Get Double from string in Java

Question

I have a text (String) an I need to get only digits from it, i mean if i have the text:

"I'm 53.2 km away", i want to get the "53.2" (not 532 or 53 or 2)

I tried the solution in Extract digits from a string in Java. it returns me "532".

Anyone have an idea for it?

Thanx

Suraj Chandran · Accepted Answer · 2012-06-10 18:18:49Z

9

You can directly use a Scanner which has a nextDouble() and hasNextDouble() methods as below:

        Scanner st = new Scanner("I'm 53.2 km away");
        while (!st.hasNextDouble())
        {
            st.next();
        }
        double value = st.nextDouble();
        System.out.println(value);

Output: 53.2

answered Jun 10, 2012 at 18:18

Suraj Chandran

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Comments

dantuch · Accepted Answer · 2012-06-10 18:10:48Z

3

Here is good regex site with tester:

http://gskinner.com/RegExr/

this works fine \d+\.?\d+

answered Jun 10, 2012 at 18:10

dantuch

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Comments

Surender Thakran · Accepted Answer · 2012-06-10 18:31:48Z

3

import java.util.regex.*;

class ExtractNumber
{
    public static void main(String[] args)
    {
        String str = "I'm 53.2 km away";
        String[] s = str.split(" ");
        Pattern p = Pattern.compile("(\\d)+\\.(\\d)+");
        double d;
        for(int i = 0; i< s.length; i++)
        {
            Matcher m = p.matcher(s[i]);
            if(m.find())
                d = Double.parseDouble(m.group());
        }
        System.out.println(d);
    }
}

edited Jun 10, 2012 at 18:31

answered Jun 10, 2012 at 18:14

Surender Thakran

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2 Comments

pb2q Over a year ago

Note that this will also match as whole 53.2km, i.e. it won't strip the km. The example does use a space, though so this is only a nit.

Surender Thakran Over a year ago

@pb2q: replaced s[i] with m.group() this should solve the problem you pointed out.

Amimo Benja · Accepted Answer · 2014-08-16 08:07:32Z

1

The best and simple way is to use a regex expression and the replaceAll string method. E.g

String a = "2.56 Kms";

String b = a.replaceAll("\\^[0-9]+(\\.[0-9]{1,4})?$","");

Double c = Double.valueOf(b);

System.out.println(c);

answered Aug 16, 2014 at 8:07

Amimo Benja

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Comments

nbarraille · Accepted Answer · 2012-06-10 18:18:09Z

0

If you know for sure your numbers are "words" (space separated) and don't want to use RegExs, you can just parse them...

String myString = "I'm 53.2 km away";
List<Double> doubles = new ArrayList<Double>();
for (String s : myString.split(" ")) {
    try {
        doubles.add(Double.valueOf(s));
    } catch (NumberFormatException e) {
    }
}

answered Jun 10, 2012 at 18:18

nbarraille

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4 Comments

dantuch Over a year ago

and create lots of useless Exception objects?

Jeffrey Over a year ago

@dantuch They're not useless: they're signalling that the String could not be parsed.

dantuch Over a year ago

@Jeffrey No! They shouldn't be created in first place. You know what "Exception driven development" stands for? Just don't use exceptions in situations that are NOT exceptional. Parsing I'm to double ending in fail isn't exception at all.

Jeffrey Over a year ago

@dantuch But the exceptions are still not useless, they are performing a task. They are, however, unnecessary.

Chu Vu Hung · Accepted Answer · 2014-11-26 08:40:19Z

I have just made a method getDoubleFromString. I think it isn't best solution but it works good!

public static double getDoubleFromString(String source) {
        if (TextUtils.isEmpty(source)) {
            return 0;
        }

        String number = "0";
        int length = source.length();

        boolean cutNumber = false;
        for (int i = 0; i < length; i++) {
            char c = source.charAt(i);
            if (cutNumber) {
                if (Character.isDigit(c) || c == '.' || c == ',') {
                    c = (c == ',' ? '.' : c);
                    number += c;
                } else {
                    cutNumber = false;
                    break;
                }
            } else {
                if (Character.isDigit(c)) {
                    cutNumber = true;
                    number += c;
                }
            }
        }
        return Double.parseDouble(number);
    }

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