Writing a more concise lambda in Python

Question

I am trying to work with Pandas on some data problems and have come to a point where I am writing code like this:

groups.segment = groups.topic.map(lambda x: 'friends' if 'friend' in str(x) else x)
groups.segment = groups.topic.map(lambda x: 'friends' if 'bro' in str(x) else x)
groups.segment = groups.topic.map(lambda x: 'friends' if 'girls' in str(x) else x)

I would like to write it in a more concise way way where I don't have to have a bunch of copy and paste code. A little bit new to python so not sure how to make it better. Any help appreciated.

Something along the lines of:

groups.segment = groups.segment.map(lambda x: 'friends' if 'bro' or 'girls' or 'friend' in str(x) else x)

Is there a way to do this ?

Thanks for any help !

Look into the any built-in function.

jonrsharpe
– jonrsharpe

2014-11-18 23:24:28 +00:00
Commented Nov 18, 2014 at 23:24 — jonrsharpe
– jonrsharpe, Commented Nov 18, 2014 at 23:24

Simeon Visser · Accepted Answer · 2014-11-18 23:24:56Z

5

It's probably better not to use lambda here:

def mapper(x)
    if any(y in str(x) for y in ('friend', 'bro', 'girls')):
        return 'friends'
    return x

groups.segment = groups.topic.map(mapper)

You can use any here which returns True when at least one of the passed values is True.

answered Nov 18, 2014 at 23:24

Simeon Visser

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gefei · Accepted Answer · 2014-11-18 23:23:51Z

1

lambda x: 'friends' if 'bro' in str(x) or 'girls' in str(x) or 'friend' in str(x) else x

answered Nov 18, 2014 at 23:23

gefei

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Marius · Accepted Answer · 2014-11-18 23:34:37Z

1

You can avoid map and lambda altogether and use isin(), which ends up working more like the typical Python idiom of if item in ['a', 'b', 'c']:

import pandas as pd

df = pd.DataFrame({'relationship': ['friends', 'friend', 'bro', 'girls']})
df
Out[3]: 
  relationship
0      friends
1       friend
2          bro
3        girls

is_synonym = df.relationship.isin(['friend', 'bro', 'girls'])
df['relationship2'] = df['relationship'].copy()
df.loc[is_synonym, 'relationship2'] = 'friends'
df
Out[15]: 
  relationship relationship2
0      friends       friends
1       friend       friends
2          bro       friends
3        girls       friends

edited Nov 18, 2014 at 23:34

answered Nov 18, 2014 at 23:29

Marius

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