Java：找出百分比差异

Question

我想弄清楚如何找到原始（无 space）文本字符串和去元音（无 space）文本字符串之间的百分比差异。我试图通过使用等式 ((newAmount-reducedAmount)/reducedAmount) 来做到这一点，但我没有运气，最终得到的值为零，如下所示。

谢谢！

我的代码：

import java.util.Scanner;

public class Prog5 {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner console = new Scanner(System.in);

        System.out.println("Welcome to the disemvoweling utility!"); // Initially typed "disemboweling" xD
        System.out.print("Enter text to be disemvoweled: ");
        String inLine = console.nextLine();
        String vowels= inLine.replaceAll("[AEIOUaeiou]", ""); // RegEx for vowel control
        System.out.println("Your disemvoweled text is: " + vowels); // Prints disemvoweled text

    // Used to count all characters without counting white space(s)
    int reducedAmount = 0;
    for (int i = 0, length = inLine.length(); i < length; i++) {
        if (inLine.charAt(i) != ' ') {
            reducedAmount++;
        }
    }

    // newAmount is the number of characters on the disemvoweled text without counting white space(s)
    int newAmount = 0;
    for (int i = 0, length = vowels.length(); i < length; i++) {
        if (vowels.charAt(i) != ' ') {
            newAmount++;
        }
    }

    int reductionRate = ((newAmount - reducedAmount) / reducedAmount); // Percentage of character reduction


    System.out.print("Reduced from " + reducedAmount + " to " + newAmount + ". Reduction rate is " + reductionRate + "%");

    }
}

我的输出：（测试字符串没有引号："Testing please"）

Welcome to the disemvoweling utility!

Enter text to be disemvoweled: Testing please

Your disemvoweled text is: Tstng pls

Reduced from 13 to 8. Reduction rate is 0%

Answer 1

您的公式给出了一个介于零和一之间的值。

整数不能包含小数，因此它始终显示为零。

乘以 100 得到常规百分比值。

int reductionRate = 100*(newAmount - reducedAmount) / reducedAmount; // Percentage of character reduction

Answer 2

您在执行整数除法计算百分比差异时使用了整数数据类型。您需要在等式右侧键入变量之一以执行双除法，然后将它们存储在双精度中。这样做的原因是 java 整数类型不能容纳实数。另外，将它乘以 100 得到百分比。

double reductionRate = 100 * ((newAmount - reducedAmount) / (double)reducedAmount);

如果你想要一个介于 0 和 1 之间的小数，那么

double reductionRate = ((newAmount - reducedAmount) / (double)reducedAmount);

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