I have the following text
Stat indicator : 01245 Values loaded
Some irrelevant data
Stat indicator : 13452 Values loaded
My output should be giving me numbers 01245, 13452
Heres what I have tried
with open('test.txt') as fd:
data = fd.read()
val_to_pattern = {'stat': r'Stat indicator : .{5}\n'}
val_dict = {}
for key, patt in val_to_pattern.items():
val_dict[key] = re.findall(patt, data)
Use
val_to_pattern = {'stat': r'Stat indicator\s*:\s*(\d+)'}
Note that \s*:\s* pattern matches a : optionally wrapped with 0+ whitespaces and the (\d+) part matches and captures into Group 1 any 0+ digits (note that re.findall only returns captured substrings if capturing groups are set in the regex).
See the regex demo (the green text will be the result).
Alternatively, if the number of digits in the indicator stats is always 5 use either of
val_to_pattern = {'stat': r'Stat indicator\s*:\s*(\d{5})\b'}
val_to_pattern = {'stat': r'Stat indicator\s*:\s*(\d{5})(?!\d)'}
\b is word boundary that will require a non-word char or end of string after 5 digits and the (?!\d) is a negative lookahead that fails the match if there is a digit immediately to the right of the current location.
You can used regex \d, and if you know the size of your number you can use {size} with it.
\d{3} Returns a match where the string contains 3 digits (numbers from 0-9)
\d{3,5} Returns a match where the string contains 3 to 5 digits (numbers from 0-9)
So you can used r'\d{3,5}' as regex pattern
import re
with open('test.txt') as fd:
data = fd.read()
val_to_pattern = {'stat': r'\d{5}'}
val_dict = {}
for key, patt in val_to_pattern.items():
re_find = re.findall(patt, data)
val_dict[key] = re.findall(patt, data)
If you don't want to use regex, you could implement a straightforward loop.
1 with open('test.txt') as fd:
2 data = fd.read()
3
4 nums = []
5 temp = ""
6 is_dig = False
7 for char in data:
8 if char.isdigit():
9 temp += char
10 is_dig = True
11 elif is_dig:
12 nums.append(temp)
13 temp = ""
14 is_dig = False
15 print(nums)