CODEWITHSUNDEEP
stringpandas
ducvu169
ducvu169

Reputation: 103

Efficient way to get email name

I had a task of extracting the email name from fullname as below ( and my code to deal with it)

list_=['peter mary','peter mary david ','pop david','peter pop ronaldo bulma']
sr=pd.Series(list_,range(4))
sr_split=sr.str.split()
first_name=sr_split.str[-1]
other_name=sr_split.str[0:-1]
other_name=other_name.str.join(' ')
other_=other_name.str.split(expand=True)
other_.fillna(' ',inplace=True)
add_name=other_[0].str[0]+other_[1].str[0]+other_[2].str[0]
email_name=pd.concat([first_name,add_name],axis=1)
email_name[2]=email_name[0]+email_name[1]

My code returned:

maryp, davidpm,davidp,bulmappr

It worked as my expectation.However,there are 2 problems with my code: 1) Use a lot of split and join 2) Can not work with long name, say : name with 10 words

Is it a way to make it better?

Upvotes: 1

Views: 71

Answers (2)

Zero
Zero

Reputation: 76947

How about using apply and string methods?

In [469]: (sr.str.strip().str.split(' ')
             .apply(lambda x: x[-1] + ''.join([y[0] for y in x[:-1]]))
          )
Out[469]:
0       maryp
1     davidpm
2      davidp
3    bulmappr
dtype: object

Upvotes: 1

greg_data
greg_data

Reputation: 2293

This should do it.

list_=['peter mary','peter mary david ','pop david','peter pop ronaldo bulma']


def mk_email_name(x):
    #get names
    names = x.split(' ')
    if len(names) == 1:
        return x
    else:
        #get first name
        fn = names[0]
        #get last_names
        lns = ''.join(map(lambda y:y[:1], names[1:]))
    email = '%s%s' % (fn, lns)
    return email


#apply without pandas
print map(mk_email_name, list_)
# ['peterm', 'petermd', 'popd', 'peterprb']
#apply on pandas df
df = pd.DataFrame(list_, columns=['full_name'])
df['email'] = df.full_name.apply(mk_email_name)

Upvotes: 0

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