Regex will kill me clear solution for Is String a Number? (Part II) by Kolia951 - python coding challenges

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Regex will kill me solution in Clear category for Is String a Number? (Part II) by Kolia951

import re

def is_number(val: str) -> bool:
    
    regex = r"^[+-]?(\d+\.{1}\d*|\d*\.{1}\d+|\d+)$"

    result = re.search(regex, val)
       
    return bool(result)


print("Example:")
print(is_number("10"))

# These "asserts" are used for self-checking
assert is_number("34") == True
assert is_number("df") == False
assert is_number("") == False
assert is_number("+10.0") == True
assert is_number("1OOO") == False
assert is_number("1.") == True
assert is_number("+.l") == False

print("The mission is done! Click 'Check Solution' to earn rewards!")

Dec. 12, 2022

Comments:

Kolia951 on Dec. 12, 2022, 9:33 a.m.
This is a third ver with using regex. Previous ones were failed after adding some new assertions into the mission.

vtflnk on Dec. 12, 2022, 3:36 p.m.
assert is_number(‘—5’) test was recently added. I consider this to be incorrect, since it contradicts the mask specified in the precondition of the problem. Mask indicates that there can be only one plus or one minus at the beginning of the number.

Kolia951 on Dec. 13, 2022, 7:12 a.m.
There is a topic on the Forum where I asked about the mask described in the mission, the author said that this is just for your reference, so this is not a strict example.

saitama-1 on Jan. 1, 2023, 4:56 a.m.
nice, very impressive RegEx

H0r4c3 on April 29, 2023, 9:13 p.m.
Regex for the win! I like this solution!

mikaeldovbnia on July 10, 2023, 5:54 p.m.
Great. I like the regular expression. Very complicated regex, but still great!

BootzenKatzen on Aug. 24, 2023, 3:17 p.m.
The fact that you can come up with your own regex means you're way ahead of me! I usually have to google it, copy it, and then reverse engineer it X-D Nice work!

Kolia951 on Aug. 25, 2023, 6:15 a.m.
This is very rare scenario that a user is able to write his own regex string fully by himself, especially a novice programmer. I was using google too to find some answers on the most complicated parts of this regex.

crispyrooster on Oct. 10, 2023, 6:09 p.m.
Kind of surprised they don't test for space in the first character class. <pre class='brush: python'>[+- ] </pre> This implies space was an optional leading character, which your solution does not allow. The regex could be simplified a little with making the decimal points optional with the "?" quantifier (0 or 1) and then no longer checking for only digits. <pre class='brush: python'>regex = r"^[+-]?(\d+\.{1}\d*|\d*\.{1}\d+|\d+)$" regex = r"^[+-]?(\d+\.?\d*|\d*\.?\d+)$" </pre>

Kolia951 on Oct. 10, 2023, 6:24 p.m.
Thank you for reviewing my solution! You are right, there is no extra tests for space. By the way, you can open Github and propose to add it.

Strings Theory

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