Capturing emails with regex in Python

Question

I will be gathering scattered emails from a larger CSV file. I am just now learning regex. I am trying to extract the emails from this example sentence. However, emails is populating with only the @ symbol and the letter immediately before that. Can you help me see what's going wrong?

import re

String = "'Jessica's email is [email protected], and Daniel's email is [email protected]. Edward's is [email protected], and his grandfather, Oscar's, is [email protected].'"

emails = re.findall(r'.[@]', String)
names = re.findall(r'[A-Z][a-z]*',String)

print(emails)
print(names)

Answer 1

your regex e-mail is not working at all: emails = re.findall(r'.[@]', String) matches anychar then @.

I would try a different approach: match the sentences and extract name,e-mails couples with the following empiric assumptions (if your text changes too much, that would break the logic)

• all names are followed by 's" and is somewhere (using non-greedy .*? to match all that is in between
• \w matches any alphanum char (or underscore), and only one dot for domain (else it matches the final dot of the sentence)

code:

import re

String = "'Jessica's email is [email protected], and Daniel's email is [email protected]. Edward's is [email protected], and his grandfather, Oscar's, is [email protected].'"

print(re.findall("(\w+)'s.*? is (\w+@\w+\.\w+)",String))

result:

[('Jessica', '[email protected]'), ('Daniel', '[email protected]'), ('Edward', '[email protected]'), ('Oscar', '[email protected]')]

converting to dict would even give you a dictionary name => address:

{'Oscar': '[email protected]', 'Jessica': '[email protected]', 'Daniel': '[email protected]', 'Edward': '[email protected]'}

The general case needs more chars (not sure I'm exhaustive):

String = "'Jessica's email is [email protected], and Daniel's email is [email protected]. Edward's is [email protected], and his grandfather, Oscar's, is [email protected].'"

print(re.findall("(\w+)'s.*? is ([\w\-.]+@[\w\-.]+\.[\w\-]+)",String))

result:

[('Jessica', '[email protected]'), ('Daniel', '[email protected]'), ('Edward', '[email protected]'), ('Oscar', '[email protected]')]

Answer 2

1. Emails

    In [1382]: re.findall(r'\S+@\w+\.\w+', text)
    Out[1382]: 
    ['[email protected]',
     '[email protected]',
     '[email protected]',
     '[email protected]']

How it works: All emails are [email protected]. One thing to note is a bunch of characters surrounding @, and the singular .. So, we use \S to demarcate anything that is not a whitespace. And + is to search for 1 or more such characters. \w+\.\w+ is just a fancy way of saying search for a string that only has one . in it.

---

2. Names

    In [1375]: re.findall('[A-Z][\S]+(?=\')', text)
    Out[1375]: ['Jessica', 'Daniel', 'Edward', 'Oscar']

How it works: Any word starting with an upper case. The (?=\') is a lookahead. As you see, all names follow the pattern Name's. We want everything before the apostrophe. Hence, the lookahead, which is not captured.

---

Now, if you want to map names to emails by capturing them together with one massive regex, you can. Jean-François Fabre's answer is a good start. But I recommend getting the basics down par first.

Answer 3

You need to find anchors, patterns to match. An improved pattern could be:

import re

String = "'Jessica's email is [email protected], and Daniel's email is 
[email protected]. Edward's is [email protected], and his 
grandfather, Oscar's, is [email protected].'"

emails = re.findall(r'[a-zA-Z0-9_.+-]+@[a-zA-Z0-9-]+\.[a-zA-Z0-9-.]+', String)
names = re.findall(r'[A-Z][a-z]*', String)

print(emails) 
print(names)

\w+ is missing '-' which are allowed in email adresses.

Answer 4

This is because you are not using the repeat operator. The below code uses the + operator which means the characters / sub patterns just before it can repeat 1 to many times.

s = '''Jessica's email is [email protected], and Daniel's email is [email protected]. Edward's is [email protected], and his grandfather, Oscar's, is [email protected].'''

p = r'[a-z0-9]+@[a-z]+\.[a-z]+'
ans = re.findall(p, s)

print(ans)