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this answer doesn't math: How to store enum to map using Java 8 stream API
I have an enum:
public enum SheetRows{
totalActive("Total active");
String value;
SheetRows(String value){
this.value = value;
}
public String getValueForTable() {
return value;
}
}
How to convert this enum to HashMap<SheetRows, String>?
I try to use:
HashMap<SheetRows, String> cellsMap = Arrays.asList(SheetRows.values()).stream()
.collect(Collectors.toMap(k -> k, v -> v.getValueForTable()));
but this code isn't compile.
The following should work, the problem is that you're trying to assign a Map to HashMap
Map<SheetRows, String> map = EnumSet.allOf(SheetRows.class)
.stream()
.collect(Collectors.toMap(Function.identity(), SheetRows::getValueForTable));
System.out.println("map = " + map); // map = {totalActive=Total active}
If you really need to return a HashMap, you can also use the overload that takes a supplier and a mergeFunction like hereunder.
The mergeFunction will never be called since your enums are unique, so just choose a random one.
HashMap<SheetRows, String> map = EnumSet.allOf(SheetRows.class)
.stream()
.collect(Collectors.toMap(Function.identity(), SheetRows::getValueForTable, (o1, o2) -> o1, HashMap::new));
System.out.println("map = " + map); // map = {totalActive=Total active}
I am new to java 8 and would like to write a function which sorts hashmap by values and if values are the same sort by keys.
To sort the hashmap by values:
Map<String, Integer> map1 = new LinkedHashMap<>();
map.entrySet()
.stream()
.sorted(Map.Entry.<String, Integer>comparingByValue().reversed())
.forEachOrdered(x -> map1.put(x.getKey(), x.getValue()));
map1.forEach((k,v) ->{ System.out.println(k +" "+v);} );
I have worked with Python 3 and it has sorting for both keys and values using mapSorted = sorted(map.items(), key = lambda item : (item[1], item[0])). Is there something similar in Java 8?
The API you are looking forward to is Comparator#thenComparing. The reason why the implementation for such sorting is not straightforward is because of the type-inference.
The type inference needs some help which can be provided such as :
Comparator.comparing(Map.Entry<String, Integer>::getValue)
.reversed().thenComparing(Map.Entry::getKey)
Apart from which you shall ideally collect the output to a Map that preserves the order, otherwise, the sorting is a waste of computations. Hence something like this shall work:
LinkedHashMap<String, Integer> sortedMap = map.entrySet()
.stream()
.sorted(Comparator.comparing(Map.Entry<String, Integer>::getValue)
.reversed().thenComparing(Map.Entry::getKey))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,
(a, b) -> a, LinkedHashMap::new));
define your map
HashMap<String, Integer>map1 = new HashMap();
map1.put("aa",5);
map1.put("bbb",2);
map1.put("ccccc",2);
map1.put("dddddd",3);
sort,if you want to compare with string your need defining it by yourself
List<Entry<String, Integer>> collect = map1.entrySet().stream().sorted(new Comparator<Entry<String, Integer>>(){
#Override
public int compare(Entry<String, Integer> o1, Entry<String, Integer> o2) {
int ll=0;
if (o1.getValue()>o2.getValue()){
ll=-1;
}
else if(o1.getValue()<o2.getValue()){
ll=1;
}
else if (o1.getKey().length()>o2.getKey().length()) {
ll=-1;
}
else if (o1.getKey().length()<o2.getKey().length()) {
ll=1;
};
return ll;
}
}).collect(Collectors.toList());
the result like [aa=5, dddddd=3, ccccc=2, bbb=2]
I am not so familiar with Java 8 (still learning) and looking to see if I could find something equivalent of the below code using streams.
The below code mainly tries to get corresponding double value for each value in String and then sums it up. I could not find much help anywhere on this format. I am not sure if using streams would clean up the code or would make it messier.
// safe assumptions - String/List (Key/Value) cannot be null or empty
// inputMap --> Map<String, List<String>>
Map<String, Double> finalResult = new HashMap<>();
for (Map.Entry<String, List<String>> entry : inputMap.entrySet()) {
Double score = 0.0;
for (String current: entry.getValue()) {
score += computeScore(current);
}
finalResult.put(entry.getKey(), score);
}
private Double computeScore(String a) { .. }
Map<String, Double> finalResult = inputMap.entrySet()
.stream()
.collect(Collectors.toMap(
Entry::getKey,
e -> e.getValue()
.stream()
.mapToDouble(str -> computeScore(str))
.sum()));
Above code iterates over the map and creates a new map with same keys & before putting the values, it first iterates over each value - which is a list, computes score via calling computeScore() over each list element and then sums the scores collected to be put in the value.
You could also use the forEach method along with the stream API to yield the result you're seeking.
Map<String, Double> resultSet = new HashMap<>();
inputMap.forEach((k, v) -> resultSet.put(k, v.stream()
.mapToDouble(s -> computeScore(s)).sum()));
s -> computeScore(s) could be changed to use a method reference i.e. T::computeScore where T is the name of the class containing computeScore.
How about this one:
Map<String, Double> finalResult = inputMap.entrySet()
.stream()
.map(entry -> new AbstractMap.SimpleEntry<String, Double>( // maps each key to a new
// Entry<String, Double>
entry.getKey(), // the same key
entry.getValue().stream()
.mapToDouble(string -> computeScore(string)).sum())) // List<String> mapped to
// List<Double> and summed
.collect(Collectors.toMap(Entry::getKey, Entry::getValue)); // collected by the same
// key and a newly
// calulcated value
The version above could be merged to the single collect(..) method:
Map<String, Double> finalResult = inputMap.entrySet()
.stream()
.collect(Collectors.toMap(
Entry::getKey, // keeps the same key
entry -> entry.getValue()
.stream() // List<String> -> Stream<String>
// then Stream<String> -> Stream<Double>
.mapToDouble(string -> computeScore(string))
.sum())); // and summed
The key parts:
collect(..) performs a reduction on the elements using a certain strategy with a Collector.
Entry::getKey is a shortcut for entry -> entry.getKey. A function for mapping the key.
entry -> entry.getValue().stream() returns the Stream<String>
mapToDouble(..) returns the DoubleStream. This has an aggregating operation sum(..) which sums the elements - together creates a new value for the Map.
Regardless of whether you use the stream-based or the loop-based solution, it would be beneficial and add some clarity and structure to extract the inner loop into a method:
private double computeScore(Collection<String> strings)
{
return strings.stream().mapToDouble(this::computeScore).sum();
}
Of course, this could also be implemented using a loop, but ... that's exactly the point: This method can now be called, either in the outer loop, or on the values of a stream of map entries.
The outer loop or stream could also be pulled into a method. In the example below, I generalized this a bit: The type of the keys of the map does not matter. Neither does whether the values are List or Collection instances.
As an alternative to the currently accepted answer, the stream-based solution here does not fill a new map that is created manually. Instead, it uses a Collector.
(This is similar to other answers, but I think that the extracted computeScore method greatly simplifies the otherwise rather ugly lambdas that are necessary for the nested streams)
import java.util.Arrays;
import java.util.Collection;
import java.util.HashMap;
import java.util.LinkedHashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.stream.Collectors;
public class ToStreamOrNotToStream
{
public static void main(String[] args)
{
ToStreamOrNotToStream t = new ToStreamOrNotToStream();
Map<String, List<String>> inputMap =
new LinkedHashMap<String, List<String>>();
inputMap.put("A", Arrays.asList("1.0", "2.0", "3.0"));
inputMap.put("B", Arrays.asList("2.0", "3.0", "4.0"));
inputMap.put("C", Arrays.asList("3.0", "4.0", "5.0"));
System.out.println("Result A: " + t.computeA(inputMap));
System.out.println("Result B: " + t.computeB(inputMap));
}
private <T> Map<T, Double> computeA(
Map<T, ? extends Collection<String>> inputMap)
{
Map<T, Double> finalResult = new HashMap<>();
for (Entry<T, ? extends Collection<String>> entry : inputMap.entrySet())
{
double score = computeScore(entry.getValue());
finalResult.put(entry.getKey(), score);
}
return finalResult;
}
private <T> Map<T, Double> computeB(
Map<T, ? extends Collection<String>> inputMap)
{
return inputMap.entrySet().stream().collect(
Collectors.toMap(Entry::getKey, e -> computeScore(e.getValue())));
}
private double computeScore(Collection<String> strings)
{
return strings.stream().mapToDouble(this::computeScore).sum();
}
private double computeScore(String a)
{
return Double.parseDouble(a);
}
}
I found it somewhat shorter:
value = startDates.entrySet().stream().mapToDouble(Entry::getValue).sum();
Let's say I have a HashMap with String keys and Integer values:
map = {cat=1, kid=3, girl=3, adult=2, human=5, dog=2, boy=2}
I want to switch the keys and values by putting this information into another HashMap. I know that a HashMap cannot have duplicate keys, so I tried to put the information into a HashMap with the Integer for the keys that would map to a String ArrayList so that I could potentially have one Integer mapping to multiple Strings:
swap = {1=[cat], 2=[adult, dog, boy], 3=[kid, girl], 5=[human]}
I tried the following code:
HashMap<Integer, ArrayList<String>> swap = new HashMap<Integer, ArrayList<String>>();
for (String x : map.keySet()) {
for (int i = 0; i <= 5; i++) {
ArrayList<String> list = new ArrayList<String>();
if (i == map.get(x)) {
list.add(x);
swap.put(i, list);
}
}
}
The only difference in my code is that I didn't hard code the number 5 into my index; I have a method that finds the highest integer value in the original HashMap and used that. I know it works correctly because I get the same output even if I hard code the 5 in there, I just didn't include it to save space.
My goal here is to be able to do this 'reversal' with any set of data, otherwise I could just hard code the value. The output I get from the above code is this:
swap = {1=[cat], 2=[boy], 3=[girl], 5=[human]}
As you can see, my problem is that the value ArrayList is only keeping the last String that was put into it, instead of collecting all of them. How can I make the ArrayList store each String, rather than just the last String?
With Java 8, you can do the following:
Map<String, Integer> map = new HashMap<>();
map.put("cat", 1);
map.put("kid", 3);
map.put("girl", 3);
map.put("adult", 2);
map.put("human", 5);
map.put("dog", 2);
map.put("boy", 2);
Map<Integer, List<String>> newMap = map.keySet()
.stream()
.collect(Collectors.groupingBy(map::get));
System.out.println(newMap);
The output will be:
{1=[cat], 2=[adult, dog, boy], 3=[kid, girl], 5=[human]}
you are recreating the arrayList for every iteration and i can't figure out a way to do it with that logic, here is a good way though and without the need to check for the max integer:
for (Map.Entry<String, Integer> entry : map.entrySet()) {
String key = entry.getKey();
Integer value = entry.getValue();
List<String> get = swap.get(value);
if (get == null) {
get = new ArrayList<>();
swap.put(value, get);
}
get.add(key);
}
Best way is to iterate over the key set of the original map.
Also you have to asure that the List is present for any key in the target map:
for (Map.Entry<String,Integer> inputEntry : map.entrySet())
swap.computeIfAbsent(inputEntry.getValue(),()->new ArrayList<>()).add(inputEntry.getKey());
This is obviously not the best solution, but approaches the problem the same way you did by interchanging inner and outer loops as shown below.
Map<String, Integer> map = new HashMap<String, Integer>();
map.put("cat", 1);
map.put("kid", 3);
map.put("girl", 3);
map.put("adult", 2);
map.put("human", 5);
map.put("dog", 2);
map.put("boy", 2);
HashMap<Integer, ArrayList<String>> swap = new HashMap<Integer, ArrayList<String>>();
for (Integer value = 0; value <= 5; value++) {
ArrayList<String> list = new ArrayList<String>();
for (String key : map.keySet()) {
if (map.get(key) == value) {
list.add(key);
}
}
if (map.containsValue(value)) {
swap.put(value, list);
}
}
Output
{1=[cat], 2=[adult, dog, boy], 3=[kid, girl], 5=[human]}
Best way I can think of is using Map.forEach method on existing map and Map.computeIfAbsent method on new map:
Map<Integer, List<String>> swap = new HashMap<>();
map.forEach((k, v) -> swap.computeIfAbsent(v, k -> new ArrayList<>()).add(k));
As a side note, you can use the diamond operator <> to create your new map (there's no need to repeat the type of the key and value when invoking the map's constructor, as the compiler will infer them).
As a second side note, it's good practice to use interface types instead of concrete types, both for generic parameter types and for actual types. This is why I've used List and Map instead of ArrayList and HashMap, respectively.
Using groupingBy like in Jacob's answer but with Map.entrySet for better performance, as suggested by Boris:
// import static java.util.stream.Collectors.*
Map<Integer, List<String>> swap = map.entrySet()
.stream()
.collect(groupingBy(Entry::getValue, mapping(Entry::getKey, toList())));
This uses two more methods of Collectors: mapping and toList.
If it wasn't for these two helper functions, the solution could look like this:
Map<Integer, List<String>> swap = map.entrySet()
.stream()
.collect(
groupingBy(
Entry::getValue,
Collector.of(
ArrayList::new,
(list, e) -> {
list.add(e.getKey());
},
(left, right) -> { // only needed for parallel streams
left.addAll(right);
return left;
}
)
)
);
Or, using toMap instead of groupingBy:
Map<Integer, List<String>> swap = map.entrySet()
.stream()
.collect(
toMap(
Entry::getValue,
(e) -> new ArrayList<>(Arrays.asList(e.getKey())),
(left, right) -> {
left.addAll(right);
return left;
}
)
);
It seams you override the values instrad of adding them to the already creared arraylist. Try this:
HashMap<Integer, ArrayList<String>> swapedMap = new HashMap<Integer, ArrayList<String>>();
for (String key : map.keySet()) {
Integer swappedKey = map.get(key);
ArrayList<String> a = swapedMap.get(swappedKey);
if (a == null) {
a = new ArrayList<String>();
swapedMap.put(swappedKey, a)
}
a.add(key);
}
I didn't have time to run it (sorry, don't have Java compiler now), but should be almost ok :)
You could use the new merge method in java-8 from Map:
Map<Integer, List<String>> newMap = new HashMap<>();
map.forEach((key, value) -> {
List<String> values = new ArrayList<>();
values.add(key);
newMap.merge(value, values, (left, right) -> {
left.addAll(right);
return left;
});
});
I want to write a comparator that will let me sort a TreeMap by value instead of the default natural ordering.
I tried something like this, but can't find out what went wrong:
import java.util.*;
class treeMap {
public static void main(String[] args) {
System.out.println("the main");
byValue cmp = new byValue();
Map<String, Integer> map = new TreeMap<String, Integer>(cmp);
map.put("de",10);
map.put("ab", 20);
map.put("a",5);
for (Map.Entry<String,Integer> pair: map.entrySet()) {
System.out.println(pair.getKey()+":"+pair.getValue());
}
}
}
class byValue implements Comparator<Map.Entry<String,Integer>> {
public int compare(Map.Entry<String,Integer> e1, Map.Entry<String,Integer> e2) {
if (e1.getValue() < e2.getValue()){
return 1;
} else if (e1.getValue() == e2.getValue()) {
return 0;
} else {
return -1;
}
}
}
I guess what am I asking is: Can I get a Map.Entry passed to the comparator?
You can't have the TreeMap itself sort on the values, since that defies the SortedMap specification:
A Map that further provides a total ordering on its keys.
However, using an external collection, you can always sort Map.entrySet() however you wish, either by keys, values, or even a combination(!!) of the two.
Here's a generic method that returns a SortedSet of Map.Entry, given a Map whose values are Comparable:
static <K,V extends Comparable<? super V>>
SortedSet<Map.Entry<K,V>> entriesSortedByValues(Map<K,V> map) {
SortedSet<Map.Entry<K,V>> sortedEntries = new TreeSet<Map.Entry<K,V>>(
new Comparator<Map.Entry<K,V>>() {
#Override public int compare(Map.Entry<K,V> e1, Map.Entry<K,V> e2) {
int res = e1.getValue().compareTo(e2.getValue());
return res != 0 ? res : 1;
}
}
);
sortedEntries.addAll(map.entrySet());
return sortedEntries;
}
Now you can do the following:
Map<String,Integer> map = new TreeMap<String,Integer>();
map.put("A", 3);
map.put("B", 2);
map.put("C", 1);
System.out.println(map);
// prints "{A=3, B=2, C=1}"
System.out.println(entriesSortedByValues(map));
// prints "[C=1, B=2, A=3]"
Note that funky stuff will happen if you try to modify either the SortedSet itself, or the Map.Entry within, because this is no longer a "view" of the original map like entrySet() is.
Generally speaking, the need to sort a map's entries by its values is atypical.
Note on == for Integer
Your original comparator compares Integer using ==. This is almost always wrong, since == with Integer operands is a reference equality, not value equality.
System.out.println(new Integer(0) == new Integer(0)); // prints "false"!!!
Related questions
When comparing two Integers in Java does auto-unboxing occur? (NO!!!)
Is it guaranteed that new Integer(i) == i in Java? (YES!!!)
polygenelubricants answer is almost perfect. It has one important bug though. It will not handle map entries where the values are the same.
This code:...
Map<String, Integer> nonSortedMap = new HashMap<String, Integer>();
nonSortedMap.put("ape", 1);
nonSortedMap.put("pig", 3);
nonSortedMap.put("cow", 1);
nonSortedMap.put("frog", 2);
for (Entry<String, Integer> entry : entriesSortedByValues(nonSortedMap)) {
System.out.println(entry.getKey()+":"+entry.getValue());
}
Would output:
ape:1
frog:2
pig:3
Note how our cow dissapeared as it shared the value "1" with our ape :O!
This modification of the code solves that issue:
static <K,V extends Comparable<? super V>> SortedSet<Map.Entry<K,V>> entriesSortedByValues(Map<K,V> map) {
SortedSet<Map.Entry<K,V>> sortedEntries = new TreeSet<Map.Entry<K,V>>(
new Comparator<Map.Entry<K,V>>() {
#Override public int compare(Map.Entry<K,V> e1, Map.Entry<K,V> e2) {
int res = e1.getValue().compareTo(e2.getValue());
return res != 0 ? res : 1; // Special fix to preserve items with equal values
}
}
);
sortedEntries.addAll(map.entrySet());
return sortedEntries;
}
In Java 8:
LinkedHashMap<Integer, String> sortedMap = map.entrySet().stream()
.sorted(Map.Entry.comparingByValue(/* Optional: Comparator.reverseOrder() */))
.collect(Collectors.toMap(Map.Entry::getKey,
Map.Entry::getValue,
(e1, e2) -> e1, LinkedHashMap::new));
A TreeMap is always sorted by the keys, anything else is impossible. A Comparator merely allows you to control how the keys are sorted.
If you want the sorted values, you have to extract them into a List and sort that.
This can't be done by using a Comparator, as it will always get the key of the map to compare. TreeMap can only sort by the key.
Olof's answer is good, but it needs one more thing before it's perfect. In the comments below his answer, dacwe (correctly) points out that his implementation violates the Compare/Equals contract for Sets. If you try to call contains or remove on an entry that's clearly in the set, the set won't recognize it because of the code that allows entries with equal values to be placed in the set. So, in order to fix this, we need to test for equality between the keys:
static <K,V extends Comparable<? super V>> SortedSet<Map.Entry<K,V>> entriesSortedByValues(Map<K,V> map) {
SortedSet<Map.Entry<K,V>> sortedEntries = new TreeSet<Map.Entry<K,V>>(
new Comparator<Map.Entry<K,V>>() {
#Override public int compare(Map.Entry<K,V> e1, Map.Entry<K,V> e2) {
int res = e1.getValue().compareTo(e2.getValue());
if (e1.getKey().equals(e2.getKey())) {
return res; // Code will now handle equality properly
} else {
return res != 0 ? res : 1; // While still adding all entries
}
}
}
);
sortedEntries.addAll(map.entrySet());
return sortedEntries;
}
"Note that the ordering maintained by a sorted set (whether or not an explicit comparator is provided) must be consistent with equals if the sorted set is to correctly implement the Set interface... the Set interface is defined in terms of the equals operation, but a sorted set performs all element comparisons using its compareTo (or compare) method, so two elements that are deemed equal by this method are, from the standpoint of the sorted set, equal."
(http://docs.oracle.com/javase/6/docs/api/java/util/SortedSet.html)
Since we originally overlooked equality in order to force the set to add equal valued entries, now we have to test for equality in the keys in order for the set to actually return the entry you're looking for. This is kinda messy and definitely not how sets were intended to be used - but it works.
I know this post specifically asks for sorting a TreeMap by values, but for those of us that don't really care about implementation but do want a solution that keeps the collection sorted as elements are added, I would appreciate feedback on this TreeSet-based solution. For one, elements are not easily retrieved by key, but for the use case I had at hand (finding the n keys with the lowest values), this was not a requirement.
TreeSet<Map.Entry<Integer, Double>> set = new TreeSet<>(new Comparator<Map.Entry<Integer, Double>>()
{
#Override
public int compare(Map.Entry<Integer, Double> o1, Map.Entry<Integer, Double> o2)
{
int valueComparison = o1.getValue().compareTo(o2.getValue());
return valueComparison == 0 ? o1.getKey().compareTo(o2.getKey()) : valueComparison;
}
});
int key = 5;
double value = 1.0;
set.add(new AbstractMap.SimpleEntry<>(key, value));
A lot of people hear adviced to use List and i prefer to use it as well
here are two methods you need to sort the entries of the Map according to their values.
static final Comparator<Entry<?, Double>> DOUBLE_VALUE_COMPARATOR =
new Comparator<Entry<?, Double>>() {
#Override
public int compare(Entry<?, Double> o1, Entry<?, Double> o2) {
return o1.getValue().compareTo(o2.getValue());
}
};
static final List<Entry<?, Double>> sortHashMapByDoubleValue(HashMap temp)
{
Set<Entry<?, Double>> entryOfMap = temp.entrySet();
List<Entry<?, Double>> entries = new ArrayList<Entry<?, Double>>(entryOfMap);
Collections.sort(entries, DOUBLE_VALUE_COMPARATOR);
return entries;
}
import java.util.*;
public class Main {
public static void main(String[] args) {
TreeMap<String, Integer> initTree = new TreeMap();
initTree.put("D", 0);
initTree.put("C", -3);
initTree.put("A", 43);
initTree.put("B", 32);
System.out.println("Sorted by keys:");
System.out.println(initTree);
List list = new ArrayList(initTree.entrySet());
Collections.sort(list, new Comparator<Map.Entry<String, Integer>>() {
#Override
public int compare(Map.Entry<String, Integer> e1, Map.Entry<String, Integer> e2) {
return e1.getValue().compareTo(e2.getValue());
}
});
System.out.println("Sorted by values:");
System.out.println(list);
}
}
//convert HashMap into List
List<Entry<String, Integer>> list = new LinkedList<Entry<String, Integer>>(map.entrySet());
Collections.sort(list, (o1, o2) -> o1.getValue().compareTo(o2.getValue()));
If you want to use a Hash map you can add a condition in Comparator to check by values first & if values are equal perform a sort on keys
HashMap<String , Integer> polpularity = new HashMap<>();
List<String> collect = popularity.entrySet().stream().sorted((t2, t1) -> {
if (t2.getValue() > t1.getValue()) {
return -1;
} else if (t2.getValue() < t1.getValue()) {
return +1;
} else {
return t2.getKey().compareTo(t1.getKey());
}
}).map(entry -> entry.getKey()).collect(Collectors.toList());
If you don't want to take care of the latter condition then use a Treemap which will offer you sorting by itself, this can be done in an elegant single line of code:
TreeMap<String, Integer> popularity = new TreeMap<>();
List<String> collect = popularity.entrySet().stream().sorted(Collections.reverseOrder(Map.Entry.comparingByValue())).map(entry -> entry.getKey()).collect(Collectors.toList());
TreeMap is always sorted by the keys.
If you want TreeMap to be sorted by the values, so you can simply construct it also.
Example:
// the original TreeMap which is sorted by key
Map<String, Integer> map = new TreeMap<>();
map.put("de",10);
map.put("ab", 20);
map.put("a",5);
// expected output:
// {a=5, ab=20, de=10}
System.out.println(map);
// now we will constrcut a new TreeSet which is sorted by values
// [original TreeMap values will be the keys for this new TreeMap]
TreeMap<Integer, String> newTreeMapSortedByValue = new TreeMap();
treeMapmap.forEach((k,v) -> newTreeMapSortedByValue.put(v, k));
// expected output:
// {5=a, 10=de, 20=ab}
System.out.println(newTreeMapSortedByValue);
Only 1 Line Of Code Solution
Normal Order
map.entrySet().stream().sorted(Map.Entry.comparingByValue()).forEach(x->{});
Reverse Order
map.entrySet().stream().sorted(Map.Entry.comparingByValue(Comparator.reverseOrder())).forEachOrdered(x -> {});