在熊猫中做到这一点的最有效方法是什么

Question

I'm trying to figure out the most efficient way to join two dataframes such as below.

I've tried pd.merge and maybe using the rank function but cannot seem to figure a way.

Thanks in advance

df1

| A        | B              | C          |
| -------- | -------------- |------------|
| TBK1     | 2022-01-01     |2022-04-04  |
| TBK1     | 2022-02-02     |2021-01-09  | 
| TBK3     | 2022-05-07     |2023-02-04  |

What I'm trying to achieve is this

df2

| A        | B              | C          | D              | E          |
| -------- | -------------- |------------|----------------|------------|
| TBK1     | 2022-01-01     |2022-04-04  | 2022-02-02     |2021-01-09  | 
| TBK3     | 2022-05-07     |2023-02-04  |NaN             |NaN         |

Answer 1

You might want to use groupby with unstack as advised in this answer :

import pandas as pd
from string import ascii_uppercase

# Reproduce the data
df = pd.DataFrame()
df['A'] = ['TBK1','TBK1', 'TBK3']
df['B'] = ['2022-01-01' , '2022-02-02', '2022-05-07']
df['C'] = ['2022-04-04', '2021-01-09', '2023-02-04']

# Count how many rows exists per unique entry
s = df.groupby(['A']).cumcount() 
# Unstack
df1 = df.set_index(['A', s]).unstack().sort_index(level=1, axis=1)
# Rename columns
df1.columns = [l for l in ascii_uppercase[1:len(df1.columns)+1]]
# Flatten columns names (aesthetics)
df1 = df1.reset_index()

print(df1)

      A           B           C           D           E
0  TBK1  2022-01-01  2022-04-04  2022-02-02  2021-01-09
1  TBK3  2022-05-07  2023-02-04         NaN         NaN