在Python中使用熊猫在两个DataFrame之间进行值匹配

嗨,我有两个像下面的DataFrames

 DF1

 Alpha   |  Numeric  |  Special

 and     |   1        |  @
 or      |   2       |  #
 lol ok  |   4       |  &






DF2 with single column

Content          

boy or girl  
school @ morn
pyc LoL ok student
Chandra

我想搜索DF1列中的任何人是否在DF2内容列中的任何关键字,并且输出应该在新的DF中

 `df11 = (df1.unstack()
      .reset_index(level=2,drop=True)
      .rename_axis(('col_order','col_name'))
      .dropna()
      .reset_index(name='val_low'))

 df22 = (df2['Content'].str.split(expand=True)
                 .stack()
                 .rename('val')
                 .reset_index(level=1,drop=True)
                 .rename_axis('idx')
                 .reset_index())`

 df22['val_low'] = df22['val'].str.lower()                    

 df = (pd.merge(df22, df11, on='val_low', how='left')
   .dropna(subset=['col_name'])
   .sort_values(['idx','col_order'])
   .drop_duplicates(['idx']))


 df = (pd.concat([df2, df.set_index('idx')], axis=1)
   .fillna({'col_name':'Other'})[['val','col_name','Content']])

但它没有考虑大声笑之间的空格

 expected_output_DF

     val      col_name          Content
 0   or       Alpha             boy or girl
 1    @      Special            school @ morn
 2   lol ok  Alpha              pyc LoL ok student
 3  NaN      Other              Chandra

有人帮我这个

解决方法:

使用str.cat str.extract.然后,使用map作为列名,并使用pd.concat加入.

i = df.stack().astype(str)
j = i.reset_index(level=0, drop=1)

m = dict(zip(j.values, j.index))
v = i.str.cat(sep='|')  

df2['val'] = df2.Content.str.extract(r'\s(' + v + r')\s', flags=re.I, expand=False)
df2['col_name'] = df2['val'].str.lower().map(m).fillna('Other')

df2

              Content     val col_name
0         boy or girl      or    Alpha
1       school @ morn       @  Special
2  pyc LoL ok student  LoL ok    Alpha
3             Chandra     NaN    Other

细节

> i和j是用于创建映射的设置变量
> m是值到列名称的映射
> v是发送到str.extract进行关键字提取的正则表达式模式.我用re.I忽略了这种情况
>我使用map fillna通过m将提取的值映射到列名

v
'and|1|@|or|2|#|lol ok|4|&'

m
{'#': 'Special',
 '&': 'Special',
 '1': 'Numeric',
 '2': 'Numeric',
 '4': 'Numeric',
 '@': 'Special',
 'and': 'Alpha',
 'lol ok': 'Alpha',
 'or': 'Alpha'}


df['val']

0        or
1         @
2    LoL ok
3       NaN
Name: val, dtype: object

df['col_name']

0      Alpha
1    Special
2      Alpha
3      Other
Name: col_name, dtype: object

标签：data-analysis,python,pandas,dataframe
来源： https://codeday.me/bug/20191009/1880572.html