How print out the contents of a HashMap<String, String> in ascending order based on its values?

Question

I have this HashMap that I need to print out in ascending order according to the values contained in it (not the keys).

But the order when I print it out is seemingly random.

What's the best way to print it out in ascending value order?

Map<String, String> codes = new HashMap<String, String>();

codes.put("A1", "Aania");
codes.put("X1", "Abatha");
codes.put("C1", "Acathan");
codes.put("S1", "Adreenas");

In other words, the example above should print out as this:

A1, Aania
X1, Abatha
C1, Acathan
S1, Adreenas

Just curious. What is the use case? My guess is you have chosen a wrong data structure and trying to tweak it to suit your specs. — chedine
– chedine, Commented Aug 31, 2010 at 5:25

catch32 · Accepted Answer · 2016-10-12 12:50:06Z

41

You aren't going to be able to do this from the HashMap class alone.

I would take the Map<String, String> codes, construct a reverse map of TreeMap<String, String> reversedMap where you map the values of the codes Map to the keys (this would require your original Map to have a one-to-one mapping from key-to-value). Since the TreeMap provides Iterators which returns entries in ascending key order, this will give you the value/key combination of the first map in the order (sorted by values) you desire.

Map<String, String> reversedMap = new TreeMap<String, String>(codes);

//then you just access the reversedMap however you like...
for (Map.Entry entry : reversedMap.entrySet()) {
    System.out.println(entry.getKey() + ", " + entry.getValue());
}

There are several collections libraries (commons-collections, Google Collections, etc) which have similar bidirectional Map implementations.

edited Oct 12, 2016 at 12:50

catch32

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answered Aug 31, 2010 at 0:53

matt b

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7 Comments

TheLQ Over a year ago

Why is the second loop necessary? Just create a temporary map and add from codes switching the key and value?

matt b Over a year ago

it's not necessary, it's an example of accessing the reversedMap and doing something with the entries

Ankit Bansal Over a year ago

What if two entries in map has different keys but same value? this code will remove one of them...

matt b Over a year ago

@akb this is why I said this would require your original map to have a one-to-one mapping from key-to-value

Mr_and_Mrs_D Over a year ago

System.out.println(entry.getValue() + ", " + entry.getKey()); to reproduce OP's original print outs

|

Sean · Accepted Answer · 2010-08-31 01:05:26Z

You'll need to make a list of the keys, sort them according to the corresponding values, then iterate over the sorted keys.

Map<String, String> map = getMyMap();
List<String> keys = new ArrayList<String>(map.keySet());
Collections.sort(keys, someComparator);
for (String key: keys) {
    System.out.println(key + ": " + map.get(key));
}

As for what to use for someComparator, here are some handy, generic Comparator-creating routines I often find useful. The first one sorts by the values according to their natural ordering, and the second allows you to specify any arbitrary Comparator to sort the values:

public static <K, V extends Comparable<? super V>>
        Comparator<K> mapValueComparator(final Map<K, V> map) {
    return new Comparator<K>() {
        public int compare(K key1, K key2) {
            return map.get(key1).compareTo(map.get(key2));
        }
    };
}

public static <K, V>
        Comparator<K> mapValueComparator(final Map<K, V> map,
                                         final Comparator<V> comparator) {
    return new Comparator<K>() {
        public int compare(K key1, K key2) {
            return comparator.compare(map.get(key1), map.get(key2));
        }
    };
}

AlexW · Accepted Answer · 2015-04-27 08:22:47Z

6

It's time to add some lambdas:

codes.entrySet()
    .stream()
    .sorted(Comparator.comparing(Map.Entry::getValue))
    .forEach(System.out::println);

answered Apr 27, 2015 at 8:22

AlexW

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6 Comments

almost a beginner Over a year ago

This does not print the String value, unless I'm missing something here.

AlexW Over a year ago

@almostabeginner Code forEach(System.out::println) will print every entry. Check this

almost a beginner Over a year ago

I think this only prints the address, rather than the value? I tried it but didn't work. My map contains two objects, the key being a "person" object, which has a firstname, lastname field etc, I want to print firstname, I don't see how this code would do that? Thanks for your time.

AlexW Over a year ago

@almostabeginner To be precise, this code prints all Entry's in map, calling their toString method implicitly. If you want to print keys, or some field in key, you can do something like this:

codes.entrySet().stream().sorted(Comparator.comparing(Map.Entry::getValue)).map(Map.Entry::getKey).map(Person::getLastName).forEach(System.out::println);

lmo Over a year ago

How to use this with a logger ? I don't want to Syso ! Thanks :)

|

akjoshi · Accepted Answer · 2012-11-20 07:04:22Z

5

the for loop of for(Map.Entry entry: codes.entrySet()) didn't work for me. Used Iterator instead.

Iterator<Map.Entry<String, String>> i = codes.entrySet().iterator(); 
while(i.hasNext()){
    String key = i.next().getKey();
    System.out.println(key+", "+codes.get(key));
}

edited Nov 20, 2012 at 7:04

akjoshi

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answered Nov 20, 2012 at 6:45

psun

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1 Comment

Dan Getz Over a year ago

This doesn't answer the question, because you're iterating on the original HashMap, which isn't sorted.

David · Accepted Answer · 2016-04-21 08:52:11Z

5

you just need to use:

 Map<>.toString().replace("]","\n");

and replaces the ending square bracket of each key=value set with a new line.

edited Apr 21, 2016 at 8:52

David

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answered Apr 21, 2016 at 7:38

ole scharling

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Comments

bluehallu · Accepted Answer · 2017-04-24 15:29:19Z

3

Java 8

map.entrySet().stream().sorted(Map.Entry.comparingByValue()).forEach(System.out::println);

answered Apr 24, 2017 at 15:29

bluehallu

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Comments

bmargulies · Accepted Answer · 2010-08-31 11:26:34Z

2

Create a TreeMap<String,String>
Add each of the HashMap entries with the value as the key.
iterate the TreeMap

If the values are nonunique, you would need a list in the second position.

edited Aug 31, 2010 at 11:26

answered Aug 31, 2010 at 0:37

bmargulies

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2 Comments

user166390 Over a year ago

Beautiful. Had to delete my answer :)

I82Much Over a year ago

Not quite right... "Red-Black tree based implementation of the SortedMap interface. This class guarantees that the map will be in ascending key order, sorted according to the natural order for the key's class (see Comparable), or by the comparator provided at creation time, depending on which constructor is used." You'll have it sorted by the keys, not the values.

Dan Getz · Accepted Answer · 2015-07-02 01:42:15Z

You can use a list of the entry set rather than the key set and it is a more natural choice given you are sorting based on the value. This avoids a lot of unneeded lookups in the sorting and printing of the entries.

Map<String, String> map = ...
List<Map.Entry<String, String>> listOfEntries = new ArrayList<Map.Entry<String, String>>(map.entrySet());
Collections.sort(listOfEntries, new SortByValueComparator());
for(Map.Entry<String, String> entry: listOfEntries)
   System.out.println(entry);

static class SortByValueComparator implements Comparator<Map.Entry<String, String>> {
   public int compareTo(Map.Entry<String, String> e1, Map.Entry<String, String> e2) {
       return e1.getValue().compateTo(e2.getValue());
   }
}

Tomek Szpakowicz · Accepted Answer · 2018-11-02 14:56:52Z

0

the simplest and shortest code i think is this:

public void listPrinter(LinkedHashMap<String, String> caseList) {

    for(Entry entry:caseList.entrySet()) {
        System.out.println("K: \t"+entry.getKey()+", V: \t"+entry.getValue());
    }
}

edited Nov 2, 2018 at 14:56

Tomek Szpakowicz

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answered Jun 6, 2017 at 9:45

Eric Jalal

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Comments

aar · Accepted Answer · 2010-08-31 19:00:12Z

-1

The simplest solution would be to use a sorted map like TreeMap instead of HashMap. If you do not have control over the map construction, then the minimal solution would be to construct a sorted set of keys. You don't really need a new map.

Set<String> sortedKeys = new TreeSet<String>();
sortedKeys.addAll(codes.keySet());

for(String key: sortedKeys){
    println(key  + ":" + codes.get(key));
}

answered Aug 31, 2010 at 19:00

aar

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2 Comments

jackrabbit Over a year ago

OP wants the output to be sorted by value, not by key.

Mr_and_Mrs_D Over a year ago

Delete this - does not answer the question

enb081 · Accepted Answer · 2015-04-27 09:28:47Z

-2

Try:

try
{
    int cnt= m.getSmartPhoneCount("HTC",true);      
    System.out.println("total count of HTC="+cnt);
}  
catch (NoSuchBrandSmartPhoneAvailableException e)
{
    // TODO Auto-generated catch 
    e.printStackTrace();
}

edited Apr 27, 2015 at 9:28

enb081

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answered Apr 27, 2015 at 8:12

deeps

1

1 Comment

Dan Getz Over a year ago

Was this intended to be posted as an answer to a different question? If so, please delete it here.

deeps · Accepted Answer · 2015-04-27 08:20:12Z

-3

 SmartPhone[] sp=new SmartPhone[4];
 sp[0]=new SmartPhone(1,"HTC","desire","black",20000,10,true,true);
 sp[1]=new SmartPhone(2,"samsung","grand","black",5000,10,false,true);
 sp[2]=new SmartPhone(14,"google nexus","desire","black",2000,30,true,false);
 sp[3]=new SmartPhone(13,"HTC","desire","white",50000,40,false,false);

answered Apr 27, 2015 at 8:20

deeps

1

2 Comments

Markus W Mahlberg Over a year ago

Mind to explain your solution a bit? You might want to read How Do I Write A Good Answer.

Dan Getz Over a year ago

Was this intended to be posted as an answer to a different question? If so, please delete it here.

luk2302 · Accepted Answer · 2015-04-27 09:38:37Z

-3

while (itr.hasNext()) {
    Vehicle vc=(Vehicle) itr.next();
    if(vc.getVehicleType().equalsIgnoreCase(s)) {
        count++;
    }
}

edited Apr 27, 2015 at 9:38

luk2302

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answered Apr 27, 2015 at 8:21

Roks

1

1 Comment

Dan Getz Over a year ago

Was this intended to be posted as an answer to a different question? If so, please delete it here.

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