round BigDecimal to nearest 5 cents

I'm trying to figure out how to round a monetary amount upwards to the nearest 5 cents. The following shows my expected results

1.03     => 1.05
1.051    => 1.10
1.05     => 1.05
1.900001 =&g开发者_Go百科t; 1.10

I need the result to be have a precision of 2 (as shown above).

Update

Following the advice below, the best I could do is this

    BigDecimal amount = new BigDecimal(990.49)

    // To round to the nearest .05, multiply by 20, round to the nearest integer, then divide by 20
   def result =  new BigDecimal(Math.ceil(amount.doubleValue() * 20) / 20)
   result.setScale(2, RoundingMode.HALF_UP)

I'm not convinced this is 100% kosher - I'm concerned precision could be lost when converting to and from doubles. However, it's the best I've come up with so far and seems to work.

Using BigDecimal without any doubles (improved on the answer from marcolopes):

public static BigDecimal round(BigDecimal value, BigDecimal increment,
                               RoundingMode roundingMode) {
    if (increment.signum() == 0) {
        // 0 increment does not make much sense, but prevent division by 0
        return value;
    } else {
        BigDecimal divided = value.divide(increment, 0, roundingMode);
        BigDecimal result = divided.multiply(increment);
        return result;
    }
}

The rounding mode is e.g. RoundingMode.HALF_UP. For your examples, you actually want RoundingMode.UP (bd is a helper which just returns new BigDecimal(input)):

assertEquals(bd("1.05"), round(bd("1.03"), bd("0.05"), RoundingMode.UP));
assertEquals(bd("1.10"), round(bd("1.051"), bd("0.05"), RoundingMode.UP));
assertEquals(bd("1.05"), round(bd("1.05"), bd("0.05"), RoundingMode.UP));
assertEquals(bd("1.95"), round(bd("1.900001"), bd("0.05"), RoundingMode.UP));

Also note that there is a mistake in your last example (rounding 1.900001 to 1.10).

I'd try multiplying by 20, rounding to the nearest integer, then dividing by 20. It's a hack, but should get you the right answer.

I wrote this in Java a few years ago: https://github.com/marcolopes/dma/blob/master/org.dma.java/src/org/dma/java/math/BusinessRules.java

/**
 * Rounds the number to the nearest<br>
 * Numbers can be with or without decimals<br>
 */
public static BigDecimal round(BigDecimal value, BigDecimal rounding, RoundingMode roundingMode){

    return rounding.signum()==0 ? value :
        (value.divide(rounding,0,roundingMode)).multiply(rounding);

}


/**
 * Rounds the number to the nearest<br>
 * Numbers can be with or without decimals<br>
 * Example: 5, 10 = 10
 *<p>
 * HALF_UP<br>
 * Rounding mode to round towards "nearest neighbor" unless
 * both neighbors are equidistant, in which case round up.
 * Behaves as for RoundingMode.UP if the discarded fraction is >= 0.5;
 * otherwise, behaves as for RoundingMode.DOWN.
 * Note that this is the rounding mode commonly taught at school.
 */
public static BigDecimal roundUp(BigDecimal value, BigDecimal rounding){

    return round(value, rounding, RoundingMode.HALF_UP);

}


/**
 * Rounds the number to the nearest<br>
 * Numbers can be with or without decimals<br>
 * Example: 5, 10 = 0
 *<p>
 * HALF_DOWN<br>
 * Rounding mode to round towards "nearest neighbor" unless
 * both neighbors are equidistant, in which case round down.
 * Behaves as for RoundingMode.UP if the discarded fraction is > 0.5;
 * otherwise, behaves as for RoundingMode.DOWN.
 */
public static BigDecimal roundDown(BigDecimal value, BigDecimal rounding){

    return round(value, rounding, RoundingMode.HALF_DOWN);

}

Here are a couple of very simple methods in c# I wrote to always round up or down to any value passed.

public static Double RoundUpToNearest(Double passednumber, Double roundto)
    {

        // 105.5 up to nearest 1 = 106
        // 105.5 up to nearest 10 = 110
        // 105.5 up to nearest 7 = 112
        // 105.5 up to nearest 100 = 200
        // 105.5 up to nearest 0.2 = 105.6
        // 105.5 up to nearest 0.3 = 105.6

        //if no rounto then just pass original number back
        if (roundto == 0)
        {
            return passednumber;
        }
        else
        {
            return Math.Ceiling(passednumber / roundto) * roundto;
        }
    }
    public static Double RoundDownToNearest(Double passednumber, Double roundto)
    {

        // 105.5 down to nearest 1 = 105
        // 105.5 down to nearest 10 = 100
        // 105.5 down to nearest 7 = 105
        // 105.5 down to nearest 100 = 100
        // 105.5 down to nearest 0.2 = 105.4
        // 105.5 down to nearest 0.3 = 105.3

        //if no rounto then just pass original number back
        if (roundto == 0)
        {
            return passednumber;
        }
        else
        {
            return Math.Floor(passednumber / roundto) * roundto;
        }
    }

In Scala I did the following (Java below)

import scala.math.BigDecimal.RoundingMode

def toFive(
   v: BigDecimal,
   digits: Int,
   roundType: RoundingMode.Value= RoundingMode.HALF_UP
):BigDecimal = BigDecimal((2*v).setScale(digits-1, roundType).toString)/2

And in Java

import java.math.BigDecimal;
import java.math.RoundingMode;

public static BigDecimal toFive(BigDecimal v){
    return new BigDecimal("2").multiply(v).setScale(1, RoundingMode.HALF_UP).divide(new BigDecimal("2"));
}

Based on your edit, another possible solution would be:

BigDecimal twenty = new BigDecimal(20);
BigDecimal amount = new BigDecimal(990.49)

// To round to the nearest .05, multiply by 20, round to the nearest integer, then divide by 20
BigDecimal result =  new BigDecimal(amount.multiply(twenty)
                                          .add(new BigDecimal("0.5"))
                                          .toBigInteger()).divide(twenty);

This has the advantage, of being guaranteed not to lose precision, although it could potentially be slower of course...

And the scala test log:

scala> var twenty = new java.math.BigDecimal(20) 
twenty: java.math.BigDecimal = 20

scala> var amount = new java.math.BigDecimal("990.49");
amount: java.math.BigDecimal = 990.49

scala> new BigDecimal(amount.multiply(twenty).add(new BigDecimal("0.5")).toBigInteger()).divide(twenty)
res31: java.math.BigDecimal = 990.5

For this test to pass :

assertEquals(bd("1.00"), round(bd("1.00")));
assertEquals(bd("1.00"), round(bd("1.01")));
assertEquals(bd("1.00"), round(bd("1.02")));
assertEquals(bd("1.00"), round(bd("1.024")));
assertEquals(bd("1.05"), round(bd("1.025")));
assertEquals(bd("1.05"), round(bd("1.026")));
assertEquals(bd("1.05"), round(bd("1.049")));

assertEquals(bd("-1.00"), round(bd("-1.00")));
assertEquals(bd("-1.00"), round(bd("-1.01")));
assertEquals(bd("-1.00"), round(bd("-1.02")));
assertEquals(bd("-1.00"), round(bd("-1.024")));
assertEquals(bd("-1.00"), round(bd("-1.0245")));
assertEquals(bd("-1.05"), round(bd("-1.025")));
assertEquals(bd("-1.05"), round(bd("-1.026")));
assertEquals(bd("-1.05"), round(bd("-1.049")));

Change ROUND_UP in ROUND_HALF_UP :

private static final BigDecimal INCREMENT_INVERTED = new BigDecimal("20");
public BigDecimal round(BigDecimal toRound) {
    BigDecimal divided = toRound.multiply(INCREMENT_INVERTED)
                                .setScale(0, BigDecimal.ROUND_HALF_UP);
    BigDecimal result = divided.divide(INCREMENT_INVERTED)
                               .setScale(2, BigDecimal.ROUND_HALF_UP);
    return result;
}

  public static BigDecimal roundTo5Cents(BigDecimal amount)
  {
    amount = amount.multiply(new BigDecimal("2"));
    amount = amount.setScale(1, RoundingMode.HALF_UP);
    // preferred scale after rounding to 5 cents: 2 decimal places
    amount = amount.divide(new BigDecimal("2"), 2, RoundingMode.HALF_UP);
    return amount;
  }

Note that this is basically the same answer as John's.

public static void roundUp()
{
    try
    {
        System.out.println("Enter the currency : $");
        Scanner keyboard = new Scanner(System.in);
        String myint = keyboard.next();
        if (!isEmptyOrBlank(myint).booleanValue())
        {
            BigDecimal d = new BigDecimal(myint);
            System.out.println("Enter the round up factor: $");
            String roundUpFactor = keyboard.next();
            if (!isEmptyOrBlank(roundUpFactor).booleanValue())
            {
                BigDecimal scale = new BigDecimal(roundUpFactor);
                BigDecimal y = d.divide(scale, MathContext.DECIMAL128);
                BigDecimal q = y.setScale(0, 0);
                BigDecimal z = q.multiply(scale);
                System.out.println("Final price after rounding up to " + roundUpFactor + " is : $" + z);
                System.out.println("Want to try with other price Y/N :");

                String exit = keyboard.next();
                if ((!isEmptyOrBlank(exit).booleanValue()) && ("y".equalsIgnoreCase(exit))) {
                    roundUp();
                } else {
                    System.out.println("See you take care");
                }
            }
        }
        else
        {
            System.out.println("Please be serious u r dealing with critical Tx Pricing");
        }
    }
    catch (Exception e)
    {
        System.out.println("Please be serious u r dealing with critical Tx Pricing enter correct rounding off value");
    }
}

Tom has the right idea, but you need to use BigDecimal methods, since you ostensibly are using BigDecimal because your values are not amenable to a primitive datatype. Something like:

BigDecimal num = new BigDecimal(0.23);
BigDecimal twenty = new BigDecimal(20);
//Might want to use RoundingMode.UP instead,
//depending on desired behavior for negative values of num.
BigDecimal numTimesTwenty = num.multiply(twenty, new MathContext(0, RoundingMode.CEILING)); 
BigDecimal numRoundedUpToNearestFiveCents
  = numTimesTwenty.divide(twenty, new MathContext(2, RoundingMode.UNNECESSARY));

You can use plain double to do this.

double amount = 990.49;
double rounded = ((double) (long) (amount * 20 + 0.5)) / 20;

EDIT: for negative numbers you need to subtract 0.5