Python Pandas 正则表达式：在列中搜索带有通配符的字符串并 return 匹配

Question

我在一个列中有一个搜索列表，其中可能包含一个键：'keyword1*keyword2' 以尝试在单独的数据框列中找到匹配项。如何包含正则表达式通配符类型 'keyword1.*keyword2' #using str.extract, extractall or findall?

使用 .str.extract 可以很好地匹配完全匹配的子字符串，但我还需要它来匹配关键字之间带有通配符的子字符串。

# dataframe column or series list as keys to search for: 
dfKeys = pd.DataFrame()
dfKeys['SearchFor'] = ['this', 'Something', 'Second', 'Keyword1.*Keyword2', 'Stuff', 'One' ]

# col_next_to_SearchFor_col
dfKeys['AdjacentCol'] = ['this other string', 'SomeString Else', 'Second String Player', 'Keyword1 Keyword2', 'More String Stuff', 'One More String Example' ]

# dataframe column to search in: 
df1['Description'] = ['Something Here','Second Item 7', 'Something There', 'strng KEYWORD1 moreJARGON 06/0 010 KEYWORD2 andMORE b4END', 'Second Item 7', 'Even More Stuff']]

# I've tried:
df1['Matched'] = df1['Description'].str.extract('(%s)' % '|'.join(key['searchFor']), flags=re.IGNORECASE, expand=False)

我也尝试用 'extractall' 和 'findall' 替换上面代码中的 'extract' 但它仍然没有给我需要的结果。我希望 'Keyword1*Keyword2' 匹配 "strng KEYWORD1 moreJARGON 06/0 010 KEYWORD2 andMORE b4END"

更新：“.*”有效！我还尝试在 'SearchFor' 列中匹配键旁边的单元格中添加值，即 dfKeys['AdjacentCol'].

我试过： df1['From_AdjacentCol'] = df1['Description'].str.extract('(%s)' % '|'.join(key['searchFor']), flags=re.IGNORECASE, expand=False).map(dfKeys.set_index('SearchFor')['AdjacentCol'].to_dict()).fillna('') 除了带通配符的键外，它适用于所有内容。

# expected:
  Description                                      Matched            From_AdjacentCol
0 'Something Here'                                 'Something'         'this other string'
1 'Second Item 7'                                  'Second'            'Second String Player'
2 'Something There'                                'Something'         'this other string'  
3 'strng KEYWORD1 moreJARGON 06/0 010 KEYWORD2...' 'Keyword1*Keyword2' 'Keyword1 Keyword2'
4 'Second Item 7'                                  'Second'            'Second String Player'
5 'Even More Stuff'                                'Stuff'             'More String Stuff'

非常感谢对此的任何帮助。谢谢！

Answer 1

解决方案

您已接近解决方案，只需将 * 更改为 .*。阅读 docs:

. (Dot.) In the default mode, this matches any character except a newline. If the DOTALL flag has been specified, this matches any character including a newline.

* Causes the resulting RE to match 0 or more repetitions of the preceding RE, as many repetitions as are possible. ab* will match ‘a’, ‘ab’, or ‘a’ followed by any number of ‘b’s.

在正则表达式中星号 * 单独没有任何意义。它与 Unix/Windows 文件系统中的常用 glob 运算符 * 具有不同的含义。

星号是一个量词（即gready量词），它必须与某种模式相关联（这里.匹配任何字符）才有意义。

MCVE

重塑您的 MCVE：

import re
import pandas as pd

keys = ['this', 'Something', 'Second', 'Keyword1.*Keyword2', 'Stuff', 'One' ]

df1 = pd.DataFrame()
df1['Description'] = ['Something Here','Second Item 7', 'Something There',
                      'strng KEYWORD1 moreJARGON 06/0 010 KEYWORD2 andMORE b4END',
                      'Second Item 7', 'Even More Stuff']


regstr = '(%s)' % '|'.join(keys)

df1['Matched'] = df1['Description'].str.extract(regstr, flags=re.IGNORECASE, expand=False)

正则表达式现在是：

(this|Something|Second|Keyword1.*Keyword2|Stuff|One)

并匹配缺失的大小写：

                                         Description                                Matched
0                                     Something Here                              Something
1                                      Second Item 7                                 Second
2                                    Something There                              Something
3  strng KEYWORD1 moreJARGON 06/0 010 KEYWORD2 an...  KEYWORD1 moreJARGON 06/0 010 KEYWORD2
4                                      Second Item 7                                 Second
5                                    Even More Stuff                                  Stuff

Python Pandas 正则表达式：在列中搜索带有通配符的字符串并 return 匹配

Python Pandas Regex: Search for strings with a wildcard in a column and return matches

python

regex

pandas

wildcard-expansion

解决方案

MCVE