精确的单词匹配和列显示

Question

我有以下数据帧（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，我需要搜索一个完全匹配的

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  

我试过一些方法，比如芬德尔街你知道吗

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

还有其他一些，但我似乎无法得到确切的匹配我的话。你知道吗

任何帮助解决这个问题都将不胜感激。你知道吗

谢谢

Answer 1

你需要每个单词的边界。将^{}、^{}和^{}转换为原始DataFrame的一种可能的解决方案：

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add^{}中的小写数据：

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              

您的解决方案应该按照相同的模式进行更改，将值转换为lists和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add .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(''))

Answer 2

您可以使用pandas的apply函数。 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