Find a decimal with a precision of 2 using yes pattern|no pattern

Question

Can someone please point out the mistake I am making, why 123 is not matching here?

def is_decimal(num):
    import re
    pattern = re.compile(r"(\b\d+\.)(?(1)\d{1,2}\b|\d+)")
    result = pattern.search(num)
    return bool(result)


print(is_decimal('123.11'))
print(is_decimal('123'))
print(is_decimal('0.21'))

print(is_decimal('123.1214'))
print(is_decimal('e666.86'))

expected output: True, True, True, False, False

actual output : True, False, True, False, False

It's not even clear what type of decimal you wish to match and which you do not. You should also mention the pattern you want the desired numbers to follow . — Robo Mop, Sep 20 '20 at 16:41
As @mousetail said, `(\b\d+\.)` matches `123.` and not `123`, `.` is included in the capture group — Alex Mandelias, Sep 20 '20 at 16:48

score 0 · Answer 1 · answered Sep 20 '20 at 19:17

0

try pattern = re.compile(r"(\b\d+\.*)(?(1)\d{1,2}\b|\d+)")

answered Sep 20 '20 at 19:17

Poe Dator

4,535
2
14
35

score 0 · Answer 2 · answered Sep 22 '20 at 04:13

You may try the following pattern:

^(\d+)(\.\d{1,2})?$

One of the key is to not use \b since the dot (.) is also considered as a word boundary which causes false parsing. Instead, you may use ^ to match the start of the string and $ the end of the string.

The above pattern can be tested interactively at the following website:

https://regex101.com/r/CixCHT/4

Find a decimal with a precision of 2 using yes pattern|no pattern

2 Answers2