圆形圆形达到最接近5美分

Question

我试图弄清楚如何将货币量到最接近的5美分。以下显示了我的预期结果

1.03     => 1.05
1.051    => 1.10
1.05     => 1.05
1.900001 => 1.10

我需要结果的精度为2（如上所示）。

更新

遵循以下建议，我能做的最好的就是

    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)

我不相信这是100％犹太洁食 - 我担心在转换为双打时可能会丢失精度。但是，这是我到目前为止提出的最好的 似乎 去工作。

Answer 1

您可以使用普通的双重来做到这一点。

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

编辑：对于负数，您需要减去0.5

Answer 2

使用 BigDecimal 没有任何双打（在Marcolopes的答案中得到改善）：

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;
    }
}

圆形模式是 RoundingMode.HALF_UP. 。对于您的示例，您实际上想要 RoundingMode.UP (bd 是一个刚刚返回的助手 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));

另请注意，您的上一个示例中存在一个错误（将1.900001舍入至1.10）。

Answer 3

我会尝试将20乘以20，然后将其舍入到最近的整数，然后除以20。这是一个黑客，但应该为您提供正确的答案。

Answer 4

几年前，我在Java写了这篇文章： https://github.com/marcolopes/dma/blob/master/org.dma.java/src/org/dma/java/math/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);

}

Answer 5

这是C＃中的几种非常简单的方法，我写的是总是向上或向下汇总到传递的任何值。

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;
        }
    }

Answer 6

在Scala中，我做了以下（下面的Java）

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

和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"));
}

Answer 7

根据您的编辑，另一个可能的解决方案将是：

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);

这有一个优势，可以保证不要失去精度，尽管当然可能会更慢...

和Scala测试日志：

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

Answer 8

为了通过此测试：

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")));

改变 ROUND_UP 在 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;
}

Answer 9

  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;
  }

请注意，这基本上是相同的答案 约翰.

Answer 10

汤姆（Tom）有一个正确的想法，但是您需要使用BigDecimal方法，因为表面上您使用的是BigDecimal，因为您的值不适合原始数据类型。就像是：

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));