如何在条件变化的函数中匹配DataFrame列上的值？

Question

I have one panda DataFrame match with NaN values in Col 3 and i'm trying to replace it with the correct value present only once in the Col 3 in function of the associated string in Col 2 :

What I have:

Col 1 | Col 2 | Col 3

1 | 'GE1' | '1.2'

2 | 'GE1' | NaN

3 | 'GE1' |NaN

4 |'GE2' |'1.3'

5 |'GE2'|NaN

I've tried this : (match_without_nan is the same DF than match but with only the rows without NaN)

`` `

mylist=[[match_without_nan['Col 2'],match_without_nan['Col 3']]]

for i in range(0,len(mylist)):
    match.loc[match['Col 2'] == mylist[0][i], 'Col 3'] = mylist[1][i]

` ``

This code is running without any errors but nothing changed in Col 3.

What I would like to get :

Col 1 | Col 2 | Col 3

1 | 'GE1' | '1.2'

2 | 'GE1' | '1.2'

3 | 'GE1' | '1.2'

4 |'GE2' |'1.3'

5 |'GE2'|'1.3'

Answer 1

I don't understand fully your code, and why it doesn't work, but i would just use "sql-like" operators to achieve your goal:

df = pd.DataFrame({
  "col2": ['GE1', 'GE1', 'GE1', 'GE2', 'GE2'],
  "col3": [np.nan, '1.2', np.nan, '1.3', np.nan]
})

unique_values = df.groupby("col2").first()
final_df = df.join(unique_values, on="col2", lsuffix="_orig")

The groupby will extract the mapping from your col2 keys to your unique values present in col3, and the join will generate the final table you want.

Answer 2

IIUC, you want to do ffill with groupby :

df['Col 3'] = df.groupby('Col 2')['Col 3'].transform(lambda x: x.ffill())
print(df)

Output:

   Col 1  Col 2  Col 3
0      1  'GE1'  '1.2'
1      2  'GE1'  '1.2'
2      3  'GE1'  '1.2'
3      4  'GE2'  '1.3'
4      5  'GE2'  '1.3'