Question

我有以下数据框（df）

process_response

和一些我需要搜索完全匹配的单词

   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

这是我想要的输出

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

我已经尝试了诸如str.findall

   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

和其他一些，但我似乎无法完全匹配我的单词。

任何帮助解决此问题的方法将不胜感激。

谢谢

Answer 1

您可以使用熊猫的apply功能。示例：

import pandas as pd

my_dataframe = pd.DataFrame({'Comments': [10, 11, 12, 13, 14, 15],
                             'ID': [
                                 'Looking for help',
                                 'Look at him but be nice',
                                 'Be calm',
                                 'Being good',
                                 'Him and Her',
                                 'Himself']
                             })

print(my_dataframe)

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


word_list = ['look','be','him']
for index, word in enumerate(word_list):
    def match_word(val):
        """
        Under-optimized pattern matching
        :param val:
        :type val:
        :return:
        :rtype:
        """
        if word.lower() in val.lower():
            return word
        return None
    my_dataframe['Word_{}'.format(index)] = my_dataframe['ID'].apply(match_word)

print(my_dataframe)

输出：

   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

   Comments                       ID Word_0 Word_1 Word_2
0        10         Looking for help   look   None   None
1        11  Look at him but be nice   look     be    him
2        12                  Be calm   None     be   None
3        13               Being good   None     be   None
4        14              Him and Her   None   None    him
5        15                  Himself   None   None    him

Answer 2

每个单词都需要单词边界。使用Series.str.extractall，DataFrame.add_prefix和DataFrame.join到原始DataFrame的一种可能解决方案：

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_'))

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

您的解决方案应以相同的方式更改，将值转换为list s，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(''))

或使用列表理解（应该最快）：

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(''))

精确的单词匹配并按列显示

2 个答案: