CODEWITHSUNDEEP
pythonpandasstring-parsing
Kalimantan
Kalimantan

Reputation: 771

Keeping domain of Email but removing TLD

I am using python and I want to be able to keep the domain of the email but remove the 'com', or '.co.uk', or 'us', etc

So basically if I have an email, say [email protected]. I want to have only @gmail left in string format, but I want to do this for any email. So [email protected] would leave me with @yahoo, or [email protected], would leave me with @aol

so far I have:

 domain = re.search("@[\w.]+", val)
 domain = domain.group()

That returns the domain but with the TLD . So @gmail.com, or @aol.co

Upvotes: 3

Views: 1161

Answers (4)

Lex Scarisbrick
Lex Scarisbrick

Reputation: 1570

For posterity and completeness, this can also be done via index and slice:

email = '[email protected]'
at = email.index('@')
dot = email.index('.', at)
domain = email[at:dot]

Using split()and re seems like overkill when the goal is to extract a single sub-string.

Upvotes: 0

jezrael
jezrael

Reputation: 863166

With pandas functions use split:

df = pd.DataFrame({'a':['[email protected]','[email protected]','[email protected]']})

print (df)
                  a
0  [email protected]
1     [email protected]
2  [email protected]

print ('@' + df.a.str.split('@').str[1].str.split('.', 1).str[0] )
0    @yahoo
1      @aol
2      @aol
Name: a, dtype: object

But faster is use apply, if in column are not NaN values:

df = pd.concat([df]*10000).reset_index(drop=True)

print ('@' + df.a.str.split('@').str[1].str.split('.', 1).str[0] )
print (df.a.apply(lambda x: '@' + x.split('@')[1].split('.')[0]))

In [363]: %timeit ('@' + df.a.str.split('@').str[1].str.split('.', 1).str[0] )
10 loops, best of 3: 79.1 ms per loop

In [364]: %timeit (df.a.apply(lambda x: '@' + x.split('@')[1].split('.')[0]))
10 loops, best of 3: 27.7 ms per loop

Another solution with extract is faster as split, it can be used if NaN values in column:

#not sure with all valid characters in email address
print ( '@' + df.a.str.extract(r"\@([A-Za-z0-9_]+)\.", expand=False))
In [365]: %timeit ( '@' + df.a.str.extract(r"\@([A-Za-z0-9 _]+)\.", expand=False))
10 loops, best of 3: 39.7 ms per loop

Upvotes: 1

joel goldstick
joel goldstick

Reputation: 4493

First split on "@", take the part after "@". Then split on "." and take the first part

email = "[email protected]"
'@' + email.split("@")[1].split(".")[0]
'@gmail'

Upvotes: 2

Scott Stainton
Scott Stainton

Reputation: 394

If you do

val = string.split('@')[1].split('.')[0]

Change 'string' for your email string variable name.

This will take everything after the '@' symbol, then everything up to the first '.'

Using on '[email protected]' gives 'gmail'

If you require the '@' symbol you can add it back with;

full = '@' + val

Upvotes: 3

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