How to sort Map values by key in Java?
JavaDictionaryHashmapJava Problem Overview
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.
Java Solutions
Solution 1 - Java
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 1.3 times slower (assuming Log2). Moving from 1000 to 100,000 will be about 1.6 times slower for every element lookup.
Solution 2 - Java
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.
Solution 3 - Java
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);
}
Solution 4 - Java
Just use TreeMap
new TreeMap<String, String>(unsortMap);
Be aware that the TreeMap is sorted according to the natural ordering of its 'keys'
Solution 5 - Java
Use a TreeMap!
Solution 6 - Java
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
Solution 7 - Java
Provided you cannot use TreeMap
, in Java 8 we can make use of toMap() method in Collectors
which 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));
Solution 8 - Java
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));
Solution 9 - Java
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());
Solution 10 - Java
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
Solution 11 - Java
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);
}
Solution 12 - Java
Just in case you don't wanna use a TreeMap
public static Map<Integer, Integer> sortByKey(Map<Integer, Integer> map) {
List<Map.Entry<Integer, Integer>> list = new ArrayList<>(map.entrySet());
list.sort(Comparator.comparingInt(Map.Entry::getKey));
Map<Integer, Integer> sortedMap = new LinkedHashMap<>();
list.forEach(e -> sortedMap.put(e.getKey(), e.getValue()));
return sortedMap;
}
Also, in-case you wanted to sort your map on the basis of values
just change Map.Entry::getKey
to Map.Entry::getValue
Solution 13 - Java
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);
}
);
Solution 14 - Java
Use LinkedHashMap
, which provide the key ordering. It's also gives same performance with HashMap
. They both implements Map
interface, so you can just replace the initialization object HashMap
to LinkedHashMap
.
Solution 15 - Java
How about below tree map:
Map<String, String> sortedMap = new TreeMap<>(Comparator.comparingInt(String::length)
.thenComparing(Function.identity()));
What ever you put in this sortedMap it will be sorted automatically.
First of all TreeMap
is sorted implementation of Map
Interface.
There is a but as it sorts keys on [natural order fashion][https://docs.oracle.com/javase/tutorial/collections/interfaces/order.html]. As Java doc says String
type is a Lexicographic Natural order type. Imagine the below list of numbers with String type. means below list will be sorted not as expected.
List<String> notSortedList = List.of("78","0", "24", "39", "4","53","32");
If you just you the default TreeMap
constructor like below and push each element one-by-one like below:
Map<String, String> map = new TreeMap<>();
for (String s : notSortedList) {
map.put(s, s);
}
System.out.println(map);
output is: {0=0, 14=14, 24=24, 32=32, 39=39, 4=4, 48=48, 53=53, 54=54, 78=78}
As you see number 4 for example comes after '39'. This is the nature of the Lexicographic data types like String. If that one was an Integer data type then that was okay though.
To fix this use argument to first check the length of the String and then compare them. In java 8 is done like this:
Map<String, String> sortedMap = new TreeMap<>(Comparator.comparingInt(String::length)
.thenComparing(Function.identity()));
It first compare each element by length then apply check by compareTo
as the input the same as the element to compare with.
If you prefer to use a more understandable method, the above code will be equivalent with below code:
Map<String, String> sortedMap = new TreeMap<>(
new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
int lengthDifference = o1.length() - o2.length();
if (lengthDifference != 0) return lengthDifference;
return o1.compareTo(o2);
}
}
);
Because the TreeMap
constructor accepts the comparator Interface you can build up any even more complex implementation of Composite classes.
This is also another form more simplified version.
Map<String,String> sortedMap = new TreeMap<>(
(Comparator<String>) (o1, o2) ->
{
int lengthDifference = o1.length() - o2.length();
if (lengthDifference != 0) return lengthDifference;
return o1.compareTo(o2);
}
);
Solution 16 - Java
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);
}
Solution 17 - Java
A good solution is provided here. We have a HashMap
that stores values in unspecified order. We define an auxiliary TreeMap
and we copy all data from HashMap into TreeMap using the putAll
method. The resulting entries in the TreeMap are in the key-order.