Regex to truncate trailing zeroes

Question

I'm trying to construct a single regex (for Java) to truncate trailing zeros past the decimal point. e.g.

50.000 → 50
50.500 → 50.5
50.0500 → 50.05
-5 → -5
50 → 50
5.5 → 5.5

Idea is to represent the real number (or integer) in the most compact form possible.

Here's what I've constructed:

^(-?[.0-9]+?)\.?0+$

I'm using $1 to capture the truncated number string.

The problem with the pattern above is that 50 gets truncated to 5. I need some way to express that the 0+ must follow a . (decimal point).
I've tried using negative-behind, but couldn't get any matches.

Note that all regex approaches will break down if the String represents a number in a format that uses a different decimal separator (or uses the dot character as a grouping separator). — Hulk
– Hulk, Commented Jun 15, 2016 at 9:24

Wiktor Stribiżew · Accepted Answer · 2016-06-15 08:12:22Z

4

The best solution could be using built-in language-specific methods for that task.

If you cannot use them, you may use

^(-?\d+)(?:\.0+|(\.\d*?)0+|\.+)?$

And replace with $1$2.

See the regex demo. Adjust the regex accordingly. Here is the explanation:

^ - start of string
(-?\d+) -Group 1 capturing 1 or 0 minus symbols and then 1 or more digits
(?:\.0+|(\.\d*?)0+|\.+)? - An optional (matches 1 or 0 times due to the trailing ?) non-capturing group matching 3 alternatives:
- \.0+ - a decimal point followed with 1+ zeros
- (\.\d*?)0+ - Group 2 capturing a dot with any 0+ digits but as few as possible and matching 1+ zeros
- \.+ - (optional branch, you may remove it if not needed) - matches the trailing dot(s)
$ - end of string.

Java demo:

String s = "50.000\n50\n50.100\n50.040\n50.\n50.000\n50.500\n50\n-5";
System.out.println(s.replaceAll("(?m)^(-?\\d+)(?:\\.0+|(\\.\\d*?)0+|\\.+)?$", "$1$2"));
// => [50, 50, 50.1, 50.04, 50, 50, 50.5, 50, -5]

edited Jun 15, 2016 at 8:12

answered Jun 15, 2016 at 7:21

Wiktor Stribiżew

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2 Comments

Wiktor Stribiżew Over a year ago

@WingPoon: What do you expect to get if the input is -5? There is no decimal part in it. As far as I remember, you need to remove trailing zeros in the decimal part, or did you change your mind and now want to match all up to the trailing zeros?

Wiktor Stribiżew Over a year ago

You must add the requirement to the question. Use ^(-?\d+)(?:\.0+|(\.\d*?)0+|\.+)?$. What is the method/code you are using?

engineer14 · Accepted Answer · 2016-06-15 09:39:40Z

1

For a general regex which should do the trick:

^\d+?0*\.??\d*?(?=0*?[^\d]*$)

You can replace the caret and dollar sign with whatever your boundaries should be. Those could be replaced by whatever you would expect around your number.

basically:

/d+? non-greedy match for any number (needs at least 1 number to start the match)

\.*?? optional match for a decimal. Prefers to match 0 occurrences

\d*? (?=0*?[^\d]*$) - non-greedy match for a number, but would stop at the 0 which is proceeded by a non-number

EDIT: I just realized the original expression also trimmed the last zero on integers, this should work. I added the option 0 match to catch that

edited Jun 15, 2016 at 9:39

answered Jun 15, 2016 at 8:30

engineer14

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