Question

您好我有两个DataFrame，如下所示

 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中的任何列是否包含DF2的内容列中的任何关键字，并且输出应该是新的DF

 `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lol ok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

有人用这个帮助我

Answer 1

融合数据帧1并将其转换为dict。然后根据数据帧2中的模式匹配获得的密钥得到dict的值，即

vals = df.melt()
di = dict(zip(vals['value'],vals['variable']))
# {'or': 'Alpha', 1: 'Numeric', 2: 'Numeric', 'and': 'Alpha', 4: 'Numeric', '@': 'Special', '#': 'Special', '&': 'Special', 'Special': 'new', 'Alpha': 'new', 'lol ok': 'Alpha'}

#Create a regex pattern based on dict keys. 
pat = '|'.join(r"\s{}\s".format(x) for x in di.keys())

#Find the words that match the pattern 
df2['val'] = df2['Content'].str.lower().str.findall(pat).apply(lambda x : x[0].strip() if len(x)>=1 else np.nan)

# Map the values with di and fill nan with other. 
df2['new'] = df2['val'].map(di).fillna('other')

Ouptut：


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

Answer 2

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

<强>详情

i和j是设置变量以创建映射
m是值到列名的映射
v是发送到str.extract以进行关键字提取的正则表达式模式。我使用re.I忽略案例
我使用map + fillna通过m

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

python中使用pandas的两个DataFrame之间的值匹配

2 个答案: