英文:
analysing the proportion of names in a set of names. my output is incorrect. ("John") should return 0.25. and mary should be 0.5
问题
以下是翻译好的内容：
public class Problem
{
private int count;
private final HashMap<String, Integer> counts;
public Problem() {
this.counts = new HashMap<>();
}
public void addName(String name)
{
Integer nameCount = counts.get(name);
if (nameCount == null)
{
nameCount = 1;
counts.put(name, nameCount);
nameCount++;
}
nameCount++;
count++;
}
// 这个函数应该计算比例
public double nameProportion(String name)
{
return counts.get(name) / (double) count;
}
public static void main(String[] args)
{
Problem namesCount = new Problem();
namesCount.addName("James");
namesCount.addName("John");
namesCount.addName("Mary");
namesCount.addName("Mary");
System.out.println("约翰的比例: " + namesCount.nameProportion("John"));
System.out.println("玛丽的比例: " + namesCount.nameProportion("Mary"));
}
}
目前输出结果是不正确的，由于某种原因没有将两个玛丽视为同一人。
英文:
my output is currently ("john") returns 0.25 and ("mary") also returns 0.25
public class Problem
{
private int count;
private final HashMap<String, Integer> counts;
public Problem() {
this.counts = new HashMap<>();
}
public void addName(String name)
{
Integer nameCount = counts.get(name);
if (nameCount == null)
{
nameCount = 1;
counts.put(name, nameCount);
nameCount++;
}
nameCount++;
count++;
}
this function should calculate the proportion
public double nameProportion(String name)
{
return counts.get(name) / (double) count;
}
public static void main(String[] args)
{
Problem namesCount = new Problem();
namesCount.addName("James");
namesCount.addName("John");
namesCount.addName("Mary");
namesCount.addName("Mary");
System.out.println("Fraction of Johns: " + namesCount.nameProportion("John"));
System.out.println("Fraction of Marys: " + namesCount.nameProportion("Mary"));
}
}
right now the output is incorrect, for some reason its not seeing the two marys as the same
答案1
得分: 0
原始类型和原始包装类型在执行nameCount++操作时并不是不可变的，它不会改变哈希映射中键的值。它会创建另一个整数对象，并将nameCount + 1的值放入该对象中，因此在您的代码中，哈希映射中所有键的值都是1。
英文:
Primitive and primitive wrapper are not immutable when you do nameCount++, it doesn't change the value of key in hashmap. It create another Integer object and put the value of nameCount +1 into that object, so the value of all key in hashmap in your code is 1.
答案2
得分: 0
这一切都在你的函数 Problem#addName 中。实际上，你是在函数的末尾增加了 nameCount，但是当函数到达块的末尾时，增加后的值就丢失了（nameCount 是一个局部变量，不是成员变量）。
解决方法是，如果 nameCount 不为 null，就将其递增，并将其放入 counts 映射中，示例如下：
public void addName(String name) {
Integer nameCount = counts.get(name);
if (nameCount == null) {
counts.put(name, 1);
} else {
counts.put(name, ++nameCount);
}
count++;
}
如果为 null，则只需将 1 放入映射中，对应于相应的 name。
递增 count 是可以的，因为那是一个成员变量，并且在创建的 namesCount 引用到 Problem 对象中得以保存。
英文:
It's all in your function Problem#addName. You are actually incrementing your nameCount at the end of the function, but the incremented value is lost when the function reaches the end of the block (nameCount is a local variable, not a member variable).
The solution to this is to increment nameCount if it is not null and put that into the counts map, like so for example:
public void addName(String name) {
Integer nameCount = counts.get(name);
if (nameCount == null) {
counts.put(name, 1);
} else {
counts.put(name, ++nameCount);
}
count++;
}
If it is null, then simply a 1 is put into the map for the corresponding name.
Incrementing the count is fine, because that is a member variable and is saved within the created namesCount reference to the Problem object.
答案3
得分: 0
你如果名称已经存在于Map中，则不会对其进行更新。需要注意的是，Integer对象是不可变的；如果你不想不断丢弃先前的对象，可以使用AtomicInteger来代替。 <sup>演示</sup>
public void addName(String name) {
Integer nameCount = counts.get(name);
if (nameCount == null) {
nameCount = 1;
counts.put(name, nameCount);
} else {
counts.put(name, nameCount + 1);
}
count++;
}
你可以使用Map#getOrDefault来简化这个过程。 <sup>演示</sup>
public void addName(String name) {
counts.put(name, counts.getOrDefault(name, 0) + 1);
count++;
}
英文:
You never update the Map if the name already exists in it. It is important to note that Integer objects are immutable; you can use AtomicInteger instead if you do not want to keep discarding previous objects. <sup>Demo</sup>
public void addName(String name) {
Integer nameCount = counts.get(name);
if (nameCount == null) {
nameCount = 1;
counts.put(name, nameCount);
} else {
counts.put(name, nameCount + 1);
}
count++;
}
This can be simplified using Map#getOrDefault. <sup>Demo</sup>
public void addName(String name) {
counts.put(name, counts.getOrDefault(name, 0) + 1);
count++;
}
专注分享java语言的经验与见解，让所有开发者获益！
评论