Question

我有一个名为HashMap的{{1}}变量map是key而Integer是value。

例如，String中有100个值。我想搜索map的100个值，然后我想在"Donkey"中返回Integer的{{1}}，如果没有，则"Donkey" 0.我尝试使用map循环return Integer但没有运气。

有人可以给我一个暗示吗？

Answer 1

试试这个：

int count = Collections.frequency(mapVar.values(), "Donkey");
System.out.println(count);

让我知道它是否有效：）

Answer 2

Collections.frequency方法很不错。但是，map.values()不起作用的断言有点奇怪，因为它应该有效。我想添加它特别奇怪，因为传递给Collections.frequency 的集合是 map.values()。

// method to do the counting of a particular animal in a map
static Integer countAnimal(String animal, Map<Integer, String> map)
{
    int cnt = 0;
    for (String val : map.values()) {
        if (val.equals(animal)) {
            ++cnt;
        }
    }

    return new Integer(cnt);
}

public static void main(String[] args)
{
    String[] animals = new String[] { "cat", "dog", "pig", "goat", "donkey", "horse", "cow" };

    Map<Integer, String> map = new HashMap<>();

    // load map for test
    Random rnd = new Random();
    for (int i = 0; i < 100; ++i) {
        String animal = animals[rnd.nextInt(animals.length)];
        map.put(new Integer(i), animal);
    }

    // count how many there are
    Map<String, Integer> numAnimals = new HashMap<>();

    for (String animal : animals) {
        numAnimals.put(animal, countAnimal(animal, map));
    }

    System.out.println(numAnimals);

    // show the cool Collections.frequency approach
    Integer count = Collections.frequency(map.values(), "dog");
    System.out.println("dog: " + count);
}

示例输出：

{horse = 18，cat = 13，donkey = 23，cow = 15，goat = 17，dog = 3，pig = 11}
狗：3

编辑：允许拆分字符串以查找计数的更新。基本上countAnimal将拆分从地图中检索到的String，然后检查每个标记以查看它是否是动物。稍微更改了测试用例。它基于更新的评论工作。但是，它没有考虑复数。 “猫”和“猫”的琐碎案例很容易处理，但“鼠标”和“老鼠”会更加困难，需要额外的工作。

public static void main(String[] args)
{
    String[] animals = new String[] { "cat", "dog", "pig", "goat",
            "donkey", "horse", "cow" };

    Map<Integer, String> map = new HashMap<>();

    // load map for test
    map.put(1, "My friend has a horse");
    map.put(2, "A bear can be big"); // will not be found
    map.put(3, "A cat is a loner");
    map.put(4, "A dog is man's best friend");
    map.put(5, "The cat and the dog are peacefully sleeping");

    // count how many there are
    Map<String, Integer> numAnimals = new HashMap<>();

    for (String animal : animals) {
        numAnimals.put(animal, countAnimal(animal, map));
    }

    System.out.println(numAnimals);
}



static Integer countAnimal(String animal, Map<Integer, String> map)
{
    int cnt = 0;
    for (String val : map.values()) {
        // split the val by space
        String[] tokens = val.split("[\\s]+");
        for (String token : tokens) {
            if (token.equalsIgnoreCase(animal)) {
                ++cnt;
            }
        }
    }

    return new Integer(cnt);
}

示例输出：

{horse = 1，cat = 2，donkey = 0，cow = 0，goat = 0，dog = 2，pig = 0}

HashMap重复值 - 查找重复项的数量

2 个答案: