java - 考虑最接近的十分之一将华氏度计算为摄氏度

Question

这个问题已经有答案了:

How to round a number to n decimal places in Java (39 个回答)

已关闭 3 年前。

我正在做以下编程练习:Grasshopper Debug.声明如下:

Debug celsius converter

Your friend is traveling abroad to the United States so he wrote a program to convert fahrenheit to celsius. Unfortunately his code has some bugs.

Find the errors in the code to get the celsius converter working properly.

To convert fahrenheit to celsius:

celsius = (fahrenheit - 32) * (5/9)

Remember that typically temperatures in the current weather conditions are given in whole numbers. It is possible for temperature sensors to report temperatures with a higher accuracy such as to the nearest tenth. Instrument error though makes this sort of accuracy unreliable for many types of temperature measuring sensors.

我尝试过以下代码:

public class GrassHopper {

    public static String weatherInfo(int temp) {
        double c=convertToCelsius(temp);
        if (c < 0)
            return (c + " is freezing temperature");
        else
            return (c + " is above freezing temperature");
    }

    public static double convertToCelsius(int temperature) {
        double celsius = (temperature-32)*(5/9.0);
        return celsius;
    }
}

我们发现了以下案例:

test5
expected:<31.11111111111111[] is above freezing t...> but was:<31.11111111111111[4] is above freezing t...>

我们认为这可能是因为使用了 double。然后我们使用 BigDecimal‽ 编写了相同的解决方案:

import java.math.BigDecimal;
import java.text.*;
public class GrassHopper {

    public static String weatherInfo(int temp) {
        BigDecimal c=convertToCelsius(BigDecimal.valueOf(temp));
        NumberFormat oneDecimal = new DecimalFormat("#0.0");
        c=c.toString().contains(".00") ? BigDecimal.valueOf(Double.parseDouble(oneDecimal.format(c))) : c;
        if (c.compareTo(BigDecimal.ZERO) < 0)
            return (c + " is freezing temperature");
        else
            return (c + " is above freezing temperature");
    }

    public static BigDecimal convertToCelsius(BigDecimal temperature) {
        BigDecimal celsius = (temperature.subtract(BigDecimal.valueOf(32))).multiply(BigDecimal.valueOf(5/9.0));
        return celsius;
    }
}

测试结果如下:

test1
expected:<13.333333333333334[] is above freezing t...> but was:<13.333333333333334[4] is above freezing t...>



test5
expected:<-2.222222222222222[3] is freezing tempera...> but was:<-2.222222222222222[4] is freezing tempera...>

要了解发生了什么并尝试解决它，我们已阅读:

• Double decimal formatting in Java

• https://docs.oracle.com/javase/7/docs/api/java/text/NumberFormat.html

• Comparing double values for equality in Java.

• How do i compare values of BigInteger to be used as a condition in a loop?

你能帮助我们吗‽

编辑:根据@GovindaSakhare提供的答案，我们可以写:

public class GrassHopper {

    public static String weatherInfo(int temp) {
        System.out.println("original celsius temp: "+temp);
        double c=convertToCelsius(temp);
        double roundedValue=roundoff(c,15);
        String result = "";
        if (roundedValue < 0){
            result=roundedValue + " is freezing temperature";
        }else{
            result=roundedValue + " is above freezing temperature";
        }
        System.out.println("result: "+result);
        return result;
    }

    public static double convertToCelsius(int temperature) {
        double celsius = (temperature-32)*(5/9.0);
        return celsius;
    }

    public static double roundoff(double temperature, int precisionLevel) {
      double prec = Math.pow(10, precisionLevel);
      return Math.round(temperature * prec) / prec;
    }
}

但是，我们如何知道应该使用什么精度级别‽

我问这个问题是因为，通过前面的代码，我们观察到以下行为:

一些测试确实通过了:

 test1
Log
temp: 56
result: 13.333333333333334 is above freezing temperature

test2
Log
temp: 23
result: -5.0 is freezing temperature

 test4
Log
temp: 5
result: -15.0 is freezing temperature

还有一些没有通过:

test3
original celsius temp: temp: 33

expected:<0.55555555555555[5]6 is above freezing ...> but was:<0.55555555555555[]6 is above freezing ...>

test5
original celsius temp: 54

expected:<12.22222222222222[1] is above freezing t...> but was:<12.22222222222222[3] is above freezing t...>

我们认为它必须与最后一个练习的陈述部分相关:

"...Remember that typically temperatures in the current weather conditions are given in whole numbers. It is possible for temperature sensors to report temperatures with a higher accuracy such as to the nearest tenth. Instrument error though makes this sort of accuracy unreliable for many types of temperature measuring sensors. "

你会如何思考和行动来解决这个困难‽‽

编辑2:在@GovindaSakhare讨论链接之后，我们发现Java中的测试用例与我们预期的不同。我们尝试了用户 vztot 给出的提示，我们使用了 MathContext.DECIMAL128

我们尝试了以下方法:

import java.math.*;
import java.text.*;
public class GrassHopper {

    public static String weatherInfo(int temp) {
        BigDecimal c=convertToCelsius(BigDecimal.valueOf(temp));
        if (c.compareTo(BigDecimal.ZERO) <= 0)
            return (c + " is freezing temperature");
        else
            return (c + " is above freezing temperature");
    }

    public static BigDecimal convertToCelsius(BigDecimal temperature) {
        BigDecimal division = BigDecimal.valueOf(5).divide(BigDecimal.valueOf(9.0),MathContext.DECIMAL128);
        BigDecimal subtraction = temperature.subtract(BigDecimal.valueOf(32),MathContext.DECIMAL128);
        BigDecimal celsius = subtraction.multiply(division,MathContext.DECIMAL128);
        return celsius;
    }
}

但是它在所有测试中都给出了错误:

test1
expected:<13.33333333333333[4] is above freezing t...> but was:<13.33333333333333[333333333333333333] is above freezing t...>

test2
expected:<-5.0[] is freezing tempera...> but was:<-5.0[00000000000000000000000000000000] is freezing tempera...>

test3
expected:<0.555555555555555[]6 is above freezing ...> but was:<0.555555555555555[555555555555555555]6 is above freezing ...>

test4
expected:<-15.0[] is freezing tempera...> but was:<-15.0[0000000000000000000000000000000] is freezing tempera...>

test5
expected:<34.44444444444444[] is above freezing t...> but was:<34.44444444444444[444444444444444445] is above freezing t...>

所以，我们决定更好地思考，我们应该使用 double 作为输入参数而不是 int:

public class GrassHopper {

    public static String weatherInfo /*🌡️ℹ️*/ (double temp) {
        double c = convertToCelsius(temp);
        if (c <= 0)
            return (c + " is freezing temperature");
        else
            return (c + " is above freezing temperature");
    }

    public static double convertToCelsius(double temperature) {
        return (temperature - 32) * 5/9.0;
    }
}

因此，最后一个版本确实通过了测试。

Answer 1

您可以使用org.apache.commons.math3.util.Precision.equals()来自 Commons Math3。