Question

您好我有两个DataFrame，如下所示

 DF1

 Alpha   |  Numeric  |  Special

 and, or |  1,2,3,4,5|  @,$,&

和

DF2 with single column

Content      |

boy or girl  |
school @ morn|

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

 output_DF

 output_column|
 Alpha        |
 Special      |

有人用这个帮助我

Answer 1

解决方案有点复杂，因为多重匹配（第2行）只需匹配第一列df1：

df1 = pd.DataFrame({'Alpha':['and','or', None, None,None],
                    'Numeric':['1','2','3','4','5'],
                    'Special':['@','$','&', None, None]})
print (df1)
  Alpha Numeric Special
0   and       1       @
1    or       2       $
2  None       3       &
3  None       4    None
4  None       5    None


df2 = pd.DataFrame({'Content':['boy or girl','school @ morn', 
                               '1 school @ morn', 'Pechi']})
print (df2)
           Content
0      boy or girl
1    school @ morn
2  1 school @ morn
3            Pechi

#reshape df1
df1.columns = [np.arange(len(df1.columns)), df1.columns]
df11 = df1.unstack()
          .reset_index(level=2,drop=True)
          .rename_axis(('col_order','col_name'))
          .dropna()
          .reset_index(name='val')
print (df11)
   col_order col_name  val
0          0    Alpha  and
1          0    Alpha   or
2          1  Numeric    1
3          1  Numeric    2
4          1  Numeric    3
5          1  Numeric    4
6          1  Numeric    5
7          2  Special    @
8          2  Special    $
9          2  Special    &

#split column by whitespaces, reshape
df22 = df2['Content'].str.split(expand=True)
                     .stack()
                     .rename('val')
                     .reset_index(level=1,drop=True)
                     .rename_axis('idx').reset_index()
print (df22)
    idx     val
0     0     boy
1     0      or
2     0    girl
3     1  school
4     1       @
5     1    morn
6     2       1
7     2  school
8     2       @
9     2    morn
10    3   Pechi

#left join dataframes, remove non match values by dropna
#also for multiple match get always first - use sorting with drop_duplicates
df = pd.merge(df22, df11, on='val', how='left')
       .dropna(subset=['col_name'])
       .sort_values(['idx','col_order'])
       .drop_duplicates(['idx'])

#if necessary get values from df2
#if no value matched add Other category
df = pd.concat([df2, df.set_index('idx')], axis=1)
       .fillna({'col_name':'Other'})[['val','col_name','Content']]
print (df)
   val col_name          Content
0   or    Alpha      boy or girl
1    @  Special    school @ morn
2    1  Numeric  1 school @ morn
3  NaN    Other            Pechi

编辑：

df1 = pd.DataFrame({'Alpha':['and','or', None, None,None],
                    'Numeric':['1','2','3','4','5'],
                    'Special':['@','$','&', None, None]})


df2 = pd.DataFrame({'Content':['boy OR girl','school @ morn', 
                               '1 school @ morn', 'Pechi']})

#If df1 Alpha values are not lower
#df1['Alpha'] = df1['Alpha'].str.lower()
df1.columns = [np.arange(len(df1.columns)), df1.columns]

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())

#convert columns values to lower to new column
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']])
print (df)
   val col_name          Content
0   OR    Alpha      boy OR girl
1    @  Special    school @ morn
2    1  Numeric  1 school @ morn
3  NaN    Other            Pechi

使用python pandas在两个DataFrame之间搜索关键字

1 个答案: