用于有条件地捕获逗号分隔字符串的 Python 正则表达式

Question

I have a list of person names which can have 3 different styles:

1. {last name}, {first name} {middle name} (Example: Bob, Dylan Tina)
2. {last name}, {first name} {middle initial}. (Example: Bob, Dylan T.)
3. {last name}, {first name} (Example: Bob, Dylan)

And this is the regex which I wrote:

^[a-zA-Z]+(([' ,.-][a-zA-Z ])?[a-zA-Z]*)*$

But it doesn't work.

Answer 1

You could write the regex like this

^(\w+),\s(\w+)\s*(\w*\.?)$

Here is the demo .

Update the regex to like this and you can get three different groups for your three cases

^(\w+,\s\w+\s\w+)$|^(\w+,\s\w+\s\w+\.)$|^(\w+,\s\w+)$

Here is the demo .

Here is the python code

import re
s2 = "Bob, Dylan"
out = re.findall(r"^(\w+),\s(\w+)\s*(\w*\.?)$",s2)
print(out)

OUTPUT

[('Bob', 'Dylan', '')]

Answer 2

You should use this regex:

(\w+),\s*(\w+)\s*(\w{0,}\.*)

This is the result you'll get:

>>> import re
>>> s1 = "Bob, Dylan Tina"
>>> s2 = "Bob, Dylan"
>>> s3 = "Bob, Dylan T."
>>> p = re.compile(r"(\w+),\s*(\w+)\s*(\w{0,}\.*)")
>>> re.findall(p, s1)
[('Bob', 'Dylan', 'Tina')]
>>> re.findall(p, s2)
[('Bob', 'Dylan', '')]
>>> re.findall(p, s3)
[('Bob', 'Dylan', 'T.')]