We can leverage the point that True is 1 and False is 0 when used in multiplication.

3 * True  -> 3
3 * False -> 0

We have to check if values are equal to online in the status column.

df['cost'] = df['price'] * df['status'].eq('online') * metric

Wherever, status is offline cost value is 0.

The above solution relies on the fact you want to set offline values to 0. If you want to set offline to let's 999. Then we can use Series.where here.

df['cost'] = df['price'].mul(metric).where(df['status'].eq('online'), 999)

Now, every offline value to set to 999.

Useful links:

• Series.where
• Series.eq
• Series.mul
• Multiplying boolean with float

You can use np.where:

import numpy as np
df['cost'] = np.where(df['status']=='online', df['price']*metric, 0)

If you want plain pandas you can use loc:

df['cost'] = 0
df.loc[df['status']== "online" , 'cost'] = df['price'] * metric

Or with a list comprehension:

df['cost'] = [df['price']*metric if status == 'online' else 0 for status in df['status']]