在python中使用re findall返回任意数量的匹配组

Question

I have a relatively complex string that contains a bunch of data. I am trying to extract the relevant pieces of the string using a regex command. The portions I am interested in are contained in square brackets, like this:

s = '"data":["value":3.44}] lol haha "data":["value":55.34}] 
                "data":["value":2.44}] lol haha "data":["value":56.34}]'

And the regex expression I have built is as follows:

l = re.findall(r'\"data\"\:.*(\[.*\])', s)

I was expecting this to return

['["value":3.44}]', '["value":55.34}]', '["value":2.44}]', '["value":56.34}]']

But instead all I get is the last one, ie,

['["value":56.34}]']

How can I catch 'em all?

Answer 1

It's because quantifiers are greedy by default. So .* will match everything between the first "data": and the last [ , so there's only one [...] left to match.

Use non-greedy quantifiers by adding ? .

l = re.findall(r'\"data\"\:.*?(\[.*?\])', s)

Answer 2

You can also use finditer to extract the relevant content iteratively:

import re

s = '"data":["value":3.44}] lol haha "data":["value":55.34}] "data":["value":2.44}] lol haha "data":["value":56.34}]'
for m in re.finditer(r'(\[.*?\])', s):
    print m.group(1)

OUTPUT

["value":3.44}]
["value":55.34}]
["value":2.44}]
["value":56.34}]