102

This question already has an answer here:

I need to sort my HashMap according to the values stored in it. The HashMap contains the contacts name stored in phone.

Also I need that the keys get automatically sorted as soon as I sort the values, or you can say the keys and values are bound together thus any changes in values should get reflected in keys.

HashMap<Integer,String> map = new HashMap<Integer,String>();
map.put(1,"froyo");
map.put(2,"abby");
map.put(3,"denver");
map.put(4,"frost");
map.put(5,"daisy");

Required output:

2,abby;
5,daisy;
3,denver;
4,frost;
1,froyo;

12 답변


72

Assuming Java, you could sort hashmap just like this:

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

    LinkedHashMap<Integer, String> sortedMap =
        new LinkedHashMap<>();

    Iterator<String> valueIt = mapValues.iterator();
    while (valueIt.hasNext()) {
        String val = valueIt.next();
        Iterator<Integer> keyIt = mapKeys.iterator();

        while (keyIt.hasNext()) {
            Integer key = keyIt.next();
            String comp1 = passedMap.get(key);
            String comp2 = val;

            if (comp1.equals(comp2)) {
                keyIt.remove();
                sortedMap.put(key, val);
                break;
            }
        }
    }
    return sortedMap;
}

Just a kick-off example. This way is more useful as it sorts the HashMap and keeps the duplicate values as well.


  • see the edited version.I don't think collections.sort(mapvalues) will solve the problem - prof_jack
  • this code has arranged hashmap according to the keys.what I wanted was:(2,abby; 5,daisy; 3,denver; 4,frost; 1,froyo;)i.e values are arranged according to their initials and the change get reflected in the keys... - prof_jack
  • see the edited version - prof_jack
  • I edited your code a bit and it is working as desired.thanks a tonne. - prof_jack
  • Please be aware that the given algorithm has a time complexity of O(n^2) due to repeatedly looking up in values in two while loops. Converting the Entry Set to a List and then sorting the List based on a comparator would be a more efficient solution. - picmate 涅

144

Try below code it works fine for me. You can choose both Ascending as well as descending order

import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.LinkedHashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;

public class SortMapByValue
{
    public static boolean ASC = true;
    public static boolean DESC = false;

    public static void main(String[] args)
    {

        // Creating dummy unsorted map
        Map<String, Integer> unsortMap = new HashMap<String, Integer>();
        unsortMap.put("B", 55);
        unsortMap.put("A", 80);
        unsortMap.put("D", 20);
        unsortMap.put("C", 70);

        System.out.println("Before sorting......");
        printMap(unsortMap);

        System.out.println("After sorting ascending order......");
        Map<String, Integer> sortedMapAsc = sortByComparator(unsortMap, ASC);
        printMap(sortedMapAsc);


        System.out.println("After sorting descindeng order......");
        Map<String, Integer> sortedMapDesc = sortByComparator(unsortMap, DESC);
        printMap(sortedMapDesc);

    }

    private static Map<String, Integer> sortByComparator(Map<String, Integer> unsortMap, final boolean order)
    {

        List<Entry<String, Integer>> list = new LinkedList<Entry<String, Integer>>(unsortMap.entrySet());

        // Sorting the list based on values
        Collections.sort(list, new Comparator<Entry<String, Integer>>()
        {
            public int compare(Entry<String, Integer> o1,
                    Entry<String, Integer> o2)
            {
                if (order)
                {
                    return o1.getValue().compareTo(o2.getValue());
                }
                else
                {
                    return o2.getValue().compareTo(o1.getValue());

                }
            }
        });

        // Maintaining insertion order with the help of LinkedList
        Map<String, Integer> sortedMap = new LinkedHashMap<String, Integer>();
        for (Entry<String, Integer> entry : list)
        {
            sortedMap.put(entry.getKey(), entry.getValue());
        }

        return sortedMap;
    }

    public static void printMap(Map<String, Integer> map)
    {
        for (Entry<String, Integer> entry : map.entrySet())
        {
            System.out.println("Key : " + entry.getKey() + " Value : "+ entry.getValue());
        }
    }
}

Edit: Version 2

Used new java feature like stream for-each etc

Map will be sorted by keys if values are same

 import java.util.*;
 import java.util.Map.Entry;
 import java.util.stream.Collectors;

 public class SortMapByValue

 {
    private static boolean ASC = true;
    private static boolean DESC = false;
    public static void main(String[] args)
    {

        // Creating dummy unsorted map
        Map<String, Integer> unsortMap = new HashMap<>();
        unsortMap.put("B", 55);
        unsortMap.put("A", 20);
        unsortMap.put("D", 20);
        unsortMap.put("C", 70);

        System.out.println("Before sorting......");
        printMap(unsortMap);

        System.out.println("After sorting ascending order......");
        Map<String, Integer> sortedMapAsc = sortByValue(unsortMap, ASC);
        printMap(sortedMapAsc);


        System.out.println("After sorting descending order......");
        Map<String, Integer> sortedMapDesc = sortByValue(unsortMap, DESC);
        printMap(sortedMapDesc);
    }

    private static Map<String, Integer> sortByValue(Map<String, Integer> unsortMap, final boolean order)
    {
        List<Entry<String, Integer>> list = new LinkedList<>(unsortMap.entrySet());

        // Sorting the list based on values
        list.sort((o1, o2) -> order ? o1.getValue().compareTo(o2.getValue()) == 0
                ? o1.getKey().compareTo(o2.getKey())
                : o1.getValue().compareTo(o2.getValue()) : o2.getValue().compareTo(o1.getValue()) == 0
                ? o2.getKey().compareTo(o1.getKey())
                : o2.getValue().compareTo(o1.getValue()));
        return list.stream().collect(Collectors.toMap(Entry::getKey, Entry::getValue, (a, b) -> b, LinkedHashMap::new));

    }

    private static void printMap(Map<String, Integer> map)
    {
        map.forEach((key, value) -> System.out.println("Key : " + key + " Value : " + value));
    }
}


  • Nice answer, written it in a proper way using the collections utils. - Aditya
  • Tried for my problem and found that there needs to be a slight modification to your comparator logic, where you are not checking for null checks which should be added as map values can contain null elements. - Aditya
  • Useful and clear answer. - Emalka
  • I found this solutions simplest and easy to understand. - vikramvi
  • What if I want to sort it by the length of the value, which for example is a String? - Srujan Barai

54

In Java 8:

Map<Integer, String> sortedMap = 
     unsortedMap.entrySet().stream()
    .sorted(Entry.comparingByValue())
    .collect(Collectors.toMap(Entry::getKey, Entry::getValue,
                              (e1, e2) -> e1, LinkedHashMap::new));


  • Can we use something like this? Arrays.sort(hm.values().toArray()); - Hengameh
  • this sorting can be inverted? the values I need for from bigger to smaller - Aquarius Power
  • @AquariusPower See the reversed method at docs.oracle.com/javase/8/docs/api/java/util/… You can apply it to the comparator returned by comparingByValue() - Vitalii Fedorenko
  • the return of Entry.comparingByValue().reversed() is incompatible with .sorted(...) expected params :(, I ended up with a reversed loop for over .entrySet().toArray(), may be a cast could solve it, I need more time to test :) - Aquarius Power
  • @AquariusPower you don't need casting, just give the compiler a hint on what the generic type is: Entry.<Integer, String>comparingByValue().reversed() - Vitalii Fedorenko

24

You don't, basically. A HashMap is fundamentally unordered. Any patterns you might see in the ordering should not be relied on.

There are sorted maps such as TreeMap, but they traditionally sort by key rather than value. It's relatively unusual to sort by value - especially as multiple keys can have the same value.

Can you give more context for what you're trying to do? If you're really only storing numbers (as strings) for the keys, perhaps a SortedSet such as TreeSet would work for you?

Alternatively, you could store two separate collections encapsulated in a single class to update both at the same time?


  • see the edited version - prof_jack
  • For example, sorting the colors that appear on an image. It has to be fast, because we can have max_int colors. - Rafael Sanches
  • @RafaelSanches: It's not clear what the context for your comment is. What would the map be in this case anyway? You may want to ask a new question. - Jon Skeet
  • I'm just giving an example that would be useful to order the hashmap by values, in the most performant way. - Rafael Sanches
  • Can we use something like this? Arrays.sort(hm.values().toArray()); - Hengameh

11

package com.naveen.hashmap;

import java.util.*;
import java.util.Map.Entry;

public class SortBasedonValues {

    /**
     * @param args
     */
    public static void main(String[] args) {

        HashMap<String, Integer> hm = new HashMap<String, Integer>();
        hm.put("Naveen", 2);
        hm.put("Santosh", 3);
        hm.put("Ravi", 4);
        hm.put("Pramod", 1);
        Set<Entry<String, Integer>> set = hm.entrySet();
        List<Entry<String, Integer>> list = new ArrayList<Entry<String, Integer>>(
                set);
        Collections.sort(list, new Comparator<Map.Entry<String, Integer>>() {
            public int compare(Map.Entry<String, Integer> o1,
                    Map.Entry<String, Integer> o2) {
                return o2.getValue().compareTo(o1.getValue());
            }
        });

        for (Entry<String, Integer> entry : list) {
            System.out.println(entry.getValue());

        }

    }
}


  • I tried executing this code, runs fine with import java.util.Map.Entry;' But the same code does not run with import java.util.*; The later includes the former according to my knowledge. Then why it gives an error? - NaveeNeo
  • Simple and elegant - Caveman

9

map.entrySet().stream()
                .sorted((k1, k2) -> -k1.getValue().compareTo(k2.getValue()))
                .forEach(k -> System.out.println(k.getKey() + ": " + k.getValue()));


  • most readable solution for me and does the job, e.g. running over a HashMap sorted by values - user1813222

7

As a kind of simple solution you can use temp TreeMap if you need just a final result:

TreeMap<String, Integer> sortedMap = new TreeMap<String, Integer>();
for (Map.Entry entry : map.entrySet()) {
    sortedMap.put((String) entry.getValue(), (Integer)entry.getKey());
}

This will get you strings sorted as keys of sortedMap.


  • This only works if all the integer values are unique -- otherwise, you get strings overwritten. - Kyzderp

4

I extends a TreeMap and override entrySet() and values() methods. Key and value need to be Comparable.

Follow the code:

public class ValueSortedMap<K extends Comparable, V extends Comparable> extends TreeMap<K, V> {

    @Override
    public Set<Entry<K, V>> entrySet() {
        Set<Entry<K, V>> originalEntries = super.entrySet();
        Set<Entry<K, V>> sortedEntry = new TreeSet<Entry<K, V>>(new Comparator<Entry<K, V>>() {
            @Override
            public int compare(Entry<K, V> entryA, Entry<K, V> entryB) {
                int compareTo = entryA.getValue().compareTo(entryB.getValue());
                if(compareTo == 0) {
                    compareTo = entryA.getKey().compareTo(entryB.getKey());
                }
                return compareTo;
            }
        });
        sortedEntry.addAll(originalEntries);
        return sortedEntry;
    }

    @Override
    public Collection<V> values() {
        Set<V> sortedValues = new TreeSet<>(new Comparator<V>(){
            @Override
            public int compare(V vA, V vB) {
                return vA.compareTo(vB);
            }
        });
        sortedValues.addAll(super.values());
        return sortedValues;
    }
}

Unit Tests:

public class ValueSortedMapTest {

    @Test
    public void basicTest() {
        Map<String, Integer> sortedMap = new ValueSortedMap<>();
        sortedMap.put("A",3);
        sortedMap.put("B",1);
        sortedMap.put("C",2);

        Assert.assertEquals("{B=1, C=2, A=3}", sortedMap.toString());
    }

    @Test
    public void repeatedValues() {
        Map<String, Double> sortedMap = new ValueSortedMap<>();
        sortedMap.put("D",67.3);
        sortedMap.put("A",99.5);
        sortedMap.put("B",67.4);
        sortedMap.put("C",67.4);

        Assert.assertEquals("{D=67.3, B=67.4, C=67.4, A=99.5}", sortedMap.toString());
    }

}


  • This doesn't adhere to the Map interface. A proper implementation of entrySet() is: "Returns a Set view of the mappings contained in this map. The set is backed by the map, so changes to the map are reflected in the set, and vice-versa." Same thing for values(). - Radiodef
  • Awesome just what i was looking for - Shashank

2

found a solution but not sure the performance if the map has large size, useful for normal case.

   /**
     * sort HashMap<String, CustomData> by value
     * CustomData needs to provide compareTo() for comparing CustomData
     * @param map
     */

    public void sortHashMapByValue(final HashMap<String, CustomData> map) {
        ArrayList<String> keys = new ArrayList<String>();
        keys.addAll(map.keySet());
        Collections.sort(keys, new Comparator<String>() {
            @Override
            public int compare(String lhs, String rhs) {
                CustomData val1 = map.get(lhs);
                CustomData val2 = map.get(rhs);
                if (val1 == null) {
                    return (val2 != null) ? 1 : 0;
                } else if (val1 != null) && (val2 != null)) {
                    return = val1.compareTo(val2);
                }
                else {
                    return 0;
                }
            }
        });

        for (String key : keys) {
            CustomData c = map.get(key);
            if (c != null) {
                Log.e("key:"+key+", CustomData:"+c.toString());
            } 
        }
    }


  • typos in it should be else if ((val1 != null) && (val2 != null)) { return val1.compareTo(val2); } in compare method - Tanuj Verma

0

package SortedSet;

import java.util.*;

public class HashMapValueSort {
public static void main(String[] args){
    final Map<Integer, String> map = new HashMap<Integer,String>();
    map.put(4,"Mango");
    map.put(3,"Apple");
    map.put(5,"Orange");
    map.put(8,"Fruits");
    map.put(23,"Vegetables");
    map.put(1,"Zebra");
    map.put(5,"Yellow");
    System.out.println(map);
    final HashMapValueSort sort = new HashMapValueSort();
    final Set<Map.Entry<Integer, String>> entry = map.entrySet();
    final Comparator<Map.Entry<Integer, String>> comparator = new Comparator<Map.Entry<Integer, String>>() {
        @Override
        public int compare(Map.Entry<Integer, String> o1, Map.Entry<Integer, String> o2) {
            String value1 = o1.getValue();
            String value2 = o2.getValue();
            return value1.compareTo(value2);
        }
    };
    final SortedSet<Map.Entry<Integer, String>> sortedSet = new TreeSet(comparator);
    sortedSet.addAll(entry);
    final Map<Integer,String> sortedMap =  new LinkedHashMap<Integer, String>();
    for(Map.Entry<Integer, String> entry1 : sortedSet ){
        sortedMap.put(entry1.getKey(),entry1.getValue());
    }
    System.out.println(sortedMap);
}
}



-1

public static TreeMap<String, String> sortMap(HashMap<String, String> passedMap, String byParam) {
    if(byParam.trim().toLowerCase().equalsIgnoreCase("byValue")) {
        // Altering the (key, value) -> (value, key)
        HashMap<String, String> newMap =  new HashMap<String, String>();
        for (Map.Entry<String, String> entry : passedMap.entrySet()) {
            newMap.put(entry.getValue(), entry.getKey());
        }
        return new TreeMap<String, String>(newMap);
    }
    return new TreeMap<String, String>(passedMap);
}


-1

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map.Entry;

public class CollectionsSort {

    /**
     * @param args
     */`enter code here`
    public static void main(String[] args) {
        // TODO Auto-generated method stub

        CollectionsSort colleciotns = new CollectionsSort();

        List<combine> list = new ArrayList<combine>();
        HashMap<String, Integer> h = new HashMap<String, Integer>();
        h.put("nayanana", 10);
        h.put("lohith", 5);

        for (Entry<String, Integer> value : h.entrySet()) {
            combine a = colleciotns.new combine(value.getValue(),
                    value.getKey());
            list.add(a);
        }

        Collections.sort(list);
        for (int i = 0; i < list.size(); i++) {
            System.out.println(list.get(i));
        }
    }

    public class combine implements Comparable<combine> {

        public int value;
        public String key;

        public combine(int value, String key) {
            this.value = value;
            this.key = key;
        }

        @Override
        public int compareTo(combine arg0) {
            // TODO Auto-generated method stub
            return this.value > arg0.value ? 1 : this.value < arg0.value ? -1
                    : 0;
        }

        public String toString() {
            return this.value + " " + this.key;
        }
    }

}


  • Please be sure your code runs correctly before answering. Also consider adding comments so the questioner can better understand your answer. - Daniel F. Thornton

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