How to sort Map values by key in Java?

问题:

I have a Map that has strings for both keys and values.

Data is like following:

\"question1\", \"1\"
\"question9\", \"1\"
\"question2\", \"4\"
\"question5\", \"2\"

I want to sort the map based on its keys. So, in the end, I will have question1, question2, question3....and so on.

Eventually, I am trying to get two strings out of this Map.

• First String: Questions ( in order 1 ..10)
• Second String: Answers (in the same order as the question)

Right now I have the following:

Iterator it = paramMap.entrySet().iterator();
while (it.hasNext()) {
    Map.Entry pairs = (Map.Entry) it.next();
    questionAnswers += pairs.getKey() + \",\";
}

This gets me the questions in a string but they are not in order.

回答1:

Short answer

Use a TreeMap. This is precisely what it\'s for.

If this map is passed to you and you cannot determine the type, then you can do the following:

SortedSet<String> keys = new TreeSet<>(map.keySet());
for (String key : keys) { 
   String value = map.get(key);
   // do something
}

This will iterate across the map in natural order of the keys.

---

Longer answer

Technically, you can use anything that implements SortedMap, but except for rare cases this amounts to TreeMap, just as using a Map implementation typically amounts to HashMap.

For cases where your keys are a complex type that doesn\'t implement Comparable or you don\'t want to use the natural order then TreeMap and TreeSet have additional constructors that let you pass in a Comparator:

// placed inline for the demonstration, but doesn\'t have to be a lambda expression
Comparator<Foo> comparator = (Foo o1, Foo o2) -> {
        ...
    }

SortedSet<Foo> keys = new TreeSet<>(comparator);
keys.addAll(map.keySet());

Remember when using a TreeMap or TreeSet that it will have different performance characteristics than HashMap or HashSet. Roughly speaking operations that find or insert an element will go from O(1) to O(Log(N)).

In a HashMap, moving from 1000 items to 10,000 doesn\'t really affect your time to lookup an element, but for a TreeMap the lookup time will be about 3 times slower (assuming Log₂). Moving from 1000 to 100,000 will be about 6 times slower for every element lookup.

回答2:

Assuming TreeMap is not good for you (and assuming you can\'t use generics):

List sortedKeys=new ArrayList(yourMap.keySet());
Collections.sort(sortedKeys);
// Do what you need with sortedKeys.

回答3:

Using the TreeMap you can sort the map.

Map<String, String> map = new HashMap<>();        
Map<String, String> treeMap = new TreeMap<>(map);
for (String str : treeMap.keySet()) {
    System.out.println(str);
}

回答4:

Use a TreeMap!

回答5:

If you already have a map and would like to sort it on keys, simply use :

Map<String, String> treeMap = new TreeMap<String, String>(yourMap);

A complete working example :

import java.util.HashMap;
import java.util.Set;
import java.util.Map;
import java.util.TreeMap;
import java.util.Iterator;

class SortOnKey {

public static void main(String[] args) {
   HashMap<String,String> hm = new HashMap<String,String>();
   hm.put(\"3\",\"three\");
   hm.put(\"1\",\"one\");
   hm.put(\"4\",\"four\");
   hm.put(\"2\",\"two\");
   printMap(hm);
   Map<String, String> treeMap = new TreeMap<String, String>(hm);
   printMap(treeMap);
}//main

public static void printMap(Map<String,String> map) {
    Set s = map.entrySet();
    Iterator it = s.iterator();
    while ( it.hasNext() ) {
       Map.Entry entry = (Map.Entry) it.next();
       String key = (String) entry.getKey();
       String value = (String) entry.getValue();
       System.out.println(key + \" => \" + value);
    }//while
    System.out.println(\"========================\");
}//printMap

}//class

回答6:

Just use TreeMap

new TreeMap<String, String>(unsortMap);

Be aware that the TreeMap is sorted according to the natural ordering of its \'keys\'

回答7:

Provided you cannot use TreeMap, in Java 8 we can make use of toMap() method in Collectorswhich takes following parameters:

• keymapper: mapping function to produce keys
• valuemapper: mapping function to produce values
• mergeFunction: a merge function, used to resolve collisions between values associated with the same key
• mapSupplier: a function which returns a new, empty Map into which the results will be inserted.

Java 8 Example

Map<String,String> sample = new HashMap<>();  // push some values to map  
Map<String, String> newMapSortedByKey = sample.entrySet().stream()
                    .sorted(Map.Entry.<String,String>comparingByKey().reversed())
                    .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));
Map<String, String> newMapSortedByValue = sample.entrySet().stream()
                        .sorted(Map.Entry.<String,String>comparingByValue().reversed())
                        .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1,e2) -> e1, LinkedHashMap::new));

We can modify the example to use custom comparator and to sort based on keys as:

Map<String, String> newMapSortedByKey = sample.entrySet().stream()
                .sorted((e1,e2) -> e1.getKey().compareTo(e2.getKey()))
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1,e2) -> e1, LinkedHashMap::new));

回答8:

This code can sort a key-value map in both orders i.e. ascending and descending.

<K, V extends Comparable<V>> Map<K, V> sortByValues
     (final Map<K, V> map, int ascending)
{
     Comparator<K> valueComparator =  new Comparator<K>() {         
        private int ascending;
        public int compare(K k1, K k2) {
            int compare = map.get(k2).compareTo(map.get(k1));
            if (compare == 0) return 1;
            else return ascending*compare;
        }
        public Comparator<K> setParam(int ascending)
        {
            this.ascending = ascending;
            return this;
        }
    }.setParam(ascending);

    Map<K, V> sortedByValues = new TreeMap<K, V>(valueComparator);
    sortedByValues.putAll(map);
    return sortedByValues;
}

As an example:

Map<Integer,Double> recommWarrVals = new HashMap<Integer,Double>();
recommWarrVals = sortByValues(recommWarrVals, 1);  // Ascending order
recommWarrVals = sortByValues(recommWarrVals,-1);  // Descending order

回答9:

Using Java 8:

Map<String, Integer> sortedMap = unsortMap.entrySet().stream()
            .sorted(Map.Entry.comparingByKey())
            .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,
                    (oldValue, newValue) -> oldValue, LinkedHashMap::new));

回答10:

In Java 8

To sort a Map<K, V> by key, putting keys into a List<K>:

List<K> result = map.keySet().stream().sorted().collect(Collectors.toList());

To sort a Map<K, V> by key, putting entries into a List<Map.Entry<K, V>>:

List<Map.Entry<K, V>> result =
    map.entrySet()
       .stream()
       .sorted(Map.Entry.comparingByKey())
       .collect(Collectors.toList());

Last but not least: to sort strings in a locale-sensitive manner - use a Collator (comparator) class:

Collator collator = Collator.getInstance(Locale.US);
collator.setStrength(Collator.PRIMARY); // case insensitive collator

List<Map.Entry<String, String>> result =
    map.entrySet()
       .stream()
       .sorted(Map.Entry.comparingByKey(collator))
       .collect(Collectors.toList());

回答11:

List<String> list = new ArrayList<String>();
Map<String, String> map = new HashMap<String, String>();
for (String str : map.keySet()) {
 list.add(str);
}
Collections.sort(list);
for (String str : list) {
 System.out.println(str);
}

回答12:

We can also sort the key by using Arrays.sort method.

Map<String, String> map = new HashMap<String, String>();
Object[] objArr = new Object[map.size()];
for (int i = 0; i < map.size(); i++) {
objArr[i] = map.get(i);
}
Arrays.sort(objArr);
for (Object str : objArr) {
System.out.println(str);
}

回答13:

In Java 8 you can also use .stream().sorted():

myMap.keySet().stream().sorted().forEach(key -> {
        String value = myMap.get(key);

        System.out.println(\"key: \" + key);
        System.out.println(\"value: \" + value);
    }
);