Trim all zeros before and after decimal Regex

Question

Objective-C: I want to remove all zeros which trail and lead with some exceptions. For example:

0.05000 -> 0.05
000.2300 -> 0.23
005.0 -> 5
500.0 -> 500
60.0000 -> 60
56.200000 -> 56.2
56.04000 -> 56.04

What I tried is something like this [.0]+$|(^0+). But it doesn't help what I'm trying to achieve. I am using NSString with RegularExpressionSearch as follows:

 NSString *cleaned = [someString stringByReplacingOccurrencesOfString:@"[.0]+$|(^0+)"
                                                                           withString:@""
                                                                              options:NSRegularExpressionSearch
                                                                                range:NSMakeRange(0, self.doseTextField.text.length)];

Any help would be appreciated.

What about `500`? Or `-1000`? `.50`? Do you just want to account for any possible leading/trailing zeros? — Wiktor Stribiżew, Aug 08 '16 at 21:11
Why use a regex? Just convert the string to a number and back to a string? — rmaddy, Aug 08 '16 at 21:57
Or try [`"^(0)+(?=\\.)|^0+(?=[1-9])|\\.0+$|(\\.\\d*?)0+$"`](https://regex101.com/r/xN8dN6/1) to replace with `$1$2`. Or a grouped one: [`"^(?:(0)+(?=\\.)|0+(?=[1-9]))|(?:\\.0+|(\\.\\d*?)0+)$"`](https://regex101.com/r/xN8dN6/2) — Wiktor Stribiżew, Aug 08 '16 at 22:12
You may be interested in `NSNumberFormatter` and avoid the regex? — Larme, Aug 09 '16 at 07:52
@rmaddy I tried converting from string to number and then back to string. Didn't work. — Harsh, Aug 09 '16 at 16:12
@WiktorStribiżew Your regex works perfectly fine. :) Thanks for the quick answer. — Harsh, Aug 09 '16 at 16:13
@WiktorStribiżew Why wouldn't it be met positively? Looks like a great solution to me. Learnt some new things. — Harsh, Aug 09 '16 at 17:41
@WiktorStribiżew - Change you first alternative to `^(0)+(?=\\.|$)` and you'll catch strings of just zeros as well (0000 -> 0). However as you've seen I preferred the non-APL approach ;-) — CRD, Aug 09 '16 at 23:44
@WiktorStribiżew If we want to change the .5 to 0.5. How can we do that? — Harsh, Aug 23 '16 at 19:07
If there is no text that we can match, you can only insert it in the replacement pattern. — Wiktor Stribiżew, Aug 23 '16 at 19:10
I think it can only be implemented as a replace callback method in a regex substitution. See http://stackoverflow.com/questions/3957092/is-there-an-objective-c-regex-replace-with-callback-c-matchevaluator-equivalent — Wiktor Stribiżew, Aug 23 '16 at 19:39

score 2 · Accepted Answer · answered Aug 09 '16 at 03:28

Sometimes it is best to break tasks up, try these in order with a replacement string of @"" each time:

@"^0+(?!\\.)" - an initial sequence of 0's not followed by a dot. This will trim "003.0" to "3.0" and "000.3" to "0.3".
@"(?<!\\.)0+$" - a trailing sequence of zeros not preceeded by a dot. So "3.400" goes to "3.4" and "3.000" goes to "3.0"
@\\.0$ - remove a trailing ".0". The first two REs may leave you with ".0" on purpose so that you can then remove the dot in this step.

This isn’t the only way to break it down but the three basic tasks are to remove leading zeros, remove trailing zeros, and cleanup. You also don't need to use REs at all, a simple finite state machine might be a good fit.

Important: The above assumes the string contains a decimal point, if it may not you need handle that as well (left as an exercise).

HTH

Good suggestion to break the problem into understandable, testable parts - six-months-from-now me greatly appreciates this sort of thing! — Kendall Lister, Aug 09 '16 at 03:42
As this seems to be a good practice, for now am going with Wiktor's answer for a single line regex. i.e. Or try `"^(0)+(?=\\.)|^0+(?=[1-9])|\\.0+$|(\\.\\d*?)0+$"` to replace with `$1$2`. Or a grouped one: `"^(?:(0)+(?=\\.)|0+(?=[1-9]))|(?:\\.0+|(\\.\\d*?)0+)$"` — Harsh, Aug 09 '16 at 17:48

score -1 · Answer 2 · answered Aug 08 '16 at 22:10

something like this?

let numberStrings = [
    "0.05000",
    "000.2300",
    "005.0",
    "500.0",
    "60.0000",
    "56.200000",
    "56.04000"
]

print(numberStrings.map({
    String(Double($0)!).stringByReplacingOccurrencesOfString("\\.0*$", withString: "", options: .RegularExpressionSearch, range: nil)
}))

Trim all zeros before and after decimal Regex

2 Answers2