Python pandas - lambda function that access other columns

Looks like a simple loc assignment

m = df['Col_1'].str.len().eq(0)
df.loc[m, 'Col_1'] = df.loc[m, 'Col_2']

Should work even without the rhs loc, since pandas does index matching by default when assigning pd.Series.

df.loc[m, 'Col_1'] = df['Col_2']

rafaelc's answer is really good and should be of favour in general.

However, in this case, a usual list comprehension works and might be faster:

df['Col_1'] = [a if a else b for a,b in zip(df['Col_1'], df['Col_2'])]

Performance:

# rafaelc's answer
%%timeit -n 100
df = pd.DataFrame({
    'Col_1':[['Apple', 'Coffee', 'Banana'], ['Apple'], [], []],
    'Col_2':[['Food'],['Drink'],['Clothes'],['Food']]
})
m = df['Col_1'].str.len().eq(0)
df.loc[m, 'Col_1'] = df.loc[m, 'Col_2']    
# 1.4 ms ± 49.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

# list comprehension
%%timeit -n 100
df = pd.DataFrame({
    'Col_1':[['Apple', 'Coffee', 'Banana'], ['Apple'], [], []],
    'Col_2':[['Food'],['Drink'],['Clothes'],['Food']]
})

df['Col_1'] = [a if a else b for a,b in zip(df['Col_1'], df['Col_2'])]
#485 µs ± 15.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

The above answers are great. Here is one more way to do this.

import pandas as pd

data = [
    [['Apple', 'Coffee', 'Banana'], ['Food']],
    [['Apple'], ['Drink']],
    [[], ['Clothes']],
    [[], ['Food']],
    [['Apple', 'Orange'], ['Food']],
    [['Apple', 'Orange'], ['Stuff', 'Food']]
]

df = pd.DataFrame.from_records(data, columns=['col1', 'col2'])


def copy_if(row):
    if len(row['col1']) == 0:
        row['col1'] += row['col2']
    return row['col1']


df['col1'] = df[['col1', 'col2']].apply(copy_if, axis=1)
print(df)
#                       col1           col2
# 0  [Apple, Coffee, Banana]         [Food]
# 1                  [Apple]        [Drink]
# 2                [Clothes]      [Clothes]
# 3                   [Food]         [Food]
# 4          [Apple, Orange]         [Food]
# 5          [Apple, Orange]  [Stuff, Food]