Determine Number of Decimal Place using BigDecimal

I was interested to have the following getNumberOfDecimalPlace function:

System.out.println("0 = " + Utils.getNumberOfDecimalPlace(0));          // 0
System.out.println("1.0 = " + Utils.getNumberOfDecimalPlace(1.0));      // 0
System.out.println("1.01 = " + Utils.getNumberOfDecimalPlace(1.01));    // 2
System.out.println("1.012 = " + Utils.getNumberOfDecimalPlace(1.012));  // 3
System.out.println("0.01 = " + Utils.getNumberOfDecimalPlace(0.01));    // 2
System.o开发者_如何学运维ut.println("0.012 = " + Utils.getNumberOfDecimalPlace(0.012));  // 3

May I know how can I implement getNumberOfDecimalPlace, by using BigDecimal?

The following code doesn't work as expected:

public static int getNumberOfDecimalPlace(double value) {
    final BigDecimal bigDecimal = new BigDecimal("" + value);
    final String s = bigDecimal.toPlainString();
    System.out.println(s);
    final int index = s.indexOf('.');
    if (index < 0) {
        return 0;
    }
    return s.length() - 1 - index;
}

The following get printed :

0.0
0 = 1
1.0
1.0 = 1
1.01
1.01 = 2
1.012
1.012 = 3
0.01
0.01 = 2
0.012
0.012 = 3

However, for case 0, 1.0, it doesn't work well. I expect, "0" as result. But they turned out to be "0.0" and "1.0". This will return "1" as result.

Combining Turismo, Robert and user1777653's answers, we've got:

int getNumberOfDecimalPlaces(BigDecimal bigDecimal) {
    return Math.max(0, bigDecimal.stripTrailingZeros().scale());
}

• stripTrailingZeros() ensures that trailing zeros are not counted (e.g. 1.0 has 0 decimal places).
• scale() is more efficient than String.indexOf().
• A negative scale() represents zero decimal places.

There you have it, the best of both worlds.

This code:

int getNumberOfDecimalPlaces(BigDecimal bigDecimal) {
    String string = bigDecimal.stripTrailingZeros().toPlainString();
    int index = string.indexOf(".");
    return index < 0 ? 0 : string.length() - index - 1;
}

... passes these tests:

assertThat(getNumberOfDecimalPlaces(new BigDecimal("0.001")), equalTo(3));
assertThat(getNumberOfDecimalPlaces(new BigDecimal("0.01")), equalTo(2));
assertThat(getNumberOfDecimalPlaces(new BigDecimal("0.1")), equalTo(1));
assertThat(getNumberOfDecimalPlaces(new BigDecimal("1.000")), equalTo(0));
assertThat(getNumberOfDecimalPlaces(new BigDecimal("1.00")), equalTo(0));
assertThat(getNumberOfDecimalPlaces(new BigDecimal("1.0")), equalTo(0));
assertThat(getNumberOfDecimalPlaces(new BigDecimal("1")), equalTo(0));
assertThat(getNumberOfDecimalPlaces(new BigDecimal("10")), equalTo(0));
assertThat(getNumberOfDecimalPlaces(new BigDecimal("10.1")), equalTo(1));
assertThat(getNumberOfDecimalPlaces(new BigDecimal("10.01")), equalTo(2));
assertThat(getNumberOfDecimalPlaces(new BigDecimal("10.001")), equalTo(3));

... if that is indeed what you want. The other replies are correct, you have to use BigDecimal all the way through for this rather than double/float.

Without having to convert to String, it should be more efficient to use the scale directly:

  private int getNumberOfDecimalPlaces(BigDecimal bigDecimal)
  {
    int scale =  bigDecimal.stripTrailingZeros().scale();
    return scale>0?scale:0;
  }

That should do it

int getNumberOfDecimalPlace(BigDecimal number) {
    int scale = number.stripTrailingZeros().scale();
    return scale > 0 ? scale : 0;
}

If you really get doubles i recommend formating them first as strings before creating the BigDecimal. At least that has worked for me: How to check if a double has at most n decimal places?

Depending on how many digits you expect you can either use standard formating like

String.valueOf(doubleValue);

or you could use specialised formatting to avoid exponential format

DecimalFormat decimalFormat = new DecimalFormat();
decimalFormat.setMaximumIntegerDigits(Integer.MAX_VALUE);
// don't use grouping for numeric-type cells
decimalFormat.setGroupingUsed(false);
decimalFormat.setDecimalFormatSymbols(new DecimalFormatSymbols(Locale.US));
value = decimalFormat.format(numericValue);

When you have a BigDecimal you can simply call scale() to get the number of decimal places.

It's not your code that's wrong, but your expectations. double is based on a binary floating point representation and completely unfit for accurately representing decimal fractions. Decimal 0.1 e.g. has an infinite number of digits when represented in binary, thus it gets truncated and when converted back to decimal, you get erros in the least significant digits.

If you use BigDecimal exclusively, your code will work as expected.

Try this:

Math.floor(Math.log(x) / Math.log(10))
0.001 = -3
0.01  = -2
0.1   = -1  
1     = 0  
10    = 1  
100   = 2

How about having a look at the javadoc of BigDecimal. I'm not sure, but I'd give getScale and getPercision a try.

The best way to get a BigDecimal with a specified number of decimal places is by using the setscale method over it. Personally I like to also use the rounding version of the method (see the link below): http://java.sun.com/j2se/1.5.0/docs/api/java/math/BigDecimal.html#setScale(int,%20int)

If you're wanting to get the number of decimal positions that a BigDecimal is currently set at call the associated scale() method.

Best option I have found so far (not needing toString + index):

public static int digitsStripTrailingZero(BigDecimal value)
{
    return digits(value.stripTrailingZeros());
}

public static int digits(BigDecimal value)
{
    return Math.max(0, value.scale());
}

Michael Borgwardt answer is the correct one. As soon as you use any double or float, your values are already corrupted.

To provide a code example:

System.out.println("0 = " + BigDecimalUtil.getNumberOfDecimalPlace("0")); // 0
System.out.println("1.0 = " + BigDecimalUtil.getNumberOfDecimalPlace("1.0")); // 0
System.out.println("1.01 = " + BigDecimalUtil.getNumberOfDecimalPlace(new BigDecimal("1.01"))); // 2
System.out.println("1.012 = " + BigDecimalUtil.getNumberOfDecimalPlace(new BigDecimal("1.012"))); // 3
System.out.println("0.01 = " + BigDecimalUtil.getNumberOfDecimalPlace("0.01")); // 2
System.out.println("0.012 = " + BigDecimalUtil.getNumberOfDecimalPlace("0.012")); // 3
System.out.println("0.00000000000000000012 = " + BigDecimalUtil.getNumberOfDecimalPlace("0.00000000000000000012")); // 20

And an overloaded version of getNumberOfDecimalPlace so you could use it with BigDecimal or String:

public static int getNumberOfDecimalPlace(String value) {
    final int index = value.indexOf('.');
    if (index < 0) {
        return 0;
    }
    return value.length() - 1 - index;
}

public static int getNumberOfDecimalPlace(BigDecimal value) {
    return getNumberOfDecimalPlace(value.toPlainString());
}

Why not just change your code to get a doubles decimal places?

public static int getNumberOfDecimalPlace(double value) {
    //For whole numbers like 0
    if (Math.round(value) == value) return 0;
    final String s = Double.toString(value);
    System.out.println(s);
    final int index = s.indexOf('.');
    if (index < 0) {
       return 0;
    }
    return s.length() - 1 - index;
}