如何根据另一个 dataframe 中的条件在 dataframe 中创建新列？

Question

df1:

Variables     left      right
0  AUM           -0.001    28.20
1  AUM           28.20     40.28
2  AUM           40.28     58.27
3  AUM           58.27     80.72
4  AUM           80.72     100.00
0  ABS           -88.01    200.72
1  ABS           200.72    480.72
2  ABS           480.72    800.20
0  LS            10000     200000
1  LS            200000    400000

df2:
   
    Pan_no     ABS      AUM     LS      
0   AAA        28        30    10001      
2   CCC        500       98    390000     
1   BBB        250       50    150000     
3   DDD        100       60    380000     
4   EEE         88       10    378347   
  

Conditions:

Based on the left and right value in df1, a new column should be created in df2 and the values inside the column should be the index of df1 for that particular variable.

Example: In df2 if AUM value falls under this range (-0.001 - 28.20) then the new column will have the index value of df1 as the new value. ie 0

Similarly,

In df2 if ABS value falls under this range (200.72 - 480.72) then the new column ABS_BIN will have the index value of df1 as the new value. ie 1

WHat i have tried is:

binning_vars = ['ABS','AUM','LS']
def f(row):
  for i in binning_vars:
      for j in df1[df1['Variable'] == i].index:
            if df1[i] >= df1['left'] & df1[i] >= df1['right']:
                value = j
            else:
                pass
            return value
df2[i,'_bin'] = df1.apply(f, axis=1)

but it throws an error as TypeError: unsupported operand type(s) for &: 'float' and 'float'. Any help will be much appreciated.

Expected Output:
with new columns in df2:
    
    Pan_no     ABS      AUM     LS      ABS_BIN    AUM_BIN     LS_BIN
0   AAA        28        30    10001      0          1           0
1   BBB        250       50    150000     1          2           0
2   CCC        500       98    390000     2          4           1
3   DDD        100       60    380000     0          3           1
4   EEE         88       10    378347     0          0           1

Answer 1

You can use pd.cut and avoid loops inside the binning function:

def binning(sr):
    df = df1.loc[df1['Variables'] == sr.name, ['left', 'right']]
    bins = sorted(set(df.to_numpy().ravel()))
    return pd.cut(sr, bins=bins, labels=df.index)

out = df2[binning_vars].apply(binning).add_suffix('_BIN')
df2 = pd.concat([df2, out], axis=1)

Output:

>>> df2
  Pan_no  ABS  AUM      LS ABS_BIN AUM_BIN LS_BIN
0    AAA   28   30   10001       0       1      0
2    CCC  500   98  390000       2       4      1
1    BBB  250   50  150000       1       2      0
3    DDD  100   60  380000       0       3      1
4    EEE   88   10  378347       0       0      1

Answer 2

You can use merge_asof to avoid using apply :

out = df2.merge(
 pd.merge_asof((df2.melt(id_vars='Pan_no')
                   .astype({'value': float})
                   .sort_values(by='value')
                ),
               df1.reset_index().sort_values(by='left'),
               left_by='variable', right_by='Variables',
               left_on='value', right_on='left', direction='backward')
  .pivot(index='Pan_no', columns='variable', values='index')
  .add_suffix('_BIN'),
    left_on='Pan_no', right_index=True
)

output:

  Pan_no  ABS  AUM      LS  ABS_BIN  AUM_BIN  LS_BIN
0    AAA   28   30   10001        0        1       0
2    CCC  500   98  390000        2        4       1
1    BBB  250   50  150000        1        2       0
3    DDD  100   60  380000        0        3       1
4    EEE   88   10  378347        0        0       1