Lets say I have a datavframe:

I want to print every possible combination from the two lists on a new row. The output should look something like:

1A
2A 
2B
3A 
3B 
3C

CodePudding user response：

Better use pure python here:

from itertools import product
out = [''.join(x) for a,b in zip(df['Column1'], df['Column2'])
       for x in product([str(a)], b.split(', '))]

output: ['1A', '2A', '2B', '3A', '3B', '3C']

Classical loop:

for a,b in zip(df['Column1'], df['Column2']):
    for x in product([str(a)], b.split(', ')):
        print(''.join(x))

output:

1A
2A
2B
3A
3B
3C

CodePudding user response：

Let's try split Column2 into list then explode Column2. At last join two columns and convert to list.

out = (df.assign(Column2=df['Column2'].str.split(', '))
       .explode('Column2')
       # [['Column1', 'Column2']] # uncomment this line if there are more than two target columns
       .astype(str)
       .agg(''.join, axis=1)
       .tolist())

print(out)

['1A', '2A', '2B', '3A', '3B', '3C']

CodePudding user response：

Another possible solution:

df.apply(lambda x: [str(x.Column1)   item for item in x.Column2.split(', ')], axis=1).explode()

CodePudding user response：

Another way to use itertools.product:

from itertools import product
import re
out = df.apply(
    lambda x: list(product([x["Column1"]], list(re.sub("\s*,\s*", "", x["Column2"])))),
    axis=1,
)
out = [b for a in out for b in a]

print(out):

[(1, 'A'), (2, 'A'), (2, 'B'), (3, 'A'), (3, 'B'), (3, 'C')]