Check if character is number

Question

I want to check if a character can be safely converted to a numeric by using a regex.

However, I don't see my error. Example:

stringr::str_detect("4.", pattern = "-{0,1}[0-9]+(.[0-9]+){0,1}")

This produces a TRUE. My intention was to specifiy that whenever a . follows the first sequence of numbers, there must be at least one other number, therefore (.[0-9]+){0,1}.

What's wrong here?

The regex for at least one occurrence should be `{1,}`. `{0,1}` means zero or one occurrence. — benson23, Jan 31 '23 at 10:18
Maybe use `stringr::str_detect("4.", pattern = "^-{0,1}[0-9]+(\\.[0-9]+){0,1}$")` — GKi, Jan 31 '23 at 10:27
It might be cheaper to simply convert and check if the result is not `NA`. — Roland, Jan 31 '23 at 11:36

score 0 · Answer 1 · answered Jan 31 '23 at 11:07

Note:

(.[0-9]+){0,1} is an optional pattern because {0,1} (=?) makes the .[0-9]+ pattern sequence match one or zero times. So, yes, one or more digits ([0-9]+) must follow any char other than line break chars (matched with an unescaped .), but this pattern is optional, and thus you cannot require anything with it.
. is unescaped, so it matches any char other than line break chars. Escape it to match a literal dot
Your regex is not anchored, and can match partial substrings in a longer string. Use ^ and $ to make the pattern match the whole string.

So, consider using

stringr::str_detect("4.", pattern = "^-?[0-9]+(?:\\.[0-9]+)?$")

where

^ - start of string
-? - an optional - char
[0-9]+ - one or more digits
(?:\.[0-9]+)? - a non-capturing group matching an optional sequence of a . and then one or more digits
$ - end of string.

Check if character is number

1 Answers1