Extracting an element of a list in a pandas column

Question

I have a DataFrame that contains a list on each column as shown in the example below with only two columns.

    Gamma   Beta
0   [1.4652917656926299, 0.9326935235505321, float] [91, 48.611034768515864, int]
1   [2.6008354611105995, 0.7608529935313189, float] [59, 42.38646954167245, int]
2   [2.6386970166722348, 0.9785848171888037, float] [89, 37.9011122659478, int]
3   [3.49336632573625, 1.0411524946972244, float]   [115, 36.211134224288344, int]
4   [2.193991200007534, 0.7955134305428825, float]  [128, 50.03563864975485, int]
5   [3.4574527664490997, 0.9399880977511021, float] [120, 41.841146628802875, int]
6   [3.1190582380554863, 1.0839109431114795, float] [148, 55.990072419824514, int]
7   [2.7757359940789916, 0.8889801332053203, float] [142, 51.08885697101243, int]
8   [3.23820908493237, 1.0587479742892683, float]   [183, 43.831293356668425, int]
9   [2.2509032790941985, 0.8896196407231622, float] [66, 35.9377662201882, int]

I'd like to extract for every column the first position of the list on each row to get a DataFrame looking as follows.

    Gamma   Beta
0   1.4652917656926299  91
1   2.6008354611105995  59
2   2.6386970166722348  89
...

Up to now, my solution would be like [row[1][0] for row in df_params.itertuples()], which I could iterate for every column index of the row and then compose my new DataFrame.

An alternative is new_df = df_params['Gamma'].apply(lambda x: x[0]) and then to iterate to go through all the columns.

My question is, is there a less cumbersome way to perform this operation?

Answer 1

You can use the str accessor for lists, e.g.:

df_params['Gamma'].str[0]

This should work for all columns:

df_params.apply(lambda col: col.str[0])

Answer 2

Itertuples would be pretty slow. You could speed this up with the following:

for column_name in df_params.columns:
    df_params[column_name] = [i[0] for i in df_params[column_name]]