Question

I m试图以两种方式打上HashMap。缺省方式:按价值计算,按数值计算;其次:从数字上看,由钥匙计算,人数最多。我已经搜索过,但找不到有关这个问题的任何东西,我确实发现,没有工作。如果不可能对两者进行分类(我希望最关键的人,因为人们的钥匙较低,就会减少,那么所有其他人(其关键人物为0)的脑血清分类。

在这里,我已尝试了:

private HashMap<String, Integer> userGains = new HashMap<String, Integer>();

public void sortGains(int skill, int user) {
    userGains.put(users.get(user).getUsername(), users.get(user).getGainedExperience(skill));
    HashMap<String, Integer> map = sortHashMap(userGains);
    for (int i = 0; i < map.size(); i++) {
        Application.getTrackerOutput().getOutputArea(skill).append(users.get(user).getUsername() + " gained " + map.get(users.get(user).getUsername()) + "  experience in " + getSkillName(skill) + ".
");
    }
}

public LinkedHashMap<String, Integer> sortHashMap(HashMap<String, Integer> passedMap) {
    List<String> mapKeys = new ArrayList<String>(passedMap.keySet());
    List<Integer> mapValues = new ArrayList<Integer>(passedMap.values());
    LinkedHashMap<String, Integer> sortedMap = new LinkedHashMap<String, Integer>();

    Collections.sort(mapValues);
    Collections.sort(mapKeys);

    Iterator<Integer> it$ = mapValues.iterator();
    while (it$.hasNext()) {
        Object val = it$.next();
        Iterator<String> keyIt = mapKeys.iterator();
        while (keyIt.hasNext()) {
            Object key = keyIt.next();
            String comp1 = passedMap.get(key).toString();
            String comp2 = val.toString();
            if (comp1.equals(comp2)) {
                passedMap.remove(key);
                mapKeys.remove(key);
                sortedMap.put((String) key, (Integer) val);
                break;
            }
        }
    }
    return sortedMap;
}

既然你不能这样说,那就是一个常设专家委员会:

private HashMap<String, Integer> userGains = new HashMap<String, Integer>();

private Object[][] testUsers = { { "Test user", 15 }, { "Test", 25 }, { "Hello", 11 }, { "I m a user", 21 }, { "No you re not!", 14 }, { "Yes I am!", 45 }, { "Oh, okay.  Sorry about the confusion.", 0 }, { "It s quite alright.", 0 } };

public static void main(String[] arguments) {
    new Sorting().sortGains();
}

public void sortGains() {
    for (Object[] test : testUsers) {
        userGains.put((String) test[0], (Integer) test[1]);
    }
    HashMap<String, Integer> map = sortHashMap(userGains);
    for (int i = 0; i < map.size(); i++) {
        System.out.println(testUsers[i][0] + " gained " + map.get(testUsers[i][0]) + "  experience.");
    }
}

public LinkedHashMap<String, Integer> sortHashMap(HashMap<String, Integer> passedMap) {
    List<String> mapKeys = new ArrayList<String>(passedMap.keySet());
    List<Integer> mapValues = new ArrayList<Integer>(passedMap.values());
    LinkedHashMap<String, Integer> sortedMap = new LinkedHashMap<String, Integer>();

    Collections.sort(mapValues);
    Collections.sort(mapKeys);

    Iterator<Integer> it$ = mapValues.iterator();
    while (it$.hasNext()) {
        Object val = it$.next();
        Iterator<String> keyIt = mapKeys.iterator();
        while (keyIt.hasNext()) {
            Object key = keyIt.next();
            String comp1 = passedMap.get(key).toString();
            String comp2 = val.toString();
            if (comp1.equals(comp2)) {
                passedMap.remove(key);
                mapKeys.remove(key);
                sortedMap.put((String) key, (Integer) val);
                break;
            }
        }
    }
    return sortedMap;
}

目前该方案的产出是:

Test user gained 15  experience.
Test gained 25  experience.
Hello gained 11  experience.
I m a user gained 21  experience.
No you re not! gained 14  experience.
Yes I am! gained 45  experience.
Oh, okay.  Sorry about the confusion. gained 0  experience.
It s quite alright. gained 0  experience.

当我需要时:

Yes I am! gained 45  experience. // start numeric sorting here, by highest key.
Test gained 25  experience.
I m a user gained 21  experience.
Test user gained 15  experience.
No you re not! gained 14  experience.
Hello gained 11  experience.
It s quite alright. gained 0  experience. // start alphabetical sorting here, if possible.
Oh, okay.  Sorry about the confusion. gained 0  experience.

任何见解?

Answer 1

你在显示价值观方面犯了错误。

HashMap<String, Integer> map = sortHashMap(userGains);
for (int i = 0; i < map.size(); i++) {
    System.out.println(testUsers[i][0] + " gained " + map.get(testUsers[i][0]) + "  experience.");
}

你们需要展示地图的价值观,而不是原阵列的价值观。

应:

HashMap<String, Integer> map = sortHashMap(userGains);
for (Entry<String, Integer> entry : map.entrySet()) {
    System.out.println(entry.getKey() + " gained " + entry.getValue() + "  experience.");
}

你们必须改变这一秩序。此外,我还建议在<代码>Map上而不是HashMap或LinkedHashMap上宣布,以避免 yourself和其他人混淆。您的分类也可采用<代码>。比较。情况有所改善:

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

private Object[][] testUsers = { { "Test user", 15 }, { "Test", 25 }, { "Hello", 11 }, { "I m a user", 21 }, { "No you re not!", 14 }, { "Yes I am!", 45 }, { "Oh, okay.  Sorry about the confusion.", 0 }, { "It s quite alright.", 0 } };

public static void main(String[] arguments) {
    new Sorting().sortGains();
}

public void sortGains() {
    for (Object[] test : testUsers) {
        userGains.put((String) test[0], (Integer) test[1]);
    }

    Map<String, Integer> map = createSortedMap(userGains);

    for (Entry<String, Integer> entry : map.entrySet()) {
        System.out.println(entry.getKey() + " gained " + entry.getValue() + "  experience.");
    }
}

public Map<String, Integer> createSortedMap(Map<String, Integer> passedMap) {
    List<Entry<String, Integer>> entryList = new ArrayList<Entry<String, Integer>>(passedMap.entrySet());

    Collections.sort(entryList, new Comparator<Entry<String, Integer>>() {

        @Override
        public int compare(Entry<String, Integer> e1, Entry<String, Integer> e2) {
            if (!e1.getValue().equals(e2.getValue())) {
                return e1.getValue().compareTo(e2.getValue()) * -1; // The * -1 reverses the order.
            } else {
                return e1.getKey().compareTo(e2.getKey());
            }
        }
    });

    Map<String, Integer> orderedMap = new LinkedHashMap<String, Integer>();

    for (Entry<String, Integer> entry : entryList) {
        orderedMap.put(entry.getKey(), entry.getValue());
    }

    return orderedMap;
}

Answer 2

It s not possible to sort a HashMap at all. By definition, the keys in a HashMap are unordered. If you want the keys of your Map to be ordered, then use a TreeMap with an appropriate Comparator object. You can create multiple TreeMaps with different Comparators if you want to access the same data multiple ways.

Answer 3

This question approaches what you re trying to do, by sorting on value in a TreeMap. If you take the most voted answer and modify the Comparator to sort on value then key, it should give you what you want.

有效地,你创建了一个比较器,其田地指树苗(能够研究价值)。而树马普则使用这一比较方法。在树苗中添加物品时,比较器研究价值,并对数值进行比较。

if the value a < value b, return 1
if the value a > value b, return -1
if the key a < key b, return 1
if the key a > key b, return -1
otherwise, return 0

a. 从该答复中编篡许多法典(没有检查守则是否可行,因为它只是出于这一想法):

public class Main {

    public static void main(String[] args) {

        ValueComparator<String> bvc =  new ValueComparator<String>();
        TreeMap<String,Integer> sorted_map = new TreeMap<String,Integer>(bvc);
        bvc.setBase(sorted_map);

        // add items
        // ....

        System.out.println("results");
            for (String key : sorted_map.keySet()) {
            System.out.println("key/value: " + key + "/"+sorted_map.get(key));
        }
     }

}

class ValueComparator implements Comparator<String> {
    Map base;

    public setBase(Map<String,Integer> base) {
        this.base = base;
    }

    public int compare(String a, String b) {
        Integer value_a = base.get(a);
        Integer value_b = base.get(b);

        if(value_a < value_b) {
            return 1;
        }
        if(value_a>< value_b) {
            return -1;
        }
        return a.compareTo(b);
    }
}

友情链接