Regex for removing leading zeros from neative float numbers

Question

guys. I have a problem. My input strings looks like:

1-000000.02
1-000000.00
1+000025.48
1-000025.47
1-000000.00
1+000000.00
1+000025.46

And I want to extract normilize (remove plus sign, remove leading zeros, but exclude one zero before dot) float numbers like this:

0.02
0.00
25.48
-25.47
-0.00
0.00
25.46

I use next expression: 0*([0-9]+.?[0-9]+) (http://regexr.com/3dicv), it's works fine, but I can't catch minus sign ("25.47" instead "-25.47"). So if somebody point me to right way I will very grateful. Thanks.

Does [What is a non capturing group? (?:)](http://stackoverflow.com/questions/3512471/what-is-a-non-capturing-group) give you any clues? — Andrew Morton, Jun 04 '16 at 13:21
Will, I used similar variant - in this case we lost "-" in negative numbers. — Sergey Alikin, Jun 04 '16 at 13:45
Andrew, thanks for your comment. I'm using noncapturing groups in my expressions. But the problem that when number is negative leading zeros placed INSIDE capturing group (-000012.34), and in this case leading zeros will catch too( — Sergey Alikin, Jun 04 '16 at 13:49
Can't you replace `.replace(/^1?([+-])0*([0-9]+(?:\.[0-9]+)?)$/, "$1$2")`? — Wiktor Stribiżew, Jun 04 '16 at 13:51
Wiktor, thanks. Unfortunately - no. I can't use replace, my final goal - extract number without language-specific tools. Rather, this is my last posibility - use post-proccessing function for "normalizing" any input number, but I want to skip this variant very strong. — Sergey Alikin, Jun 04 '16 at 13:56
You cannot use single regex match operation to match discontinuous texts. You have to use a replace or capture the different string parts to concatenate them later. — Wiktor Stribiżew, Jun 04 '16 at 14:10
I was able to add an optional hyphen (`-?`) to the front of your regexr expression and match the minus signs. — aghast, Jun 04 '16 at 15:20
Is there a reason for keeping `-` before `0.00` ? I'd more think of [something like this](https://regex101.com/r/fP9bE6/1) but useless if you can't use replace (: — bobble bubble, Jun 04 '16 at 17:45

LukStorms · Answer 1 · 2016-06-04T15:38:29.137

2

This works in regexr.com

\d[+]?([-]?)0*(\d+\.\d+)

Replace:

$1$2

For these values:

1+000050.93
1-000025.47
1+000000.00
1-000000.00
1+000000.02
1-000000.02
1+100025.47
1-100025.48

It returns :

50.93
-25.47
0.00
-0.00
0.02
-0.02
100025.47
-100025.48

But I don't see why the trailing zero's are an issue when you would use it in C++.
This C++ seems to extract and parse it to a double just fine.

#include <iostream>
#include <string>
#include <cstdlib>
#include <regex>
using namespace std;

int main() {
    string teststring("1+000001.62");

    regex re("1([+-][0-9]+[.][0-9]{2})");
    smatch match;

    string resultstring = regex_replace(teststring, re, "$1" );
    double value = std::atof(resultstring.c_str());
    cout << value;

    return 0;
}

edited Jun 04 '16 at 15:38

answered Jun 04 '16 at 14:07

LukStorms

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This regex requires anchors or it will might match what is not required – Wiktor Stribiżew Jun 04 '16 at 14:20
Luk, thanks for your example - it's a very good, but it produce TWO separate capturing group, not one. I agree with Wiktor from last comment in init post - this task is unsolveable with regex. – Sergey Alikin Jun 04 '16 at 14:22
So the language you would be using it for can't handle multiple capture groups? A pity if that's the case. – LukStorms Jun 04 '16 at 14:25
Luk, of course, my language (C++, boost, Qt) allow to do any actions with regexs, but I want to have unificate method for proccessing input strings from different instruments, for execute this condition I need to have one capturing group. So as we found out that my task is unsolveable, I will use post-proccessing function where will be placed replace() and all others. – Sergey Alikin Jun 04 '16 at 14:30
You could use something like this 1([+-]\d+\.\d{2}) That only has 1 capture group. but it will contain the leading zero's. But would that really be a problem to cast to a number in c++ ? – LukStorms Jun 04 '16 at 14:35

Ro Yo Mi · Answer 2 · 2016-06-06T04:41:17.837

Description

You just have to validate your string with a look ahead, then match the substrings you want removed while capturing the minus sign if it were there.

(?=^[0-9][+-][0-9]+\.[0-9]{2}$)(?:[0-9]+(?:(-)|\+))0+(?!\.)

Replace with: $1

Regular expression visualization

This regular expression will do the following:

validate your string is in the format integer plus or minus real with two decimal points
replaces everything else that is not desirable, like the leading integer, and zeros before the decimal point, not including the zero directly before the decimal point.

Example

Live Demo

https://regex101.com/r/mP4gH1/2

Sample text

1-000000.02
1-000000.00
1+000025.48
1-000025.47
1-000000.00
1+000000.00
1+000025.46

After Replacement

-0.02
-0.00
25.48
-25.47
-0.00
0.00
25.46

Explanation

NODE                     EXPLANATION
----------------------------------------------------------------------
  (?=                      look ahead to see if there is:
----------------------------------------------------------------------
    ^                        the beginning of a "line"
----------------------------------------------------------------------
    [0-9]                    any character of: '0' to '9'
----------------------------------------------------------------------
    [+-]                     any character of: '+', '-'
----------------------------------------------------------------------
    [0-9]+                   any character of: '0' to '9' (1 or more
                             times (matching the most amount
                             possible))
----------------------------------------------------------------------
    \.                       '.'
----------------------------------------------------------------------
    [0-9]{2}                 any character of: '0' to '9' (2 times)
----------------------------------------------------------------------
    $                        before an optional \n, and the end of a
                             "line"
----------------------------------------------------------------------
  )                        end of look-ahead
----------------------------------------------------------------------
  (?:                      group, but do not capture:
----------------------------------------------------------------------
    [0-9]+                   any character of: '0' to '9' (1 or more
                             times (matching the most amount
                             possible))
----------------------------------------------------------------------
    (?:                      group, but do not capture:
----------------------------------------------------------------------
      (                        group and capture to \1:
----------------------------------------------------------------------
        -                      '-' character
----------------------------------------------------------------------
      )                        end of \1
----------------------------------------------------------------------
     |                        OR
----------------------------------------------------------------------
      \+                       '+' character
----------------------------------------------------------------------
    )                        end of grouping
----------------------------------------------------------------------
  )                        end of grouping
----------------------------------------------------------------------
  0+                       '0' (1 or more times (matching the most
                           amount possible))
----------------------------------------------------------------------
  (?!                      look ahead to see if there is not:
----------------------------------------------------------------------
    \.                       '.'
----------------------------------------------------------------------
  )                        end of look-ahead
----------------------------------------------------------------------

Regex for removing leading zeros from neative float numbers

2 Answers2

Description

Example

Explanation