西，比如 str.findall

str.findall(r"\b" + '|'.join(word_list) + r"\b",flags = re.I)

还有其他一些，但我似乎无法为我的话找到完全匹配的内容。

任何解决此问题的帮助将不胜感激。

谢谢

Answer 1

您需要每个单词的单词边界。一种可能的解决方案是 Series.str.extractall , DataFrame.add_prefix和 DataFrame.join原始DataFrame:

word_list = ['look','be','him']

pat = '|'.join(r"\b{}\b".format(x) for x in word_list)
df1 = df['ID'].str.extractall('(' + pat + ')', flags = re.I)[0].unstack().add_prefix('Word_')

对于输出中的小写数据，请添加 Series.str.lower :

df1 = (df['ID'].str.lower()
               .str.extractall('(' + pat + ')')[0]
               .unstack()
               .add_prefix('Word_'))

<小时/>

df = df.join(df1).fillna('')
print (df)
   Comments                       ID Word_0 Word_1 Word_2
0        10         Looking for help                     
1        11  Look at him but be nice   Look    him     be
2        12                  Be calm     Be              
3        13               Being good                     
4        14              Him and Her    Him              
5        15                  Himself              

您的解决方案应该按相同的模式进行更改，将值转换为 list 并将 join 转换为原始值:

pat = '|'.join(r"\b{}\b".format(x) for x in word_list)
df1 = (pd.DataFrame(df['ID']
        .str.findall(pat, flags = re.I).values.tolist())
        .add_prefix('Word_')
        .fillna(''))   

或者使用列表理解(应该是最快的):

df1 = (pd.DataFrame([re.findall(pat, x, flags = re.I) for x in df['ID']])
       .add_prefix('Word_')
       .fillna(''))

对于小写字母，请添加.lower():

pat = '|'.join(r"\b{}\b".format(x) for x in word_list)
df1 = (pd.DataFrame([re.findall(pat, x.lower(), flags = re.I) for x in df['ID']])
           .add_prefix('Word_')
           .fillna(''))