Java Remove all cases of .0

Question

I want a filter to run through a string and eliminate all of the un-needed .0's from the doubles. I have already tried replaces, but cases like 8.02 turn into 82.

I have a feeling this has been done before, but I cannot find any help.

I am looking for a method that will take in a String, example: "[double] plus [double] is equal to [double]", where the [double] is a double that will need to be checked for the redundant decimal.

Thanks in advance!

Do you want to remove zeros only from the fractional part? Either leading or trailing, or any? — Lion, Jan 05 '12 at 00:00

davogotland · Accepted Answer · 2012-01-10T04:54:10.207

2

let's say you have a string called s that contains text like you described. simply do:

s = s.replaceAll("([0-9])\\.0+([^0-9]|$)", "$1$2");

and you're done.

ok, so i edited this a couple of times. now it works though!

edited Jan 10 '12 at 04:54

answered Jan 05 '12 at 00:25

davogotland

2,718
1
15
19

This is interesting. It worked fine earlier, but now it fixes all bug the last number. Eg: "10.0+1.0=11.0" becomes "10+1=11.0". Any way to fix this? – Dylan Wheeler Jan 10 '12 at 00:05
1

oh, didn't think of that :( but substringing isn't necessary. check out my edit, it takes care of that case as well! – davogotland Jan 10 '12 at 04:55

litterbugkid · Answer 2 · 2012-01-05T00:21:17.683

0

Can you not take away the trailing zeros before you construct them into your string? This way you could use the DecimalFormatter method someone posted earlier and then deleted.

NumberFormat formatter = new DecimalFormat("0.##"); 
String str = formatter.format("8.0200"); 
System.out.println(str);

Give them credit for the code if they come back.

edited Jan 05 '12 at 00:21

answered Jan 05 '12 at 00:05

litterbugkid

3,534
7
36
54

score 0 · Answer 3 · answered Jan 05 '12 at 00:15

0

You can do this with a java.text.DecimalFormat object.

answered Jan 05 '12 at 00:15

Stephen P

14,422
2
43
67

Lion · Answer 4 · 2012-01-05T00:29:22.677

The following program will remove all trailing zeros from the fractional part of a double value. If your requirements are somewhat different, you may need to modify it slightly.

final public class Main
{
    public static void main(String...args)
    {
        System.out.println("Enter how many numbers you want to check:->");
        Scanner scan = new Scanner(System.in);

        final double KEY_VALUE = 0.0;
        final Double TOLERANCE = 0.000000000001d;

        int n = scan.nextInt();
        double[] a = new double[n];
        for (int i = 0; i < n; i++)
        {
                a[i] = scan.nextDouble();
        }

        List<Double> newList = new ArrayList<Double>();
        for (int k = 0; k < n; k++)
        {
            if (Math.abs(a[k] - KEY_VALUE) < TOLERANCE)
            {
                    continue;
            }
            newList.add(a[k]);
        }

        System.out.println(newList);
    }
}

You can specifically use DecimalFormat to truncate the specified number of decimal places, if you need such as follows.

double x=10.4555600500000;
DecimalFormat df=new DecimalFormat("#.##");
System.out.println(df.format(x));

Would return 10.46.

score 0 · Answer 5 · answered Jan 05 '12 at 00:22

The other posts assume you're working with a short string containing only decimal. Assuming you're working with large strings of text, you can use Pattern/Matcher classes (I'm at work, so posting in a hurry. Check for errors)

Use this regex to replace:

/* >1 digits followed by a decimal and >1 zeros. Note capture group on first set of digits. This will only match decimals with trailing 0s, and not 8.0002 */
(\d+)\.0+

Replace with

/* First capture group */
$1

I'm unsure of the regex rules for Java, so use this as a concept to get what you want.

Java Remove all cases of .0

5 Answers5