Return value from list according to index number

You can do a simple list comprehension:

df['B'] = [s[i] for i, s in zip(df.index, df['A'])]

Or if you want only diagonal values:

df['B'] = np.diagonal([*df['A']])

---

               A   B
0   [6, 1, 1, 1]   6
1   [1, 5, 1, 1]   5
2  [1, 1, 11, 1]  11
3  [1, 1, 1, 20]  20

Solution using numpy:

import pandas as pd 
import numpy as np

df = pd.DataFrame({'A' : [[6,1,1,1], [1,5,1,1], [1,1,11,1], [1,1,1,20]]})
df['B'] = pd.Series(np.diag(np.vstack(df['A'].to_numpy())), index=df.index)

df

I have also been able to solve my own question, but terribly. I will post the solution anyway:

df = pd.DataFrame({'A' : [[6,1,1,1], [1,5,1,1], [1,1,11,1], [1,1,1,20]]})
s = df.explode('A')
i = s.groupby(level=0).cumcount()
df['B'] = s.loc[s.index == i, 'A']

               A   B
0   [6, 1, 1, 1]   6
1   [1, 5, 1, 1]   5
2  [1, 1, 11, 1]  11
3  [1, 1, 1, 20]  20