Python pattern matching with regex, findall and groups

Question

I had a script that scanned in piece of text and returned me a group which I would save to an array. The code looks like this:

pattern = re.compile(r'<span id="first_name">(.+?)</span>')
matches = pattern.findall(str(my_text_file))

This works awesome and I could scan first names in my text file and write them into an array doing this:

for firstname in matches:
    if firstname not in list_of_names:
        list_of_names.append(firstname)

But now I need to expand my pattern to retrieve two groups instead of one, and I have no idea how I am supposed to get to the second group.

When I have something like:

pattern = re.compile(r'<span id="first_name">(.+?)</span><span id="last_name">(.+?)</span>')
matches = pattern.findall(str(my_text_file))

How am I supposed to put those second group (last names) in a different array?

What have you tried? Have you tried looking at what pattern.findall is returning? Have you read the documentation to see what it should be returning? — Chris Morgan
– Chris Morgan, Commented Nov 27, 2011 at 15:29
python documentation say : "return a list of groups; this will be a list of tuples if the pattern has more than one group. Empty matches are included in the result unless they touch the beginning of another match." so i guess I will be getting a list of tuples... but i dont understand this because I do not natively speak english.... — Synbitz Prowduczions
– Synbitz Prowduczions, Commented Nov 27, 2011 at 15:32

Cito · Accepted Answer · 2011-11-27 15:31:44Z

2

for match in matches:
    first_names.append(match[0])
    last_names.append(match[1])

answered Nov 27, 2011 at 15:31

Cito

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