Regex function to find all sequence of digits

Question

I'm pretty new to Python and I just started learning regex. I'm trying to define a function that takes as input a string containing zero or more sequences of digits and returns a list of all sequences of digits in the string and sorts them. If there isn't it will print that there isn't any.

Here's my code:

def DigitSequence(string):
    pattern=re.compile(r'[0-9]+')
    k=re.findall(pattern,string)
    if len(k)!=0:
        return(sorted(k))
    else:
        return('None')

print( DigitSequence("My number is 98765 and my friend\'s number is 12345.") )
#This should return me ['12345','98765']. And it does.

But is there any better way to do it. Are there any cases I would miss if I do this. I thought of using pattern=re.compile(r'\d*') but I felt it was equivalent to what I did. Correct me if I am wrong.

\d is exactly the same as [0-9], however the quantifiers you used in those patterns are different. * matches the preceding entity zero or more times, while + matches one or more times. So yeah, in many cases those two patterns will give you the same result, but there's an important distinction. — CAustin
– CAustin, Commented Mar 10, 2020 at 23:33
What about the cases...would I be missing any. Should I put + or * so that the result won't be changed in case there's an alphanumeric term in the string. — JBlack
– JBlack, Commented Mar 10, 2020 at 23:36
Ohh yeah..you mean the spaces. I did get those so i wrote another line to clean them but i felt thats not the best way to do it. — JBlack
– JBlack, Commented Mar 10, 2020 at 23:38
@HeapOverflow So you mean this won't work for single digits? Then how can I capture both? Is there such a regex term or can i write some condition: pattern=re.compile(r'\d+|\d') . something like this? — JBlack
– JBlack, Commented Mar 10, 2020 at 23:45

Bill Huang · Accepted Answer · 2020-03-11 04:44:17Z

re.findall() is on the right track. Some additional points to note:

An uncompiled regex string can be passed directly into re.findall().
If full-width digits are wanted, use r"\d+". Otherwise use "[0-9]+".
Sorting order: 99 before 111? Or '1'11 before '9'9?
Separate printing logic and parsing logic.

Code:

import re

def digital_sequence(s, sort_literal=False, fullwidth_digits=False):
    """Get sorted list of numbers.

    Args:
        s (str): arbitrary string
        sort_literal (bool): literal sort
        fullwidth_digits (bool): accept full-width digits

    Returns:
        list
    """
    r = r"\d+" if fullwidth_digits else r"[0-9]+"
    k = None if sort_literal else lambda x: int(x)
    return sorted(re.findall(r, s), key=k)

def print_ls(ls):
    """Customized list printing: list contents or "None".

    Args:
        ls (list): input list

    Returns:
        None
    """
    print(ls) if len(ls) > 0 else print("None")

s1 = "99 and 112 and １２"
s2 = "abc"
print_ls(digital_sequence(s1))  # ['99', '112']
print_ls(digital_sequence(s2))  # None
print_ls(digital_sequence(s1, sort_literal=True))  # ['112', '99']
print_ls(digital_sequence(s1, fullwidth_digits=True))  # ['１２', '99', '112']

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