Replace values in pandas column with default value for missing keys

Question

I have multiple simple functions that need to be implemented on every row of certain columns of my dataframe. The dataframe is very like, 10 million+ rows. My dataframe is something like this:

Date      location   city        number  value
12/3/2018   NY       New York      2      500
12/1/2018   MN       Minneapolis   3      600
12/2/2018   NY       Rochester     1      800
12/3/2018   WA       Seattle       2      400

I have functions like these:

def normalized_location(row):
    if row['city'] == " Minneapolis":
        return "FCM"
    elif row['city'] == "Seattle":
        return "FCS"
    else:
        return "Other"

and then I use:

df['Normalized Location'] =df.apply (lambda row: normalized_location (row),axis=1)

This is extremely slow, how can I make this more efficient?

Answer 1

We can make this BLAZING fast using map with a defaultdict.

from collections import defaultdict

d = defaultdict(lambda: 'Other')
d.update({"Minneapolis": "FCM", "Seattle": "FCS"})

df['normalized_location'] = df['city'].map(d)

print(df)
        Date location         city  number  value normalized_location
0  12/3/2018       NY     New York       2    500               Other
1  12/1/2018       MN  Minneapolis       3    600                 FCM
2  12/2/2018       NY    Rochester       1    800               Other
3  12/3/2018       WA      Seattle       2    400                 FCS

...to circumvent a fillna call, for performance reasons. This approach generalises to multiple replacements quite easily.