TreeMap sort is not working on equal values

TreeMap sort is not working on equal values - java

I have written a method that sorts the TreeMap by its values
public TreeMap<String, Integer> sortByValues(final TreeMap<String, Integer> map) {
Comparator<String> valueComparator = new Comparator<String>() {
public int compare(String k1, String k2) {
int compare = map.get(k1).compareTo(map.get(k2));
return compare;
}
};
Map<String, Integer> sortedByValues = new TreeMap<String, Integer>(valueComparator);
sortedByValues.putAll(map);
return sortedByValues;
}
The above method works fine in normal case but fails when there is a duplicate value present in the TreeMap. Any duplicate value entry is removed from the Map.After googling it I found the solution as
public Map<String, Integer> sortByValuesTree(final Map<String, Integer> map) {
Comparator<String> valueComparator = new Comparator<String>() {
public int compare(String k1, String k2) {
int compare = map.get(k1).compareTo(map.get(k2));
if (compare == 0) return 1;
else return compare;
}
};
Map<String, Integer> sortedByValues = new TreeMap<String, Integer>(valueComparator);
sortedByValues.putAll(map);
return sortedByValues;
}
The above works fine but I am not able to understand why first method didn't work. Why did it remove duplicate value entry? Can someone please let me know

Why did it remove duplicate value entry?
Because that's the very definition of a Map: for a given key, it stores one value. If you put another value for a key that is already in the map, the new value replaces the old one for this key. And since you told the map that a key is equal to another one when their associated value are equal, the map considers two keys to be equal when their value are equal.
Note that your solution is a bad one which probably works by accident. Indeed, your comparator doesn't respect the contract of the Comparator interface. Indeed, when two values are equal, you arbitrarily decide to make the first one bigger than the second one. This means that your comparator makes A > B and B > A true at the same time, which is not correct.
Sorting a TreeMap by value just looks like an absurdity to me. You won't be able to add any new value to the map anyway, since it would require the entry to already exist in the old map. Shouldn't you simply have a sorted list of map entries?

Actually, especially in Java 7 onwards, even your second method is not going to work. (See below for why.) Anyway, the reason is that maps must have distinct keys, and when you are using your value as key, two equal values would be treated as equal keys.
The proper fix, by the way, is to sort by value, then by key:
int compare = map.get(k1).compareTo(map.get(k2));
if (compare == 0) {
compare = k1.compareTo(k2);
}
return compare;
Proper comparators must follow three rules:
compare(a, a) == 0 for all values of a.
signum(compare(a, b)) == -signum(compare(b, a)) for all values of a and b.
if signum(compare(a, b)) == signum(compare(b, c)), then signum(compare(a, c)) must also have the same value, for all values of a, b, and c.

Related

why different key with duplicated value disappear when transfer hashmap to treemap

I use the code below to sort my hashmap by its value. But the result seems confused because it only keep one entry for one value and remove another entry with duplicate value.
Here is the Comparator code:
class ValueComparator implements Comparator {
Map map;
public ValueComparator(Map map) {
this.map = map;
}
public int compare(Object keyA, Object keyB) {
Comparable valueA = (Comparable) map.get(keyA);
Comparable valueB = (Comparable) map.get(keyB);
return valueB.compareTo(valueA);
}
And here is how I use it:
TreeMap sortedMap=new TreeMap(new ValueComparator(allCandidateMap));
sortedMap.putAll(allCandidateMap);

That makes perfect sense. You've declared that if two keys map to equal values in allCandidateMap, they should be considered equal, since your compare will be returning 0.
What this comes down to is that your comparator has almost reversed the roles of key and value. If you try doing other operations you will probably find that the values of the map often behave like keys. Methods like get and containsKey will act as if they're looking up the values, not the keys (but then get will return the value you passed in, so values are still values as well). The comparator defines the behaviour of the TreeMap, and you've asked for very weird behaviour.

How do I sort the elements of an HashMap according to their values? [duplicate]

This question already has answers here:
Closed 10 years ago.
I have the following HashMap:
HashMap<String, Integer> counts = new HashMap<String, Integer>();
What is the simplest way to order it according to the values?

You can't sort a Map by the values, especially not a HashMap, which can't be sorted at all.
Instead, you can sort the entries:
List<Map.Entry<String, Integer>> entries = new ArrayList<Map.Entry<String, Integer>>(map.entrySet());
Collections.sort(entries, new Comparator<Map.Entry<String, Integer>>() {
public int compare(
Map.Entry<String, Integer> entry1, Map.Entry<String, Integer> entry2) {
return entry1.getValue().compareTo(entry2.getValue());
}
});
will sort the entries in ascending order of count.

You can get a set of entries (Set of Map.Entry) from a map, by using map.entrySet(). Just iterate over them, and check the values by getValue().

A work around, if you want to them print them in order(Not storing).
Create a new Map (tempMap) and put your value as key and key as value. To make the keys unique, please add some unique value in each of the keys e.g. key1 = value1+#0.
Get the list of values as map.values() as list myVlues
Sort the myVlues list as Collections.sort(myVlues)
Now iterate the myVlues, get the corresponding key from tempMap, restore the key e.g. key.substring(0, key.length-2) and print the key and value pair.
Hope this helps.

A TreeMap can keep its entries in an order defined by a Comparator.
We can create a comparator that will order the Map by putting the greatest value first.
Then, we will build a TreeMap that uses that Comparator.
We will then put all the entries in our counts map into the Comparator.
Finally, we will get the first key in the map, which should be the most common word (or at least one of them, if multiple words have equal counts).
public class Testing {
public static void main(String[] args) {
HashMap<String,Double> counts = new HashMap<String,Integer>();
// Sample word counts
counts.put("the", 100);
counts.put("pineapple",5);
counts.put("a", 50);
// Step 1: Create a Comparator that order by value with greatest value first
MostCommonValueFirst mostCommonValueFirst = new MostCommonValueFirst(counts);
// Step 2: Build a TreeMap that uses that Comparator
TreeMap<String,Double> sortedMap = new TreeMap<String,Integer (mostCommonValueFirst);
// Step 3: Populate TreeMap with values from the counts map
sortedMap.putAll(counts);
// Step 4: The first key in the map is the most commonly used word
System.out.println("Most common word: " + sortedMap.firstKey());
}
}
private class MostCommonValueFirst implements Comparator<String> {
Map<String, Integer> base;
public MostCommonValueFirst(Map<String, Integer> base) {
this.base = base;
}
// Note: this comparator imposes orderings that are inconsistent with equals.
public int compare(String a, String b) {
if (base.get(a) >= base.get(b)) {
return 1;
} else {
return -1;
} // returning 0 would merge keys
}
}
Source: https://stackoverflow.com/a/1283722/284685

Sorting Java TreeMap by value not working when more than two values have the same sort-property

I want to sort a Java TreeMap based on some attribute of value. To be specific, I want to sort a TreeMap<Integer, Hashset<Integer>> based on the size of Hashset<Integer>. To achieve this, I have done the following:
A Comparator class:
private static class ValueComparer implements Comparator<Integer> {
private Map<Integer, HashSet<Integer>> map = null;
public ValueComparer (Map<Integer, HashSet<Integer>> map){
super();
this.map = map;
}
#Override
public int compare(Integer o1, Integer o2) {
HashSet<Integer> h1 = map.get(o1);
HashSet<Integer> h2 = map.get(o2);
int compare = h2.size().compareTo(h1.size());
if (compare == 0 && o1!=o2){
return -1;
}
else {
return compare;
}
}
}
A usage example:
TreeMap<Integer, HashSet<Integer>> originalMap = new TreeMap<Integer, HashSet<Integer>>();
//load keys and values into map
ValueComparer comp = new ValueComparer(originalMap);
TreeMap<Integer, HashSet<Integer>> sortedMap = new TreeMap<Integer, HashSet<Integer>>(comp);
sortedMap.putAll(originalMap);
The problem:
This doesn't work when originalMap contains more than 2 values of the same size. For other cases, it works alright. When more than two values in the map are of same size, the third value in the new sorted-map is null and throws NullPointerException when I try to access it.
I can't figure out what the problem is. Woule be nice if someone could point out.
Update:
Here's an example that works when two values have the same size: http://ideone.com/iFD9c
In the above example, if you uncomment lines 52-54, this code will fail- that's what my problem is.

Update: You cannot return -1 from ValueComparator just because you want to avoid duplicate keys to not be removed. Check the contract of Comparator.compare.
When you pass a Comparator to TreeMap you compute a ("new") place to put the entry. No (computed) key can exist more than once in a TreeMap.
If you want to sort the orginalMap by size of the value you can do as follows:
public static void main(String[] args) throws Exception {
TreeMap<Integer, HashSet<Integer>> originalMap =
new TreeMap<Integer, HashSet<Integer>>();
originalMap.put(0, new HashSet<Integer>() {{ add(6); add(7); }});
originalMap.put(1, new HashSet<Integer>() {{ add(6); }});
originalMap.put(2, new HashSet<Integer>() {{ add(9); add(8); }});
ArrayList<Map.Entry<Integer, HashSet<Integer>>> list =
new ArrayList<Map.Entry<Integer, HashSet<Integer>>>();
list.addAll(originalMap.entrySet());
Collections.sort(list, new Comparator<Map.Entry<Integer,HashSet<Integer>>>(){
public int compare(Map.Entry<Integer, HashSet<Integer>> o1,
Map.Entry<Integer, HashSet<Integer>> o2) {
Integer size1 = (Integer) o1.getValue().size();
Integer size2 = (Integer) o2.getValue().size();
return size2.compareTo(size1);
}
});
System.out.println(list);
}

Your comparator logic (which I'm not sure I follow why you'd return -1 if the set sizes are equal but they keys are different) shouldn't affect what the Map itself returns when you call get(key).
Are you positive you aren't inserting null values into the initial map? What does this code look like?

Your comparator doesn't respect the Comparator contract: if compare(o1, o2) < 0, then compare(o2, o1) should be > 0. You must find a deterministic way of comparing your elements when both sizes are the same and the integers are not identical. You could perhaps use the System.identityHashCode() of the integers to compare them in this case.
That said, I really wonder what you could do with such a map: you can't create new Integers and use them to get a value out of the map, and you can't modify the sets that it holds.
Side note: your comparator code sample uses map and data to refer to the same map.

You can have TreeMap ordered only by keys. There is no way of creating TreeMap ordered by values, because you will get StackOverflowException.
Think about it. To get an element from a tree, you need to perform comparisions, but to perform comparisions, you need to get elements.
You will have to sort it in other collection or to use Tree, you will have to encapsulate the integer (from entry value) also into the entry key and define comparator using that integer taken from a key.

Assuming you cannot use a comparator that returns 0 with a Set, this might work: Add all the elements in originalMap.entrySet() to an ArrayList and then sort the ArrayList using your ValueComparer, changing it to return 0 as necessary.
Then add all the entries in the sorted ArrayList to a LinkedHashMap.

I had a similar problem as the original poster. I had a TreeMap i wanted to sort on a value. But when I made a comparator that looked at the value, i had issues because of the breaking of the comparator that JB talked about. I was able to use my custom comparator and still observe the contract. When the valuse I was looking at were equal, i fell back to comparing the keys. I didn't care about the order if values were equal.
public int compare(String a, String b) {
if(base.get(a)[0] == base.get(b)[0]){ //need to handle when they are equal
return a.compareTo(b);
}else if (base.get(a)[0] < base.get(b)[0]) {
return -1;
} else {
return 1;
} // returning 0 would merge keys

Compare 2 keys in Java HashMap

I've made a BinaryTree< HashMap<String, String> >.
How can I compare the two keys so I can correctly insert the two elements (HashMaps) into the ordered BinaryTree?
Here's what I've got so far:
public class MyMap<K extends Comparable<K>, V> extends HashMap<K, V> implements Comparable< MyMap<K, V> >
{
#override
public int compareTo(MyMap<K, V> mapTwo)
{
if ( (this.keySet().equals(mapTwo.keySet())) ) return 0;
//How can I check for greater than/less than and keep my generics?
}
EDIT: There is only one key in each HashMap (it's a very simple language translation system), so sorting the keys shouldn't be necessary. I would have liked to use the String.compareTo() method, but because of my generics, the compiler doesn't know that K is a String

I think you've picked a bad data structure.
HashMaps are not naturally ordered. The keys in the set for a HashMap have an unpredictable order that is sensitive to the sequence of operations that populated the map. This makes it unsuitable for comparing two HashMaps.
In order to compare a pair of HashMaps, you need to extract the respective key sets, sort them and then compare the sorted sets. In other words, a compareTo method for HashSet derived classes is going to be O(NlogN) on average.
FWIW, a compareTo implementation would look something like this, assuming that the method is to order the HashMaps based on the sorted lists keys in their respective key sets. Obviously, there are other orderings based on the key sets.
public int compareTo(MyMap<K, V> other) {
List<K> myKeys = new ArrayList<K>(this.keySet());
List<K> otherKeys = new ArrayList<K>(other.keySet());
Collections.sort(myKeys);
Collections.sort(otherKeys);
final int minSize = Math.min(myKeys.size(), otherKeys.size());
for (int i = 0; i < minSize; i++) {
int cmp = myKeys.get(i).compareTo(otherKeys.get(i));
if (cmp != 0) {
return cmp;
}
}
return (myKeys.size() - otherKeys.size());
}
If there is only ever one key / value pair in the map, then you should replace it with a simple Pair<K,V> class. Using a HashMap to represent a single pair is ... crazy.

TreeMap sort by value

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 -> {});

Develop Reference

Java is a programming language and computing platform first released by Sun Microsystems in 1995.

TreeMap sort is not working on equal values - java

Related

why different key with duplicated value disappear when transfer hashmap to treemap

How do I sort the elements of an HashMap according to their values? [duplicate]

Sorting Java TreeMap by value not working when more than two values have the same sort-property

Compare 2 keys in Java HashMap

TreeMap sort by value

Categories

Resources