What's the best practice to round a float to 2 decimals?

I'm using eclipse + Android SDK.

I need to round a float value to 2 decimals. I usually use the next "trick" using Math library.

float accelerometerX = accelerometerX * 100;
accelerometerX = round(accelerometerX);
Log.d("Test","" + accelerometerX/100);

But I feel it is not the best way to do it.

Is there a library to do these type of operations?

206787 次浏览
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double roundTwoDecimals(double d) {
DecimalFormat twoDForm = new DecimalFormat("#.##");
return Double.valueOf(twoDForm.format(d));
}
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最佳答案

两年前我在 Java 中研究统计学我仍然得到了一个函数的代码它允许你把一个数四舍五入到你想要的小数。现在你需要两个，但是也许你想尝试用3来比较结果，这个函数给你自由。

/**
* Round to certain number of decimals
*
* @param d
* @param decimalPlace
* @return
*/
public static float round(float d, int decimalPlace) {
BigDecimal bd = new BigDecimal(Float.toString(d));
bd = bd.setScale(decimalPlace, BigDecimal.ROUND_HALF_UP);
return bd.floatValue();
}

您需要决定是向上舍入还是向下舍入。

希望能有帮助。

剪辑

如果你想保留小数为零时的数目(我想这只是为了显示给用户) ，你只需要将函数类型从 float 改为 BigDecimal，如下所示:

public static BigDecimal round(float d, int decimalPlace) {
BigDecimal bd = new BigDecimal(Float.toString(d));
bd = bd.setScale(decimalPlace, BigDecimal.ROUND_HALF_UP);
return bd;
}

然后这样调用函数:

float x = 2.3f;
BigDecimal result;
result=round(x,2);
System.out.println(result);

这将印刷:

2.30
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//by importing Decimal format we can do...


import java.util.Scanner;
import java.text.DecimalFormat;
public class Average
{
public static void main(String[] args)
{
int sub1,sub2,sub3,total;
Scanner in = new Scanner(System.in);
System.out.print("Enter Subject 1 Marks : ");
sub1 = in.nextInt();
System.out.print("Enter Subject 2 Marks : ");
sub2 = in.nextInt();
System.out.print("Enter Subject 3 Marks : ");
sub3 = in.nextInt();
total = sub1 + sub2 + sub3;
System.out.println("Total Marks of Subjects = " + total);
res = (float)total;
average = res/3;
System.out.println("Before Rounding Decimal.. Average = " +average +"%");
DecimalFormat df = new DecimalFormat("###.##");
System.out.println("After Rounding Decimal.. Average = " +df.format(average)+"%");
}
}
/* Output
Enter Subject 1 Marks : 72
Enter Subject 2 Marks : 42
Enter Subject 3 Marks : 52
Total Marks of Subjects = 166
Before Rounding Decimal.. Average = 55.333332%
After Rounding Decimal.. Average = 55.33%
*/


/* Output
Enter Subject 1 Marks : 98
Enter Subject 2 Marks : 88
Enter Subject 3 Marks : 78
Total Marks of Subjects = 264
Before Rounding Decimal.. Average = 88.0%
After Rounding Decimal.. Average = 88%
*/


/* You can Find Avrerage values in two ouputs before rounding average
And After rounding Average..*/
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这里有一个比@Jav _ Rock 更短的实现

   /**
* Round to certain number of decimals
*
* @param d
* @param decimalPlace the numbers of decimals
* @return
*/


public static float round(float d, int decimalPlace) {
return BigDecimal.valueOf(d).setScale(decimalPlace,BigDecimal.ROUND_HALF_UP).floatValue();
}






System.out.println(round(2.345f,2));//two decimal digits, //2.35
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让我们测试三种方法:
1)

public static double round1(double value, int scale) {
return Math.round(value * Math.pow(10, scale)) / Math.pow(10, scale);
}

2)

public static float round2(float number, int scale) {
int pow = 10;
for (int i = 1; i < scale; i++)
pow *= 10;
float tmp = number * pow;
return ( (float) ( (int) ((tmp - (int) tmp) >= 0.5f ? tmp + 1 : tmp) ) ) / pow;
}

3)

public static float round3(float d, int decimalPlace) {
return BigDecimal.valueOf(d).setScale(decimalPlace, BigDecimal.ROUND_HALF_UP).floatValue();
}



Number is 0.23453f
我们将测试每个方法的100,000次迭代。


Time 1 - 18 ms
时间2-1毫秒
时间3-378毫秒


在笔记本电脑上测试
Intel i3-3310M CPU 2.4 GHz

小开

我已经尝试支持@Ivan Stin 优秀的第二种方法 (主要功劳归于伊万 · 斯汀的方法)的-ve 值

public static float round(float value, int scale) {
int pow = 10;
for (int i = 1; i < scale; i++) {
pow *= 10;
}
float tmp = value * pow;
float tmpSub = tmp - (int) tmp;


return ( (float) ( (int) (
value >= 0
? (tmpSub >= 0.5f ? tmp + 1 : tmp)
: (tmpSub >= -0.5f ? tmp : tmp - 1)
) ) ) / pow;


// Below will only handles +ve values
// return ( (float) ( (int) ((tmp - (int) tmp) >= 0.5f ? tmp + 1 : tmp) ) ) / pow;
}

下面是我试用过的测试用例。如果没有处理其他用例，请告诉我。

@Test
public void testFloatRound() {
// +ve values
Assert.assertEquals(0F, NumberUtils.round(0F), 0);
Assert.assertEquals(1F, NumberUtils.round(1F), 0);
Assert.assertEquals(23.46F, NumberUtils.round(23.4567F), 0);
Assert.assertEquals(23.45F, NumberUtils.round(23.4547F), 0D);
Assert.assertEquals(1.00F, NumberUtils.round(0.49999999999999994F + 0.5F), 0);
Assert.assertEquals(123.12F, NumberUtils.round(123.123F), 0);
Assert.assertEquals(0.12F, NumberUtils.round(0.123F), 0);
Assert.assertEquals(0.55F, NumberUtils.round(0.55F), 0);
Assert.assertEquals(0.55F, NumberUtils.round(0.554F), 0);
Assert.assertEquals(0.56F, NumberUtils.round(0.556F), 0);
Assert.assertEquals(123.13F, NumberUtils.round(123.126F), 0);
Assert.assertEquals(123.15F, NumberUtils.round(123.15F), 0);
Assert.assertEquals(123.17F, NumberUtils.round(123.1666F), 0);
Assert.assertEquals(123.46F, NumberUtils.round(123.4567F), 0);
Assert.assertEquals(123.87F, NumberUtils.round(123.8711F), 0);
Assert.assertEquals(123.15F, NumberUtils.round(123.15123F), 0);
Assert.assertEquals(123.89F, NumberUtils.round(123.8909F), 0);
Assert.assertEquals(124.00F, NumberUtils.round(123.9999F), 0);
Assert.assertEquals(123.70F, NumberUtils.round(123.7F), 0);
Assert.assertEquals(123.56F, NumberUtils.round(123.555F), 0);
Assert.assertEquals(123.00F, NumberUtils.round(123.00F), 0);
Assert.assertEquals(123.50F, NumberUtils.round(123.50F), 0);
Assert.assertEquals(123.93F, NumberUtils.round(123.93F), 0);
Assert.assertEquals(123.93F, NumberUtils.round(123.9312F), 0);
Assert.assertEquals(123.94F, NumberUtils.round(123.9351F), 0);
Assert.assertEquals(123.94F, NumberUtils.round(123.9350F), 0);
Assert.assertEquals(123.94F, NumberUtils.round(123.93501F), 0);
Assert.assertEquals(99.99F, NumberUtils.round(99.99F), 0);
Assert.assertEquals(100.00F, NumberUtils.round(99.999F), 0);
Assert.assertEquals(100.00F, NumberUtils.round(99.9999F), 0);


// -ve values
Assert.assertEquals(-123.94F, NumberUtils.round(-123.93501F), 0);
Assert.assertEquals(-123.00F, NumberUtils.round(-123.001F), 0);
Assert.assertEquals(-0.94F, NumberUtils.round(-0.93501F), 0);
Assert.assertEquals(-1F, NumberUtils.round(-1F), 0);
Assert.assertEquals(-0.50F, NumberUtils.round(-0.50F), 0);
Assert.assertEquals(-0.55F, NumberUtils.round(-0.55F), 0);
Assert.assertEquals(-0.55F, NumberUtils.round(-0.554F), 0);
Assert.assertEquals(-0.56F, NumberUtils.round(-0.556F), 0);
Assert.assertEquals(-0.12F, NumberUtils.round(-0.1234F), 0);
Assert.assertEquals(-0.12F, NumberUtils.round(-0.123456789F), 0);
Assert.assertEquals(-0.13F, NumberUtils.round(-0.129F), 0);
Assert.assertEquals(-99.99F, NumberUtils.round(-99.99F), 0);
Assert.assertEquals(-100.00F, NumberUtils.round(-99.999F), 0);
Assert.assertEquals(-100.00F, NumberUtils.round(-99.9999F), 0);
}
小开

这里有一个简单的一行解决方案

((int) ((value + 0.005f) * 100)) / 100f

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