Follow

Keep Up to Date with the Most Important News

By pressing the Subscribe button, you confirm that you have read and are agreeing to our Privacy Policy and Terms of Use
Contact

Applying Functions in Python

I am an R User that is trying to learn more about Python.

I found this Python library that I would like to use for address parsing: https://github.com/zehengl/ez-address-parser

I was able to try an example over here:

MEDevel.com: Open-source for Healthcare and Education

Collecting and validating open-source software for healthcare, education, enterprise, development, medical imaging, medical records, and digital pathology.

Visit Medevel

from ez_address_parser import AddressParser

ap = AddressParser()

result = ap.parse("290 Bremner Blvd, Toronto, ON M5V 3L9")
print(results)
[('290', 'StreetNumber'), ('Bremner', 'StreetName'), ('Blvd', 'StreetType'), ('Toronto', 'Municipality'), ('ON', 'Province'), ('M5V', 'PostalCode'), ('3L9', 'PostalCode')]

I have the following file that I imported:

df = pd.read_csv(r'C:/Users/me/OneDrive/Documents/my_file.csv',  encoding='latin-1')

   name                               address
1 name1 290 Bremner Blvd, Toronto, ON M5V 3L9
2 name2 291 Bremner Blvd, Toronto, ON M5V 3L9
3 name3 292 Bremner Blvd, Toronto, ON M5V 3L9

I tried to apply the above function and export the file:

df['Address_Parse'] = df['ADDRESS'].apply(ap.parse)

df = pd.DataFrame(df)
df.to_csv(r'C:/Users/me/OneDrive/Documents/python_file.csv', index=False, header=True)

This seems to have worked – but everything appears to be in one line!

[('290', 'StreetNumber'), ('Bremner', 'StreetName'), ('Blvd', 'StreetType'), ('Toronto', 'Municipality'), ('ON', 'Province'), ('M5V', 'PostalCode'), ('3L9', 'PostalCode')]

Is there a way in Python to make each of these "elements" (e.g. StreetNumber, StreetName, etc.) into a separate column?

Thank you!

>Solution :

Define a custom function that returns a Series and join the output:

def parse(x):
    return pd.Series({k:v for v,k in ap.parse(x)})

out = df.join(df['ADDRESS'].apply(parse))

print(out)
Add a comment

Leave a Reply

Keep Up to Date with the Most Important News

By pressing the Subscribe button, you confirm that you have read and are agreeing to our Privacy Policy and Terms of Use

Discover more from Dev solutions

Subscribe now to keep reading and get access to the full archive.

Continue reading