Question

System.out.print("This program computes the average and variance of all numbers entered. ");

    boolean finished = false;
    int total = 0;

    Scanner inputScanner = new Scanner (System.in);
    double average = inputScanner.nextDouble();
    double variance = inputScanner.nextDouble();

    while (!finished)
    {
        System.out.println("Enter a number (or type 'exit'): ");

        if (inputScanner.hasNextDouble())
        {

            total++;
            System.out.println(total);
            double currentNum = inputScanner.nextDouble();
            average = (average) + (currentNum - average) / total;
            double prevAverage = ((average * total) - currentNum) / (total - 1);
            variance = ((variance * (total - 1)) + (currentNum - prevAverage) 
                                                            * (currentNum - average)) / total;

            System.out.println("So far the average is: " + average 
                                            + "\nAnd the variance so far is: " + variance );

        }
        else if (inputScanner.hasNext("exit"))
        {
            finished = true;
            System.out.println("Thanks for using this program.");
            inputScanner.close();

我不知道为什么在它要求我输入数字之前我必须输入两个输入？方差公式也是NaN，我不确定为什么。任何帮助将不胜感激，谢谢。

修改

double var1 = (variance * (total - 1));
            double var2 = (currentNum - prevAverage) * (currentNum -         average);  
            variance = (var1 + var2) / total;

这是我对差异的计算。我如何解决它？如果我输入＆＃39; 1＆＃39; var2是NaN，如果我输入2 var1是Nan。

Answer 1

在进入循环之前调用inputScanner.nextDouble（）两次。这需要两个值才能继续。

我建议你把它改成

chmod 755 file1

Answer 2

您必须在＆＃34之前键入两个输入;输入一个数字（或键入＆＃39;退出＆＃39;）：＆＃34;声明如下面的行： double currentNum = inputScanner.nextDouble（）;

您正在使用以下行接受来自用户的两个double值： double average = inputScanner.nextDouble（）; double variance = inputScanner.nextDouble（）;

删除以上两行，只声明/初始化变量将解决您的问题。

您正如以下公式之前的差异获得NaN 你正在做整个++内部循环，它将评估为值1 所以在下一行中，total-1将变为0，并且zoro的任何内容都将是NaN

double prevAverage =（（average * total） - currentNum）/（total - 1）; //将是NaN

Answer 3

在循环之前执行此操作：

double average = inputScanner.nextDouble();
double variance = inputScanner.nextDouble();

您阻止，直到用户输入两个双倍值...

编辑：方差计算（我没有检查它的语义BTW）

double totalVar = variance * (double) (total - 1);
double currentNumMinusPreviousAvg = currentNum - prevAverage;
double currentNumMinusAvg = currentNum - average;    
variance = (totalVar + currentNumMinusPreviousAvg * currentNumMinusAvg) / total;

这样的事情应该避免对中间表达式进行int的无法强制转换，并且使用适当的变量名称使计算更加可读。

Answer 4

您经常查询用户，并且您应该只在循环开始时执行此操作，并将其存储在变量中以在if-else语句中使用它。因为它可以是一个字符串，所以我们要求一个字符串，然后再解析为一个字符串。

boolean finished = false;
int total = 0;

Scanner inputScanner = new Scanner (System.in);
double average = inputScanner.nextDouble();
double variance = inputScanner.nextDouble();

while (!finished) {
    System.out.println("Enter a number (or type 'exit'): ");
    String input = inputScanner.nextLine();


    if( input.equals( "exit" ) ) {
        finished = true;
        System.out.println("Thanks for using this program.");
        inputScanner.close();
    }
    else {
        try {
            double nextDouble = Double.parseDouble( input );

            //If an exception is trown it won't get past here, so 
            //it won't count
            total++;
            System.out.println(total);

            double currentNum = nextDouble;
            average = (average) + (currentNum - average) / total;
            double prevAverage = ((average * total) - currentNum) / (total - 1);
            variance = ((variance * (total - 1)) + (currentNum - prevAverage) 
                    * (currentNum - average)) / total;

            System.out.println("So far the average is: " + average 
                    + "\nAnd the variance so far is: " + variance );
        }
        catch ( NumberFormatException ex ) {
            System.out.println( "The number is badly writen" );
            //It can throw NumberFormatException exception if the 
            //number is badly writen.
        }
    }

}

另外，如果您打算精确使用十进制数，我建议使用BigDecimal类。

我希望我有所帮助。

祝你有个愉快的一天。：）

扫描仪直到两个输出才扫描？

4 个答案: