给定各种温度输入，如何确定归一化温度？

Question

我需要使用归一化的温度值来检查某人是否有病。

我编写了一个从摄氏温度转换为华氏温度的函数，但需要使用归一化温度编写相同的函数。

public void hasFever(int temperature1, int temperature2, int temperature3, int temperature4, String bodylocation) throws ArithmeticException, InputMismatchException {
    final int noOfTimesTempTaken = 4;
    if(bodylocation == "ear") { //Determine if someone has fever if temperature was taken by the ear
        final double feverTemperature = 98.6;   
        double temperature1InFahrenheit = (temperature1 * 9/5) + 32;
        double temperature2InFahrenheit = (temperature2 * 9/5) + 32;
        double temperature3InFahrenheit = (temperature3 * 9/5) + 32;
        double temperature4InFahrenheit = (temperature4 * 9/5) + 32;
        double avgTemp = (temperature1InFahrenheit + temperature2InFahrenheit + temperature3InFahrenheit + temperature4InFahrenheit) / noOfTimesTempTaken;
        if(avgTemp >= feverTemperature) {
            System.out.println("This person has a fever because their temperature of " + avgTemp + " is greater than normal which is 98.6");
        } else {
            System.out.println("This person does not have a fever because their temperature of " + avgTemp + " is less than normal which is 98.6");
        }
    }
}

Answer 1

首先，需要注意的是，标准化温度并不是一个具有公认/标准含义的术语，适用于此处。术语 normalize 有时用于表示“调整以处理测量不准确”，但这似乎不是您在这里尝试做的。

（必须说你还没有成功澄清你所说的归一化是什么意思，所以我们不得不猜测。这是我最好的猜测，基于你所说的。）

我认为您正在寻找这样的东西：

double VALID_FAHRENHEIT_TEMP_THRESHOLD = 60.0;  // say

public double normalizeToFahrenheit(double temp) {
    if (temp >= VALID_FAHRENHEIT_TEMP_THRESHOLD) {
        return temp;
    } else {
        return temp * (9.0 / 5.0) + 32.0;
    }
}

但问题在于，这种“标准化”在科学和数学上都是不合理的。我们假设如果一个数字小于某个特定值，它必须以摄氏度而不是华氏度为单位。但这不一定是真的。摄氏度和华氏度都下降到非常低的值。

更实际地，如果有人淹死在冰冷的水中，他们的华氏温度可能会下降很多……可能接近冰点。它可能低于我们选择的阈值。如果我们假设它“必须”是摄氏度并转换它，我们会报告不正确的温度。

一个相关的问题是，一个奇怪的温度值可能是用户误读温度计、误听测量值或误输入值的结果；例如11.3 而不是 101.3。对错误的输入应用归一化会导致更多的误导性结果。

---

因此，鉴于这种“标准化”方法无效，您应该怎么做？

我的建议是让您的应用向用户报告异常情况。例如，向他们展示一个这样的对话框：

  The temperature xy degrees is outside of the expected range for human 
  temperatures in Fahrenheit.  

  Options: 

     1) Accept as a Fahrenheit value.  
     2) Apply Centigrade to Fahrenheit conversion. 
     3) Switch device to Centigrade mode.  
     4) Reenter value.

Answer 2

希望我能正确理解您的问题。因此，您想将通用温度单位传递给您的函数。

class Temperature {
    private final static double CELSIUS_OFFSET = 273,15;
    private final static double FAHRENHEIT_OFFSET = 32;
    private final static double FAHRENHEIT_FACTOR = 9.0 / 5.0;
    private final double tempInKelvin;

    private Temperature() {
        //Private constructor, so only this class can access it
    }

    public static Temperature fromKelvin(double tempInKelvin) {
        final Temperature temperature = new Temperature();
        temperature.tempInKelvin = tempInKelvin;
        return temperature;
    }

    public static Temperature fromCelsius(double tempInCelsius) {
        return fromKelvin(tempInCelsius - CELSIUS_OFFSET);
    }

    public static Temperature fromFahrenheit(double tempInFahrenheit) {
        return fromCelsius((tempInFahrenheit - FAHRENHEIT_OFFSET) / FAHRENHEIT_FACTOR);
    }

    public double getInKelvin() {
        return tempInKelvin;
    }

    public double getInCelsius() {
        return tempInKelvin + CELSIUS_OFFSET;
    }

    public double getInFahrenheit() {
        return getInCelsius()*FAHRENHEIT_FACTOR + FAHRENHEIT_OFFSET;
    }
}

使用这个类你可以做到这一点：

//Example use (has to be in a method, of course):
final Temperature temp1 = Temperature.fromCelsius(39.2);
final Temperature temp2 = Temperature.fromFahrenheit(95.3);
final Temperature temp3 = Temperature.fromCelsius(39.1);
final Temperature temp4 = Temperature.fromCelsius(39.3);
hasFever(temp1, temp2, temp3, temp4, "ear");

//New hasFever() method:
public void hasFever(Temperature temp1, Temperature temp2, Temperature temp3, Temperature temp4, String bodylocation) {
    final int noOfTimesTempTaken = 4;
    //USE String.equals()!!!
    if(bodylocation.equals("ear")) { //Determine if someone has fever if temperature was taken by the ear
        final double feverTemperature = 98.6;
        double avgTemp = (temp1.getInFahrenheit() + temp2.getInFahrenheit() + temp3.getInFahrenheit() + temp4.getInFahrenheit()) / noOfTimesTempTaken;
        if(avgTemp >= feverTemperature) {
            System.out.println("This person has a fever because their temperature of " + avgTemp + " is greater than normal which is 98.6");
        } else {
            System.out.println("This person does not have a fever because their temperature of " + avgTemp + " is less than normal which is 98.6");
        }
    }
}

现在你有一个通用的Temperature 类，它基本上没有可访问的单元，直到你指定你想要的。这也会为您转换所有单位。