Alphanumeric sorting in Python and negative numbers

Question

I have a fairly simple list (a number followed by a sentence), here in the right order:

-347 a negative number
-100 another negative number
-25 and again, a negative number
17 some text
25 foo bar
100 two same texts
100 two same texts (almost)
350 a positive number

I need to sort that list each time a new item is added.

I searched S.O. and found that answer :

Sorting in python - how to sort a list containing alphanumeric values?

The code I used is freakish’s, which goes :

import re
def convert(str):
    return int("".join(re.findall("\d*", str)))

list1.sort(key=convert)

In order to better explain my problem, I shuffled the list and ran the code.

The result was :

17 some text
-25 and again, a negative number
25 foo bar
100 two same texts (almost)
100 two same texts
-100 another negative number
-347 a negative number
350 a positive number

What went wrong ? What is the precision in the code that would sort naturally the negative numbers?

Answer 1

Sort can accept a tuple, so first sort by the number, then by the text after

list_text = """-347 a negative number
-100 another negative number
-25 and again, a negative number
17 some text
25 foo bar
100 two same texts
100 two same texts (almost)
350 a positive number"""

list1 = list_text.split("\n")

list1.sort(key=lambda x: (int(x.split()[0]), x.split(" ",1)))

print("\n".join(list1))

Answer 2

The easiest approach, IMHO, would be to split the string to individual words, and then sort by the first word converted to an int:

list1.sort(key = lambda x : int(x.split(" ")[0]))

EDIT:
As Keatinge, the above solution won't properly handle two strings that start with the same number. Using the entire string as a secondary search term should solve this issue:

list1.sort(key = lambda x : (int(x.split(" ")[0]), x)