Java 将 BigDecimal 舍入到最接近的 5 美分
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round BigDecimal to nearest 5 cents
提问by Dónal
I'm trying to figure out how to round a monetary amount upwards to the nearest 5 cents. The following shows my expected results
我想弄清楚如何将金额向上舍入到最接近的 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
I need the result to be have a precision of 2 (as shown above).
我需要结果的精度为 2（如上所示）。
Update
更新
Following the advice below, the best I could do is this
按照下面的建议，我能做的最好的就是这个
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 seemsto work.
我不相信这是 100% kosher - 我担心在双打转换时可能会丢失精度。然而，这是迄今为止我想出的最好的并且似乎有效。
采纳答案by Peter Lawrey
You can use plain double to do this.
你可以使用普通的 double 来做到这一点。
double amount = 990.49;
double rounded = ((double) (long) (amount * 20 + 0.5)) / 20;
EDIT: for negative numbers you need to subtract 0.5
编辑：对于负数，您需要减去 0.5
回答by James Cronen
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.
我会尝试乘以 20，四舍五入到最接近的整数，然后除以 20。这是一个技巧，但应该会得到正确的答案。
回答by Matt J
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:
Tom 的想法是正确的，但您需要使用 BigDecimal 方法，因为表面上您正在使用 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));
回答by Paul Wagland
Based on your edit, another possible solution would be:
根据您的编辑，另一种可能的解决方案是：
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);
This has the advantage, of being guaranteed not to lose precision, although it could potentially be slower of course...
这有一个优点，可以保证不会失去精度，尽管它当然可能会更慢......
And the scala test log:
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
回答by NER1808
Here are a couple of very simple methods in c# I wrote to always round up or down to any value passed.
这是我在 c# 中编写的几个非常简单的方法，它们总是向上或向下舍入到传递的任何值。
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;
}
}
回答by marcolopes
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
几年前我用 Java 写了这个：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);
}
回答by robinst
Using BigDecimalwithout any doubles (improved on the answer from marcolopes):
BigDecimal不使用任何双打（改进了 marcolopes 的答案）：
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(bdis a helper which just returns new BigDecimal(input)):
舍入模式是例如RoundingMode.HALF_UP。对于您的示例，您实际上想要RoundingMode.UP(bd是一个刚返回的助手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));
Also note that there is a mistake in your last example (rounding 1.900001 to 1.10).
另请注意，上一个示例中存在错误（将 1.900001 舍入为 1.10）。
回答by Micka?l Gauvin
For this test to pass :
要通过此测试：
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")));
Change ROUND_UPin ROUND_HALF_UP:
更改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;
}
回答by John
In Scala I did the following (Java below)
在 Scala 中，我执行了以下操作（下面是 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
And in Java
在 Java 中
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"));
}
回答by Reto H?hener
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.
请注意，这与John 的答案基本相同。
