Split two columns into multiple columns in a dataframe

Just split each cell and concat the result:

df1 = df["D1"].str.split(" ", expand=True)
df2 = df["D2"].str.split(" ", expand=True)
new_df = pd.concat([df1, df2], axis=1, ignore_index=True)
new_df.columns = ['A0', 'A1', 'A2', 'A3', 'A4', 'A5', 'A6', 'A7', 'A8', 'A9', 'A10']
print(new_df)

Output:

    A0   A1 A2  A3  A4  A5  A6    A7    A8    A9   A10
0  0329  000  8  d7  a7  7f  8c    11    2f    00    10
1  0080  000  8  00  17  ea  0a    20    1a    20    43
2  0081  000  8  7f  84  60  00    00    00    00    53
3  0120  000  4  00  00  00  00  None  None  None  None
4  0153  000  8  00  80  10  ff    00    ff    40    ce

Here is another way even removing some leading zero's.

import numpy as  np
import pandas as pd

data = {
    "D1" : ["0329 000", "0080 000", "0081 000", "0120 000", "0153 000"],
    "D2" : ["8 d7 a7 7f 8c 11 2f 00 10", "8 00 17 ea 0a 20 1a 20 43", "8 7f 84 60 00 00 00 00 53", "4 00 00 00 00", "8 00 80 10 ff 00 ff 40 ce"],
}

df = pd.DataFrame(data)

df2 = pd.DataFrame(columns=['A'+str(i) for i in range(11)])

df2['A0'] = df['D1'].str.split(' ', expand=True)[0]
df2['A1'] = df['D1'].str.split(' ', expand=True)[1]
df2['A2'] = df['D2'].str.split(' ', expand=True)[0]
df2['A3'] = df['D2'].str.split(' ', expand=True)[1]
df2['A4'] = df['D2'].str.split(' ', expand=True)[2]
df2['A5'] = df['D2'].str.split(' ', expand=True)[3]
df2['A6'] = df['D2'].str.split(' ', expand=True)[4]
df2['A7'] = df['D2'].str.split(' ', expand=True)[5]
df2['A8'] = df['D2'].str.split(' ', expand=True)[6]
df2['A9'] = df['D2'].str.split(' ', expand=True)[7]
df2['A10'] = df['D2'].str.split(' ', expand=True)[8]

df2 = df2.apply(lambda x: x.astype(int) if x.name in ['A0', 'A1'] else x)

Output:

    A0  A1 A2  A3  A4  A5  A6    A7    A8    A9   A10
0  329   0  8  d7  a7  7f  8c    11    2f    00    10
1   80   0  8  00  17  ea  0a    20    1a    20    43
2   81   0  8  7f  84  60  00    00    00    00    53
3  120   0  4  00  00  00  00  None  None  None  None
4  153   0  8  00  80  10  ff    00    ff    40    ce

I can't seem to figure out the double zeros 00 and changing it to a single zero 0.

Hope this helps.

using str.strip():

tmp = df['D1'].str.cat(df['D2'], sep=' ')
out = tmp.str.split(expand=True).add_prefix('A')

output :

>>
     A0   A1 A2  A3  A4  A5  A6    A7    A8    A9   A10
0  0329  000  8  d7  a7  7f  8c    11    2f    00    10
1  0080  000  8  00  17  ea  0a    20    1a    20    43
2  0081  000  8  7f  84  60  00    00    00    00    53
3  0120  000  4  00  00  00  00  None  None  None  None
4  0153  000  8  00  80  10  ff    00    ff    40    ce

You can use chain.from_iterable and pandas.apply with axis=1 and at the end use pandas.applymap for apply function for converting each value to int.

from itertools import chain

# We can use the below function for converting all values to 'int' if we can
def to_int(x):
    try:
        return int(x)
    except ValueError:
        return x

df = df.apply(lambda row: 
         pd.Series(chain.from_iterable(row.str.split())), axis=1
        ).add_prefix('A'
                    ).applymap(to_int)
print(df)

    A0  A1  A2  A3  A4  A5  A6    A7   A8    A9  A10
0  329   0   8  d7  a7  7f  8c  11.0   2f   0.0   10
1   80   0   8   0  17  ea  0a  20.0   1a  20.0   43
2   81   0   8  7f  84  60   0   0.0    0   0.0   53
3  120   0   4   0   0   0   0   NaN  NaN   NaN  NaN
4  153   0   8   0  80  10  ff   0.0   ff  40.0   ce

What is chain.from_iterable?

from itertools import chain

list(chain.from_iterable([
    [1,2],
    [3,4]
]))

# [1, 2, 3, 4]