Match an optional sign and decimal number

You want to accept numbers such as -12 and 3.14 and reject everything else.

import re

pattern = re.compile(r"-?[0-9]+(\.[0-9]+)?")
for s in ["42", "-12", "3.14", "-0.5", ".5", "5.", "1.2.3", "+7", "1e3"]:
    print(f"{s!r:8} fullmatch={bool(pattern.fullmatch(s))}  search={pattern.search(s).group() if pattern.search(s) else None}")

'42'     fullmatch=True  search=42
'-12'    fullmatch=True  search=-12
'3.14'   fullmatch=True  search=3.14
'-0.5'   fullmatch=True  search=-0.5
'.5'     fullmatch=False  search=5
'5.'     fullmatch=False  search=5
'1.2.3'  fullmatch=False  search=1.2
'+7'     fullmatch=False  search=7
'1e3'    fullmatch=False  search=1

The optional group at the end allows a fractional part only when digits follow the dot, so 5. is refused. fullmatch ties the pattern to both ends of the string, as the re module documents, and the search column shows what an unanchored match finds: 1.2.3 yields 1.2, and .5 yields 5, which is wrong in both cases.

This pattern has no leading plus sign and no exponent, so +7 and 1e3 are rejected. Add [+-]? at the front and an optional exponent group at the end if you need them, and allow a leading dot with a second alternative. In grep, use -E and anchor with ^ and $. A regular expression checks the shape, not the range.