Is it possible to get an array as one matching group in Python regex

Question

I have the following string:

Members: {'name': A, 'age': 30, 'gender': M, 'height': 1.56}, {'name': C, 'age': 20, 'gender': M, 'height': 1.8}, {'name': H, 'age': 45, 'gender': M, 'height': 1.97}, {'name': D, 'age': 23, 'gender': M, 'height': 1.68}; Place: 1//Members: {'name': S, 'age': 33, 'gender': M, 'height': 1.4}, {'name': C, 'age': 19, 'gender': M, 'height': 1.67}, {'name': A, 'age': 44, 'gender': M, 'height': 1.92}, {'name': C, 'age': 33, 'gender': M, 'height': 1.57}; Place: 2

I would like to know if it is possible/how to have a match like:

[['30', '20', '45', '23', '1'], ['33', '19', '44', '33', '2']] or

[(['30', '20', '45', '23'], '1'), (['33', '19', '44', '33'], '2')]

Or something similar (the resulting structure doesn't really matter) I just need to have all the ages from one Place. I know that I can iterate doing split and apply regex for each part or similar solution, but my question is if there's a way to do it once (ONE single step) using regex...

I would use findall to get all the "full matches". My issue is to get the first parameter of the "tuple" as an array...

If I do:

r = re.compile("'age': (\d+).*?; Place: (\d+).*?//")
g = r.findall("Members: {'name': A, 'age': 30, 'gender': M, 'height': 1.56}, {'name': C, 'age': 20, 'gender': M, 'height': 1.8}, {'name': H, 'age': 45, 'gender': M, 'height': 1.97}, {'name': D, 'age': 23, 'gender': M, 'height': 1.68}; Place: 1//Members: {'name': S, 'age': 33, 'gender': M, 'height': 1.4}, {'name': C, 'age': 19, 'gender': M, 'height': 1.67}, {'name': A, 'age': 44, 'gender': M, 'height': 1.92}, {'name': C, 'age': 33, 'gender': M, 'height': 1.57}; Place: 2")

I am only able to get the first age, and then the place...

g
[('30', '1')]

Answer 1

As far as I know RegEx is not powerful enough to store the hits of one capturing group with a quantifier in a list, followed by another capturing group.

The following does only perform one RegEx search, and one loop, but I admit it isn't very pretty.

import re

r = re.compile("(age|Place)'?: (\d+)")

g = r.finditer("Members: {'name': A, 'age': 30, 'gender': M, 'height': 1.56}, {'name': C, 'age': 20, 'gender': M, 'height': 1.8}, {'name': H, 'age': 45, 'gender': M, 'height': 1.97}, {'name': D, 'age': 23, 'gender': M, 'height': 1.68}; Place: 1//Members: {'name': S, 'age': 33, 'gender': M, 'height': 1.4}, {'name': C, 'age': 19, 'gender': M, 'height': 1.67}, {'name': A, 'age': 44, 'gender': M, 'height': 1.92}, {'name': C, 'age': 33, 'gender': M, 'height': 1.57}; Place: 2")

ages = []
ranks = {}
for m in g:
  if m[1] == 'age':
    ages.append(m[2])
  else:
    ranks[m[2]] = ages
    ages = []

print(ranks)

Basically just capture any age or Place, iterate over the matches. Store all ages into a list until we come across a Place, in which case we use the former list as a value and the Place as a key in a dictionary. Then we reset the list and start over.

Of course the caveat is that Place always comes after the ages.

Answer 2

Here's a way to get close to a solution using re.findall and itertools.groupby:

import re, itertools
r = re.compile(r'(?:\b(?:age|place)\'?\s*:\s*(\d+))|//|\Z', re.I)
x = r.findall("Members: {'name': A, 'age': 30, 'gender': M, 'height': 1.56}, {'name': C, 'age': 20, 'gender': M, 'height': 1.8}, {'name': H, 'age': 45, 'gender': M, 'height': 1.97}, {'name': D, 'age': 23, 'gender': M, 'height': 1.68}; Place: 1//Members: {'name': S, 'age': 33, 'gender': M, 'height': 1.4}, {'name': C, 'age': 19, 'gender': M, 'height': 1.67}, {'name': A, 'age': 44, 'gender': M, 'height': 1.92}, {'name': C, 'age': 33, 'gender': M, 'height': 1.57}; Place: 2")

Output:

['30', '20', '45', '23', '1', '', '33', '19', '44', '33', '2', '']

Splitting with a second pass:

o = [list(g[1]) for g in itertools.groupby(x, lambda i: i != '')][::2]

Output:

[['30', '20', '45', '23', '1'], ['33', '19', '44', '33', '2']]