Regex issue - Replace floats, not int

Question

Im want to remove all but the float in this sting:

string = "1 south african rand is 0.11044"

Im doing it like this:

reg = /[^\d+.\d+]/g
console.log string.replace(reg, '')

that logs

10.11044

that is wrong, I want only the xxxx.xxxxx part. 1 is not a float so it should not be part of this?

How should I chage it?

`[^....]` matches any individual character that isn't one of the ones inside the brackets. It's not looking for sequences. — Barmar, Apr 16 '13 at 09:47
I've corrected my answer using a negative look-around. Tested, works correctly my side. — Evan Knowles, Apr 16 '13 at 09:50

score 4 · Accepted Answer · edited May 23 '17 at 12:18

4

Try this instead

^((?!\d+\.\d+).)*

See this answer for more details: Regular expression to match a line that doesn't contain a word?

edited May 23 '17 at 12:18

Community

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answered Apr 16 '13 at 09:49

Evan Knowles

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score 2 · Answer 2 · answered Apr 16 '13 at 09:49

2

Following regex should match floats for you (for the example provided):

/(-?\d*\.\d+)/

To replace:

console.log (string.replace(/(-?\d*\.\d+)/, ''));

answered Apr 16 '13 at 09:49

anubhava

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He was trying to match words that were not floats however, to remove them. – Evan Knowles Apr 16 '13 at 09:50
I provided him regex to match float numbers, that can be used to replace. See my edited answer. – anubhava Apr 16 '13 at 09:52

score 1 · Answer 3 · answered Apr 16 '13 at 09:51

1

I have used positive look behind (?<=..) in below regular expression

\.\d+(?<=\d)

Use this regular expression and replace below value with ''. The result will be 1 23 33 3

1 23 33.2000 3.4445

Hope it helps

answered Apr 16 '13 at 09:51

SMS

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I think his desired result is `33.2000 3.4445`. `1` and `23` should be excluded because they don't have a decimal point and fraction. – Barmar Apr 16 '13 at 09:53

score 1 · Answer 4 · answered Apr 16 '13 at 09:52

1

var reg = /\d+\.\d+/g
var str = "1 south african rand is 0.11044";
var onlyFloats = str.match(reg).join(" ");
console.log(onlyFloats)

answered Apr 16 '13 at 09:52

user2264587

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6

score 0 · Answer 5 · answered Apr 16 '13 at 09:52

0

Try this

string = "1 south african rand is 0.11044"
reg = /(\d+\.\d+)/g;
string.match(reg);

answered Apr 16 '13 at 09:52

karthick

11,998
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score 0 · Answer 6 · answered Aug 01 '19 at 07:29

/**
 * @param $string
 * @return string
 * 
 * output
 * a.a.a.a.a.a.   => 0.0
 * 1.1.1.1.1.1.   => 1.11111
 * 2a$2.45&.wer.4 => 22.454 
 */
function stringToFlout($string) {
    $parts = explode('.', $string);

    for ($i = 0; $i < count($parts); $i++) {
        $parts[$i] = preg_replace('/[^0-9]/', '', $parts[$i]);
    }

    $left = $parts[0];
    if (empty($left))
        $left = 0;

    $float[] = $left;

    unset($parts[0]);
    $right = implode('', $parts);
    if (empty($right))
        $right = 0;

    $float[] = $right;

    return implode('.', $float);
}

Hello, and welcome to Stack Overflow. please add some explanation to your code - don't just answer with only code. — Itamar Mushkin, Aug 01 '19 at 07:46

score 0 · Answer 7 · answered May 02 '21 at 00:12

I would use this regex:

(0|([1-9][0-9]*))?\.(0|([1-9][0-9]*))

The (0|([1-9][0-9]*))? matches an optional integer part, the \. matches dot, and the final (0|([1-9][0-9]*)) matches a mandatory fractional part. It succeeds for:

.5
0.5
0.0
0.1
0.5foo
31.41592

It fails for:

00.5
100
nice
2.

Anchored version:

^(0|([1-9][0-9]*))?\.(0|([1-9][0-9]*))$

Regex issue - Replace floats, not int

7 Answers7