Find the Biggest number in HashSet/HashMap java - java

I would like to find the biggest number in HashSet and HashMap. Say I have the number [22,6763,32,42,33] in my HashSet and I want to find the largest number in my current HashSet..how would i do this? and Same thing for the HashMap as well. I hope you can help me with it. Thank you.

You can use Collections.max(Collection) to find the maximum element out of any collection.
Similarly, for a HashMap, you can use the same method on its keySet() or values(), depending upon whether you want maximum key, or maximum value.
Also, if you want as such, you can use a TreeSet and TreeMap instead, that stores the elements in sorted key order.

try
int max = Collections.max(set);
int maxKey = Collections.max(map.keySet());
int maxValue Collections.max(map.values());

If you are forced to use a HashSet/HashMap, then you have to scan the whole HashSet/HashMap in order to find the maximum. Library functions like Collections.max() will do like this.
If you want O(1) retrieval of the maximum, and you are allowed to change the type of collection being used, use a sorted set/map (e.g. TreeSet/TreeMap).

Something like this:
Set<Integer> values = new HashSet<Integer>() {{
add(22);
add(6763);
add(32);
add(42);
add(33);
}};
int maxValue = Integer.MIN_VALUE;
for (int value : values) {
if (value > maxValue) {
maxValue = value;
}
}
And this:
Map<String, Integer> values = new HashMap<String, Integer>() {{
put("0", 22);
put("1", 6763);
put("2", 32);
put("3", 42);
put("4", 33);
}};
int maxValue = Integer.MIN_VALUE;
for (int value : values.values()) {
if (value > maxValue) {
maxValue = value;
}
}

In case of TreeMap, if you know the key/values are inserted randomly, the tree will be more or less balanced. Trees become unbalanced, if data is inserted in already sorted order, the capability to quickly find (or insert or delete) a given element is lost. In case of unbalanced tree, it will take time proportional to n, O(n) else O(1).

Consider using Apache Commons Math. Here is the API docs.
The class of interest is SummaryStatistics. It works with doubles and computes max, min, mean etc. on the fly (as you add values to it). The data values are not stored in memory, so this class can be used to compute statistics for very large data streams.

Here is a simple method which does what you are asking:
public String getMapKeyWithHighestValue(HashMap<String, Integer> map) {
String keyWithHighestVal = "";
// getting the maximum value in the Hashmap
int maxValueInMap = (Collections.max(map.values()));
//iterate through the map to get the key that corresponds to the maximum value in the Hashmap
for (Map.Entry<String, Integer> entry : map.entrySet()) { // Iterate through hashmap
if (entry.getValue() == maxValueInMap) {
keyWithHighestVal = entry.getKey(); // this is the key which has the max value
}
}
return keyWithHighestVal;
}

Note : If you want to find the biggest value from Map try maxEntry.get().getValue() instead of maxEntry.get().getKey()
1. Using Stream
public <K, V extends Comparable<V>> V maxUsingStreamAndLambda(Map<K, V> map) {
Optional<Entry<K, V>> maxEntry = map.entrySet()
.stream()
.max((Entry<K, V> e1, Entry<K, V> e2) -> e1.getValue()
.compareTo(e2.getValue())
);
return maxEntry.get().getKey();
}
2. Using Collections.max() with a Lambda Expression
public <K, V extends Comparable<V>> V maxUsingCollectionsMaxAndLambda(Map<K, V> map) {
Entry<K, V> maxEntry = Collections.max(map.entrySet(), (Entry<K, V> e1, Entry<K, V> e2) -> e1.getValue()
.compareTo(e2.getValue()));
return maxEntry.getKey();
}
3. Using Stream with Method Reference
public <K, V extends Comparable<V>> V maxUsingStreamAndMethodReference(Map<K, V> map) {
Optional<Entry<K, V>> maxEntry = map.entrySet()
.stream()
.max(Comparator.comparing(Map.Entry::getValue));
return maxEntry.get()
.getKey();
}
4. Using Collections.max()
public <K, V extends Comparable<V>> V maxUsingCollectionsMax(Map<K, V> map) {
Entry<K, V> maxEntry = Collections.max(map.entrySet(), new Comparator<Entry<K, V>>() {
public int compare(Entry<K, V> e1, Entry<K, V> e2) {
return e1.getValue()
.compareTo(e2.getValue());
}
});
return maxEntry.getKey();
}
5. Using Simple Iteration
public <K, V extends Comparable<V>> V maxUsingIteration(Map<K, V> map) {
Map.Entry<K, V> maxEntry = null;
for (Map.Entry<K, V> entry : map.entrySet()) {
if (maxEntry == null || entry.getValue()
.compareTo(maxEntry.getValue()) > 0) {
maxEntry = entry;
}
}
return maxEntry.getKey();
}

Related

How to write test case for below java code

public class CollectionUtils {
public static <K, V> Map<K, V> createMap(Iterable<V> values, Function<V, K> keyFunction, boolean skipNullKeys) {
Map<K, V> map = new HashMap<>();
for (V value : values) {
K key = keyFunction.apply(value);
if (key != null || !skipNullKeys) {
map.put(key, value);
}
}
return map;
}
I would use small and simple data, e. g.
List<Integer> values = Arrays.asList(1, null);
Function<Integer, String> toKey = String::valueOf;
Map<String, Integer> withNull = createMap(values, toKey, false);
Map<String, Integer> noNull = createMap(values, toKey, true);
Then you can test with Map#size, Map#containsKey (needed for null-key) and ...get("1").equals(1).
But as I mentioned in the comment, if your function may result in the same key for different values you only have the last value, e. g. Function<Integer, Integer> toKey = i -> i%2 will only procude 0 or 1 as a key (and a NullPointerException for null-values) so List<Integer> values = Arrays.asList(1, 3) will produce a Map with only one entry "1" -> 3.

Sorting Hashmap based on value Not sorting properly giving some unexpected output? [duplicate]

I am relatively new to Java, and often find that I need to sort a Map<Key, Value> on the values.
Since the values are not unique, I find myself converting the keySet into an array, and sorting that array through array sort with a custom comparator that sorts on the value associated with the key.
Is there an easier way?
Here's a generic-friendly version:
public class MapUtil {
public static <K, V extends Comparable<? super V>> Map<K, V> sortByValue(Map<K, V> map) {
List<Entry<K, V>> list = new ArrayList<>(map.entrySet());
list.sort(Entry.comparingByValue());
Map<K, V> result = new LinkedHashMap<>();
for (Entry<K, V> entry : list) {
result.put(entry.getKey(), entry.getValue());
}
return result;
}
}
Java 8 offers a new answer: convert the entries into a stream, and use the comparator combinators from Map.Entry:
Stream<Map.Entry<K,V>> sorted =
map.entrySet().stream()
.sorted(Map.Entry.comparingByValue());
This will let you consume the entries sorted in ascending order of value. If you want descending value, simply reverse the comparator:
Stream<Map.Entry<K,V>> sorted =
map.entrySet().stream()
.sorted(Collections.reverseOrder(Map.Entry.comparingByValue()));
If the values are not comparable, you can pass an explicit comparator:
Stream<Map.Entry<K,V>> sorted =
map.entrySet().stream()
.sorted(Map.Entry.comparingByValue(comparator));
You can then proceed to use other stream operations to consume the data. For example, if you want the top 10 in a new map:
Map<K,V> topTen =
map.entrySet().stream()
.sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
.limit(10)
.collect(Collectors.toMap(
Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));
The LinkedHashMap seen above iterates entries in the order in which they were inserted.
Or print to System.out:
map.entrySet().stream()
.sorted(Map.Entry.comparingByValue())
.forEach(System.out::println);
Important note:
This code can break in multiple ways. If you intend to use the code provided, be sure to read the comments as well to be aware of the implications. For example, values can no longer be retrieved by their key. (get always returns null.)
It seems much easier than all of the foregoing. Use a TreeMap as follows:
public class Testing {
public static void main(String[] args) {
HashMap<String, Double> map = new HashMap<String, Double>();
ValueComparator bvc = new ValueComparator(map);
TreeMap<String, Double> sorted_map = new TreeMap<String, Double>(bvc);
map.put("A", 99.5);
map.put("B", 67.4);
map.put("C", 67.4);
map.put("D", 67.3);
System.out.println("unsorted map: " + map);
sorted_map.putAll(map);
System.out.println("results: " + sorted_map);
}
}
class ValueComparator implements Comparator<String> {
Map<String, Double> base;
public ValueComparator(Map<String, Double> 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
}
}
Output:
unsorted map: {D=67.3, A=99.5, B=67.4, C=67.4}
results: {D=67.3, B=67.4, C=67.4, A=99.5}
Three 1-line answers...
I would use Google Collections Guava to do this - if your values are Comparable then you can use
valueComparator = Ordering.natural().onResultOf(Functions.forMap(map))
Which will create a function (object) for the map [that takes any of the keys as input, returning the respective value], and then apply natural (comparable) ordering to them [the values].
If they're not comparable, then you'll need to do something along the lines of
valueComparator = Ordering.from(comparator).onResultOf(Functions.forMap(map))
These may be applied to a TreeMap (as Ordering extends Comparator), or a LinkedHashMap after some sorting
NB: If you are going to use a TreeMap, remember that if a comparison == 0, then the item is already in the list (which will happen if you have multiple values that compare the same). To alleviate this, you could add your key to the comparator like so (presuming that your keys and values are Comparable):
valueComparator = Ordering.natural().onResultOf(Functions.forMap(map)).compound(Ordering.natural())
= Apply natural ordering to the value mapped by the key, and compound that with the natural ordering of the key
Note that this will still not work if your keys compare to 0, but this should be sufficient for most comparable items (as hashCode, equals and compareTo are often in sync...)
See Ordering.onResultOf() and Functions.forMap().
Implementation
So now that we've got a comparator that does what we want, we need to get a result from it.
map = ImmutableSortedMap.copyOf(myOriginalMap, valueComparator);
Now this will most likely work work, but:
needs to be done given a complete finished map
Don't try the comparators above on a TreeMap; there's no point trying to compare an inserted key when it doesn't have a value until after the put, i.e., it will break really fast
Point 1 is a bit of a deal-breaker for me; google collections is incredibly lazy (which is good: you can do pretty much every operation in an instant; the real work is done when you start using the result), and this requires copying a whole map!
"Full" answer/Live sorted map by values
Don't worry though; if you were obsessed enough with having a "live" map sorted in this manner, you could solve not one but both(!) of the above issues with something crazy like the following:
Note: This has changed significantly in June 2012 - the previous code could never work: an internal HashMap is required to lookup the values without creating an infinite loop between the TreeMap.get() -> compare() and compare() -> get()
import static org.junit.Assert.assertEquals;
import java.util.HashMap;
import java.util.Map;
import java.util.TreeMap;
import com.google.common.base.Functions;
import com.google.common.collect.Ordering;
class ValueComparableMap<K extends Comparable<K>,V> extends TreeMap<K,V> {
//A map for doing lookups on the keys for comparison so we don't get infinite loops
private final Map<K, V> valueMap;
ValueComparableMap(final Ordering<? super V> partialValueOrdering) {
this(partialValueOrdering, new HashMap<K,V>());
}
private ValueComparableMap(Ordering<? super V> partialValueOrdering,
HashMap<K, V> valueMap) {
super(partialValueOrdering //Apply the value ordering
.onResultOf(Functions.forMap(valueMap)) //On the result of getting the value for the key from the map
.compound(Ordering.natural())); //as well as ensuring that the keys don't get clobbered
this.valueMap = valueMap;
}
public V put(K k, V v) {
if (valueMap.containsKey(k)){
//remove the key in the sorted set before adding the key again
remove(k);
}
valueMap.put(k,v); //To get "real" unsorted values for the comparator
return super.put(k, v); //Put it in value order
}
public static void main(String[] args){
TreeMap<String, Integer> map = new ValueComparableMap<String, Integer>(Ordering.natural());
map.put("a", 5);
map.put("b", 1);
map.put("c", 3);
assertEquals("b",map.firstKey());
assertEquals("a",map.lastKey());
map.put("d",0);
assertEquals("d",map.firstKey());
//ensure it's still a map (by overwriting a key, but with a new value)
map.put("d", 2);
assertEquals("b", map.firstKey());
//Ensure multiple values do not clobber keys
map.put("e", 2);
assertEquals(5, map.size());
assertEquals(2, (int) map.get("e"));
assertEquals(2, (int) map.get("d"));
}
}
When we put, we ensure that the hash map has the value for the comparator, and then put to the TreeSet for sorting. But before that we check the hash map to see that the key is not actually a duplicate. Also, the comparator that we create will also include the key so that duplicate values don't delete the non-duplicate keys (due to == comparison).
These 2 items are vital for ensuring the map contract is kept; if you think you don't want that, then you're almost at the point of reversing the map entirely (to Map<V,K>).
The constructor would need to be called as
new ValueComparableMap(Ordering.natural());
//or
new ValueComparableMap(Ordering.from(comparator));
From http://www.programmersheaven.com/download/49349/download.aspx
private static <K, V> Map<K, V> sortByValue(Map<K, V> map) {
List<Entry<K, V>> list = new LinkedList<>(map.entrySet());
Collections.sort(list, new Comparator<Object>() {
#SuppressWarnings("unchecked")
public int compare(Object o1, Object o2) {
return ((Comparable<V>) ((Map.Entry<K, V>) (o1)).getValue()).compareTo(((Map.Entry<K, V>) (o2)).getValue());
}
});
Map<K, V> result = new LinkedHashMap<>();
for (Iterator<Entry<K, V>> it = list.iterator(); it.hasNext();) {
Map.Entry<K, V> entry = (Map.Entry<K, V>) it.next();
result.put(entry.getKey(), entry.getValue());
}
return result;
}
With Java 8, you can use the streams api to do it in a significantly less verbose way:
Map<K, V> sortedMap = map.entrySet().stream()
.sorted(Entry.comparingByValue())
.collect(Collectors.toMap(Entry::getKey, Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));
Sorting the keys requires the Comparator to look up each value for each comparison. A more scalable solution would use the entrySet directly, since then the value would be immediately available for each comparison (although I haven't backed this up by numbers).
Here's a generic version of such a thing:
public static <K, V extends Comparable<? super V>> List<K> getKeysSortedByValue(Map<K, V> map) {
final int size = map.size();
final List<Map.Entry<K, V>> list = new ArrayList<Map.Entry<K, V>>(size);
list.addAll(map.entrySet());
final ValueComparator<V> cmp = new ValueComparator<V>();
Collections.sort(list, cmp);
final List<K> keys = new ArrayList<K>(size);
for (int i = 0; i < size; i++) {
keys.set(i, list.get(i).getKey());
}
return keys;
}
private static final class ValueComparator<V extends Comparable<? super V>>
implements Comparator<Map.Entry<?, V>> {
public int compare(Map.Entry<?, V> o1, Map.Entry<?, V> o2) {
return o1.getValue().compareTo(o2.getValue());
}
}
There are ways to lessen memory rotation for the above solution. The first ArrayList created could for instance be re-used as a return value; this would require suppression of some generics warnings, but it might be worth it for re-usable library code. Also, the Comparator does not have to be re-allocated at every invocation.
Here's a more efficient albeit less appealing version:
public static <K, V extends Comparable<? super V>> List<K> getKeysSortedByValue2(Map<K, V> map) {
final int size = map.size();
final List reusedList = new ArrayList(size);
final List<Map.Entry<K, V>> meView = reusedList;
meView.addAll(map.entrySet());
Collections.sort(meView, SINGLE);
final List<K> keyView = reusedList;
for (int i = 0; i < size; i++) {
keyView.set(i, meView.get(i).getKey());
}
return keyView;
}
private static final Comparator SINGLE = new ValueComparator();
Finally, if you need to continously access the sorted information (rather than just sorting it once in a while), you can use an additional multi map. Let me know if you need more details...
The commons-collections library contains a solution called TreeBidiMap. Or, you could have a look at the Google Collections API. It has TreeMultimap which you could use.
And if you don't want to use these framework... they come with source code.
I've looked at the given answers, but a lot of them are more complicated than needed or remove map elements when several keys have same value.
Here is a solution that I think fits better:
public static <K, V extends Comparable<V>> Map<K, V> sortByValues(final Map<K, V> map) {
Comparator<K> valueComparator = new Comparator<K>() {
public int compare(K k1, K k2) {
int compare = map.get(k2).compareTo(map.get(k1));
if (compare == 0) return 1;
else return compare;
}
};
Map<K, V> sortedByValues = new TreeMap<K, V>(valueComparator);
sortedByValues.putAll(map);
return sortedByValues;
}
Note that the map is sorted from the highest value to the lowest.
Given Map
Map<String, Integer> wordCounts = new HashMap<>();
wordCounts.put("USA", 100);
wordCounts.put("jobs", 200);
wordCounts.put("software", 50);
wordCounts.put("technology", 70);
wordCounts.put("opportunity", 200);
Sort the map based on the value in ascending order
Map<String,Integer> sortedMap = wordCounts.entrySet().
stream().
sorted(Map.Entry.comparingByValue()).
collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));
System.out.println(sortedMap);
Sort the map based on value in descending order
Map<String,Integer> sortedMapReverseOrder = wordCounts.entrySet().
stream().
sorted(Map.Entry.comparingByValue(Comparator.reverseOrder())).
collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));
System.out.println(sortedMapReverseOrder);
Output:
{software=50, technology=70, USA=100, jobs=200, opportunity=200}
{jobs=200, opportunity=200, USA=100, technology=70, software=50}
To accomplish this with the new features in Java 8:
import static java.util.Map.Entry.comparingByValue;
import static java.util.stream.Collectors.toList;
<K, V> List<Entry<K, V>> sort(Map<K, V> map, Comparator<? super V> comparator) {
return map.entrySet().stream().sorted(comparingByValue(comparator)).collect(toList());
}
The entries are ordered by their values using the given comparator. Alternatively, if your values are mutually comparable, no explicit comparator is needed:
<K, V extends Comparable<? super V>> List<Entry<K, V>> sort(Map<K, V> map) {
return map.entrySet().stream().sorted(comparingByValue()).collect(toList());
}
The returned list is a snapshot of the given map at the time this method is called, so neither will reflect subsequent changes to the other. For a live iterable view of the map:
<K, V extends Comparable<? super V>> Iterable<Entry<K, V>> sort(Map<K, V> map) {
return () -> map.entrySet().stream().sorted(comparingByValue()).iterator();
}
The returned iterable creates a fresh snapshot of the given map each time it's iterated, so barring concurrent modification, it will always reflect the current state of the map.
Create customized comparator and use it while creating new TreeMap object.
class MyComparator implements Comparator<Object> {
Map<String, Integer> map;
public MyComparator(Map<String, Integer> map) {
this.map = map;
}
public int compare(Object o1, Object o2) {
if (map.get(o2) == map.get(o1))
return 1;
else
return ((Integer) map.get(o2)).compareTo((Integer)
map.get(o1));
}
}
Use the below code in your main func
Map<String, Integer> lMap = new HashMap<String, Integer>();
lMap.put("A", 35);
lMap.put("B", 75);
lMap.put("C", 50);
lMap.put("D", 50);
MyComparator comparator = new MyComparator(lMap);
Map<String, Integer> newMap = new TreeMap<String, Integer>(comparator);
newMap.putAll(lMap);
System.out.println(newMap);
Output:
{B=75, D=50, C=50, A=35}
While I agree that the constant need to sort a map is probably a smell, I think the following code is the easiest way to do it without using a different data structure.
public class MapUtilities {
public static <K, V extends Comparable<V>> List<Entry<K, V>> sortByValue(Map<K, V> map) {
List<Entry<K, V>> entries = new ArrayList<Entry<K, V>>(map.entrySet());
Collections.sort(entries, new ByValue<K, V>());
return entries;
}
private static class ByValue<K, V extends Comparable<V>> implements Comparator<Entry<K, V>> {
public int compare(Entry<K, V> o1, Entry<K, V> o2) {
return o1.getValue().compareTo(o2.getValue());
}
}
}
And here is an embarrassingly incomplete unit test:
public class MapUtilitiesTest extends TestCase {
public void testSorting() {
HashMap<String, Integer> map = new HashMap<String, Integer>();
map.put("One", 1);
map.put("Two", 2);
map.put("Three", 3);
List<Map.Entry<String, Integer>> sorted = MapUtilities.sortByValue(map);
assertEquals("First", "One", sorted.get(0).getKey());
assertEquals("Second", "Two", sorted.get(1).getKey());
assertEquals("Third", "Three", sorted.get(2).getKey());
}
}
The result is a sorted list of Map.Entry objects, from which you can obtain the keys and values.
Use a generic comparator such as:
final class MapValueComparator<K,V extends Comparable<V>> implements Comparator<K> {
private final Map<K,V> map;
private MapValueComparator() {
super();
}
public MapValueComparator(Map<K,V> map) {
this();
this.map = map;
}
public int compare(K o1, K o2) {
return map.get(o1).compareTo(map.get(o2));
}
}
The answer voted for the most does not work when you have 2 items that equals.
the TreeMap leaves equal values out.
the exmaple:
unsorted map
key/value: D/67.3
key/value: A/99.5
key/value: B/67.4
key/value: C/67.5
key/value: E/99.5
results
key/value: A/99.5
key/value: C/67.5
key/value: B/67.4
key/value: D/67.3
So leaves out E!!
For me it worked fine to adjust the comparator, if it equals do not return 0 but -1.
in the example:
class ValueComparator implements Comparator {
Map base;
public ValueComparator(Map base) {
this.base = base;
}
public int compare(Object a, Object b) {
if((Double)base.get(a) < (Double)base.get(b)) {
return 1;
} else if((Double)base.get(a) == (Double)base.get(b)) {
return -1;
} else {
return -1;
}
}
}
now it returns:
unsorted map:
key/value: D/67.3
key/value: A/99.5
key/value: B/67.4
key/value: C/67.5
key/value: E/99.5
results:
key/value: A/99.5
key/value: E/99.5
key/value: C/67.5
key/value: B/67.4
key/value: D/67.3
as a response to Aliens (2011 nov. 22):
I Am using this solution for a map of Integer Id's and names, but the idea is the same, so might be the code above is not correct (I will write it in a test and give you the correct code), this is the code for a Map sorting, based on the solution above:
package nl.iamit.util;
import java.util.Comparator;
import java.util.Map;
public class Comparators {
public static class MapIntegerStringComparator implements Comparator {
Map<Integer, String> base;
public MapIntegerStringComparator(Map<Integer, String> base) {
this.base = base;
}
public int compare(Object a, Object b) {
int compare = ((String) base.get(a))
.compareTo((String) base.get(b));
if (compare == 0) {
return -1;
}
return compare;
}
}
}
and this is the test class (I just tested it, and this works for the Integer, String Map:
package test.nl.iamit.util;
import java.util.HashMap;
import java.util.TreeMap;
import nl.iamit.util.Comparators;
import org.junit.Test;
import static org.junit.Assert.assertArrayEquals;
public class TestComparators {
#Test
public void testMapIntegerStringComparator(){
HashMap<Integer, String> unSoretedMap = new HashMap<Integer, String>();
Comparators.MapIntegerStringComparator bvc = new Comparators.MapIntegerStringComparator(
unSoretedMap);
TreeMap<Integer, String> sorted_map = new TreeMap<Integer, String>(bvc);
//the testdata:
unSoretedMap.put(new Integer(1), "E");
unSoretedMap.put(new Integer(2), "A");
unSoretedMap.put(new Integer(3), "E");
unSoretedMap.put(new Integer(4), "B");
unSoretedMap.put(new Integer(5), "F");
sorted_map.putAll(unSoretedMap);
Object[] targetKeys={new Integer(2),new Integer(4),new Integer(3),new Integer(1),new Integer(5) };
Object[] currecntKeys=sorted_map.keySet().toArray();
assertArrayEquals(targetKeys,currecntKeys);
}
}
here is the code for the Comparator of a Map:
public static class MapStringDoubleComparator implements Comparator {
Map<String, Double> base;
public MapStringDoubleComparator(Map<String, Double> base) {
this.base = base;
}
//note if you want decending in stead of ascending, turn around 1 and -1
public int compare(Object a, Object b) {
if ((Double) base.get(a) == (Double) base.get(b)) {
return 0;
} else if((Double) base.get(a) < (Double) base.get(b)) {
return -1;
}else{
return 1;
}
}
}
and this is the testcase for this:
#Test
public void testMapStringDoubleComparator(){
HashMap<String, Double> unSoretedMap = new HashMap<String, Double>();
Comparators.MapStringDoubleComparator bvc = new Comparators.MapStringDoubleComparator(
unSoretedMap);
TreeMap<String, Double> sorted_map = new TreeMap<String, Double>(bvc);
//the testdata:
unSoretedMap.put("D",new Double(67.3));
unSoretedMap.put("A",new Double(99.5));
unSoretedMap.put("B",new Double(67.4));
unSoretedMap.put("C",new Double(67.5));
unSoretedMap.put("E",new Double(99.5));
sorted_map.putAll(unSoretedMap);
Object[] targetKeys={"D","B","C","E","A"};
Object[] currecntKeys=sorted_map.keySet().toArray();
assertArrayEquals(targetKeys,currecntKeys);
}
of cource you can make this a lot more generic, but I just needed it for 1 case (the Map)
Instead of using Collections.sort as some do I'd suggest using Arrays.sort. Actually what Collections.sort does is something like this:
public static <T extends Comparable<? super T>> void sort(List<T> list) {
Object[] a = list.toArray();
Arrays.sort(a);
ListIterator<T> i = list.listIterator();
for (int j=0; j<a.length; j++) {
i.next();
i.set((T)a[j]);
}
}
It just calls toArray on the list and then uses Arrays.sort. This way all the map entries will be copied three times: once from the map to the temporary list (be it a LinkedList or ArrayList), then to the temporary array and finally to the new map.
My solution ommits this one step as it does not create unnecessary LinkedList. Here is the code, generic-friendly and performance-optimal:
public static <K, V extends Comparable<? super V>> Map<K, V> sortByValue(Map<K, V> map)
{
#SuppressWarnings("unchecked")
Map.Entry<K,V>[] array = map.entrySet().toArray(new Map.Entry[map.size()]);
Arrays.sort(array, new Comparator<Map.Entry<K, V>>()
{
public int compare(Map.Entry<K, V> e1, Map.Entry<K, V> e2)
{
return e1.getValue().compareTo(e2.getValue());
}
});
Map<K, V> result = new LinkedHashMap<K, V>();
for (Map.Entry<K, V> entry : array)
result.put(entry.getKey(), entry.getValue());
return result;
}
This is a variation of Anthony's answer, which doesn't work if there are duplicate values:
public static <K, V extends Comparable<V>> Map<K, V> sortMapByValues(final Map<K, V> map) {
Comparator<K> valueComparator = new Comparator<K>() {
public int compare(K k1, K k2) {
final V v1 = map.get(k1);
final V v2 = map.get(k2);
/* Not sure how to handle nulls ... */
if (v1 == null) {
return (v2 == null) ? 0 : 1;
}
int compare = v2.compareTo(v1);
if (compare != 0)
{
return compare;
}
else
{
Integer h1 = k1.hashCode();
Integer h2 = k2.hashCode();
return h2.compareTo(h1);
}
}
};
Map<K, V> sortedByValues = new TreeMap<K, V>(valueComparator);
sortedByValues.putAll(map);
return sortedByValues;
}
Note that it's rather up in the air how to handle nulls.
One important advantage of this approach is that it actually returns a Map, unlike some of the other solutions offered here.
Best Approach
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.Map.Entry;
public class OrderByValue {
public static void main(String a[]){
Map<String, Integer> map = new HashMap<String, Integer>();
map.put("java", 20);
map.put("C++", 45);
map.put("Unix", 67);
map.put("MAC", 26);
map.put("Why this kolavari", 93);
Set<Entry<String, Integer>> set = map.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 (o1.getValue()).compareTo( o2.getValue() );//Ascending order
//return (o2.getValue()).compareTo( o1.getValue() );//Descending order
}
} );
for(Map.Entry<String, Integer> entry:list){
System.out.println(entry.getKey()+" ==== "+entry.getValue());
}
}}
Output
java ==== 20
MAC ==== 26
C++ ==== 45
Unix ==== 67
Why this kolavari ==== 93
Late Entry.
With the advent of Java-8, we can use streams for data manipulation in a very easy/succinct way. You can use streams to sort the map entries by value and create a LinkedHashMap which preserves insertion-order iteration.
Eg:
LinkedHashMap sortedByValueMap = map.entrySet().stream()
.sorted(comparing(Entry<Key,Value>::getValue).thenComparing(Entry::getKey)) //first sorting by Value, then sorting by Key(entries with same value)
.collect(LinkedHashMap::new,(map,entry) -> map.put(entry.getKey(),entry.getValue()),LinkedHashMap::putAll);
For reverse ordering, replace:
comparing(Entry<Key,Value>::getValue).thenComparing(Entry::getKey)
with
comparing(Entry<Key,Value>::getValue).thenComparing(Entry::getKey).reversed()
Major problem. If you use the first answer (Google takes you here), change the comparator to add an equal clause, otherwise you cannot get values from the sorted_map by keys:
public int compare(String a, String b) {
if (base.get(a) > base.get(b)) {
return 1;
} else if (base.get(a) < base.get(b)){
return -1;
}
return 0;
// returning 0 would merge keys
}
There are a lot of answers for this question already, but none provided me what I was looking for, a map implementation that returns keys and entries sorted by the associated value, and maintains this property as keys and values are modified in the map. Two other questions ask for this specifically.
I cooked up a generic friendly example that solves this use case. This implementation does not honor all of the contracts of the Map interface, such as reflecting value changes and removals in the sets return from keySet() and entrySet() in the original object. I felt such a solution would be too large to include in a Stack Overflow answer. If I manage to create a more complete implementation, perhaps I will post it to Github and then to it link in an updated version of this answer.
import java.util.*;
/**
* A map where {#link #keySet()} and {#link #entrySet()} return sets ordered
* by associated values based on the the comparator provided at construction
* time. The order of two or more keys with identical values is not defined.
* <p>
* Several contracts of the Map interface are not satisfied by this minimal
* implementation.
*/
public class ValueSortedMap<K, V> extends HashMap<K, V> {
protected Map<V, Collection<K>> valueToKeysMap;
// uses natural order of value object, if any
public ValueSortedMap() {
this((Comparator<? super V>) null);
}
public ValueSortedMap(Comparator<? super V> valueComparator) {
this.valueToKeysMap = new TreeMap<V, Collection<K>>(valueComparator);
}
public boolean containsValue(Object o) {
return valueToKeysMap.containsKey(o);
}
public V put(K k, V v) {
V oldV = null;
if (containsKey(k)) {
oldV = get(k);
valueToKeysMap.get(oldV).remove(k);
}
super.put(k, v);
if (!valueToKeysMap.containsKey(v)) {
Collection<K> keys = new ArrayList<K>();
keys.add(k);
valueToKeysMap.put(v, keys);
} else {
valueToKeysMap.get(v).add(k);
}
return oldV;
}
public void putAll(Map<? extends K, ? extends V> m) {
for (Map.Entry<? extends K, ? extends V> e : m.entrySet())
put(e.getKey(), e.getValue());
}
public V remove(Object k) {
V oldV = null;
if (containsKey(k)) {
oldV = get(k);
super.remove(k);
valueToKeysMap.get(oldV).remove(k);
}
return oldV;
}
public void clear() {
super.clear();
valueToKeysMap.clear();
}
public Set<K> keySet() {
LinkedHashSet<K> ret = new LinkedHashSet<K>(size());
for (V v : valueToKeysMap.keySet()) {
Collection<K> keys = valueToKeysMap.get(v);
ret.addAll(keys);
}
return ret;
}
public Set<Map.Entry<K, V>> entrySet() {
LinkedHashSet<Map.Entry<K, V>> ret = new LinkedHashSet<Map.Entry<K, V>>(size());
for (Collection<K> keys : valueToKeysMap.values()) {
for (final K k : keys) {
final V v = get(k);
ret.add(new Map.Entry<K,V>() {
public K getKey() {
return k;
}
public V getValue() {
return v;
}
public V setValue(V v) {
throw new UnsupportedOperationException();
}
});
}
}
return ret;
}
}
Simple way to sort any map in Java 8 and above
Map<String, Object> mapToSort = new HashMap<>();
List<Map.Entry<String, Object>> list = new LinkedList<>(mapToSort.entrySet());
Collections.sort(list, Comparator.comparing(o -> o.getValue().getAttribute()));
HashMap<String, Object> sortedMap = new LinkedHashMap<>();
for (Map.Entry<String, Object> map : list) {
sortedMap.put(map.getKey(), map.getValue());
}
if you are using Java 7 and below
Map<String, Object> mapToSort = new HashMap<>();
List<Map.Entry<String, Object>> list = new LinkedList<>(mapToSort.entrySet());
Collections.sort(list, new Comparator<Map.Entry<String, Object>>() {
#Override
public int compare(Map.Entry<String, Object> o1, Map.Entry<String, Object> o2) {
return o1.getValue().getAttribute().compareTo(o2.getValue().getAttribute());
}
});
HashMap<String, Object> sortedMap = new LinkedHashMap<>();
for (Map.Entry<String, Object> map : list) {
sortedMap.put(map.getKey(), map.getValue());
}
Depending on the context, using java.util.LinkedHashMap<T> which rememebers the order in which items are placed into the map. Otherwise, if you need to sort values based on their natural ordering, I would recommend maintaining a separate List which can be sorted via Collections.sort().
Afaik the most cleaner way is utilizing collections to sort map on value:
Map<String, Long> map = new HashMap<String, Long>();
// populate with data to sort on Value
// use datastructure designed for sorting
Queue queue = new PriorityQueue( map.size(), new MapComparable() );
queue.addAll( map.entrySet() );
// get a sorted map
LinkedHashMap<String, Long> linkedMap = new LinkedHashMap<String, Long>();
for (Map.Entry<String, Long> entry; (entry = queue.poll())!=null;) {
linkedMap.put(entry.getKey(), entry.getValue());
}
public static class MapComparable implements Comparator<Map.Entry<String, Long>>{
public int compare(Entry<String, Long> e1, Entry<String, Long> e2) {
return e1.getValue().compareTo(e2.getValue());
}
}
Since TreeMap<> does not work for values that can be equal, I used this:
private <K, V extends Comparable<? super V>> List<Entry<K, V>> sort(Map<K, V> map) {
List<Map.Entry<K, V>> list = new LinkedList<Map.Entry<K, V>>(map.entrySet());
Collections.sort(list, new Comparator<Map.Entry<K, V>>() {
public int compare(Map.Entry<K, V> o1, Map.Entry<K, V> o2) {
return o1.getValue().compareTo(o2.getValue());
}
});
return list;
}
You might want to put list in a LinkedHashMap, but if you're only going to iterate over it right away, that's superfluous...
This is just too complicated. Maps were not supposed to do such job as sorting them by Value. The easiest way is to create your own Class so it fits your requirement.
In example lower you are supposed to add TreeMap a comparator at place where * is. But by java API it gives comparator only keys, not values. All of examples stated here is based on 2 Maps. One Hash and one new Tree. Which is odd.
The example:
Map<Driver driver, Float time> map = new TreeMap<Driver driver, Float time>(*);
So change the map into a set this way:
ResultComparator rc = new ResultComparator();
Set<Results> set = new TreeSet<Results>(rc);
You will create class Results,
public class Results {
private Driver driver;
private Float time;
public Results(Driver driver, Float time) {
this.driver = driver;
this.time = time;
}
public Float getTime() {
return time;
}
public void setTime(Float time) {
this.time = time;
}
public Driver getDriver() {
return driver;
}
public void setDriver (Driver driver) {
this.driver = driver;
}
}
and the Comparator class:
public class ResultsComparator implements Comparator<Results> {
public int compare(Results t, Results t1) {
if (t.getTime() < t1.getTime()) {
return 1;
} else if (t.getTime() == t1.getTime()) {
return 0;
} else {
return -1;
}
}
}
This way you can easily add more dependencies.
And as the last point I'll add simple iterator:
Iterator it = set.iterator();
while (it.hasNext()) {
Results r = (Results)it.next();
System.out.println( r.getDriver().toString
//or whatever that is related to Driver class -getName() getSurname()
+ " "
+ r.getTime()
);
}
This could be achieved very easily with java 8
public static LinkedHashMap<Integer, String> sortByValue(HashMap<Integer, String> map) {
List<Map.Entry<Integer, String>> list = new ArrayList<>(map.entrySet());
list.sort(Map.Entry.comparingByValue());
LinkedHashMap<Integer, String> sortedMap = new LinkedHashMap<>();
list.forEach(e -> sortedMap.put(e.getKey(), e.getValue()));
return sortedMap;
}
Based on #devinmoore code, a map sorting methods using generics and supporting both ascending and descending ordering.
/**
* Sort a map by it's keys in ascending order.
*
* #return new instance of {#link LinkedHashMap} contained sorted entries of supplied map.
* #author Maxim Veksler
*/
public static <K, V> LinkedHashMap<K, V> sortMapByKey(final Map<K, V> map) {
return sortMapByKey(map, SortingOrder.ASCENDING);
}
/**
* Sort a map by it's values in ascending order.
*
* #return new instance of {#link LinkedHashMap} contained sorted entries of supplied map.
* #author Maxim Veksler
*/
public static <K, V> LinkedHashMap<K, V> sortMapByValue(final Map<K, V> map) {
return sortMapByValue(map, SortingOrder.ASCENDING);
}
/**
* Sort a map by it's keys.
*
* #param sortingOrder {#link SortingOrder} enum specifying requested sorting order.
* #return new instance of {#link LinkedHashMap} contained sorted entries of supplied map.
* #author Maxim Veksler
*/
public static <K, V> LinkedHashMap<K, V> sortMapByKey(final Map<K, V> map, final SortingOrder sortingOrder) {
Comparator<Map.Entry<K, V>> comparator = new Comparator<Entry<K,V>>() {
public int compare(Entry<K, V> o1, Entry<K, V> o2) {
return comparableCompare(o1.getKey(), o2.getKey(), sortingOrder);
}
};
return sortMap(map, comparator);
}
/**
* Sort a map by it's values.
*
* #param sortingOrder {#link SortingOrder} enum specifying requested sorting order.
* #return new instance of {#link LinkedHashMap} contained sorted entries of supplied map.
* #author Maxim Veksler
*/
public static <K, V> LinkedHashMap<K, V> sortMapByValue(final Map<K, V> map, final SortingOrder sortingOrder) {
Comparator<Map.Entry<K, V>> comparator = new Comparator<Entry<K,V>>() {
public int compare(Entry<K, V> o1, Entry<K, V> o2) {
return comparableCompare(o1.getValue(), o2.getValue(), sortingOrder);
}
};
return sortMap(map, comparator);
}
#SuppressWarnings("unchecked")
private static <T> int comparableCompare(T o1, T o2, SortingOrder sortingOrder) {
int compare = ((Comparable<T>)o1).compareTo(o2);
switch (sortingOrder) {
case ASCENDING:
return compare;
case DESCENDING:
return (-1) * compare;
}
return 0;
}
/**
* Sort a map by supplied comparator logic.
*
* #return new instance of {#link LinkedHashMap} contained sorted entries of supplied map.
* #author Maxim Veksler
*/
public static <K, V> LinkedHashMap<K, V> sortMap(final Map<K, V> map, final Comparator<Map.Entry<K, V>> comparator) {
// Convert the map into a list of key,value pairs.
List<Map.Entry<K, V>> mapEntries = new LinkedList<Map.Entry<K, V>>(map.entrySet());
// Sort the converted list according to supplied comparator.
Collections.sort(mapEntries, comparator);
// Build a new ordered map, containing the same entries as the old map.
LinkedHashMap<K, V> result = new LinkedHashMap<K, V>(map.size() + (map.size() / 20));
for(Map.Entry<K, V> entry : mapEntries) {
// We iterate on the mapEntries list which is sorted by the comparator putting new entries into
// the targeted result which is a sorted map.
result.put(entry.getKey(), entry.getValue());
}
return result;
}
/**
* Sorting order enum, specifying request result sort behavior.
* #author Maxim Veksler
*
*/
public static enum SortingOrder {
/**
* Resulting sort will be from smaller to biggest.
*/
ASCENDING,
/**
* Resulting sort will be from biggest to smallest.
*/
DESCENDING
}
Here is an OO solution (i.e., doesn't use static methods):
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.LinkedHashMap;
import java.util.List;
import java.util.Map;
public class SortableValueMap<K, V extends Comparable<V>>
extends LinkedHashMap<K, V> {
public SortableValueMap() { }
public SortableValueMap( Map<K, V> map ) {
super( map );
}
public void sortByValue() {
List<Map.Entry<K, V>> list = new LinkedList<Map.Entry<K, V>>( entrySet() );
Collections.sort( list, new Comparator<Map.Entry<K, V>>() {
public int compare( Map.Entry<K, V> entry1, Map.Entry<K, V> entry2 ) {
return entry1.getValue().compareTo( entry2.getValue() );
}
});
clear();
for( Map.Entry<K, V> entry : list ) {
put( entry.getKey(), entry.getValue() );
}
}
private static void print( String text, Map<String, Double> map ) {
System.out.println( text );
for( String key : map.keySet() ) {
System.out.println( "key/value: " + key + "/" + map.get( key ) );
}
}
public static void main( String[] args ) {
SortableValueMap<String, Double> map =
new SortableValueMap<String, Double>();
map.put( "A", 67.5 );
map.put( "B", 99.5 );
map.put( "C", 82.4 );
map.put( "D", 42.0 );
print( "Unsorted map", map );
map.sortByValue();
print( "Sorted map", map );
}
}
Hereby donated to the public domain.
Some simple changes in order to have a sorted map with pairs that have duplicate values. In the compare method (class ValueComparator) when values are equal do not return 0 but return the result of comparing the 2 keys. Keys are distinct in a map so you succeed to keep duplicate values (which are sorted by keys by the way). So the above example could be modified like this:
public int compare(Object a, Object b) {
if((Double)base.get(a) < (Double)base.get(b)) {
return 1;
} else if((Double)base.get(a) == (Double)base.get(b)) {
return ((String)a).compareTo((String)b);
} else {
return -1;
}
}
}

How to copy Key Value Elements from one Hash Table to the other HT

I am trying to copy first 5 key value pair from Hash Table 1 to Hash Table 2
and Then next 6 to 10th Key value pair from Hash Table 1 to Hash Table 3.
I am not able to get this going,, can any one has any hint please share it with me
Try something like this.
Hashtable h = new Hashtable();
Hashtable h1 = new Hashtable();
Hashtable h2 = new Hashtable();
Set s = h.keySet();
int i = 0;
for (Object key : s) {
if ( i++ < 3) {
h1.put(key, h.get(key));
} else {
h2.put(key, h.get(key));
}
}
Add generics etc. as appropriate.
First of all, you need to know that a Hashtable does not maintain the order in which elements are inserted, so the first 5 can be whatever.
If you want the insertion order to be preserved, use a LinkedHashMap (perhaps wrapping it in a Collections.synchronizedMap() if you want it to be thread-safe).
Besides, there's also a nice interface called NavigableMap, implemented by TreeMap, which offers methods like subMap(), but you need to know from which key to which key (instead of using indices). And this will also hold your entries sorted by the key.
If you can rely only on indices, aside from using a LinkedHashMap, I would suggest you go for a generic solution like the following:
<K, V> Collection<Map<K, V>> splitMap(Map<K, V> map, int count) {
if (count <= 0) {
throw new IllegalArgumentException("zero or negative count");
}
List<Map<K, V>> list = new ArrayList<Map<K, V>>();
Iterator<Entry<K, V>> entries = map.entrySet().iterator();
while (entries.hasNext()) {
list.add(subMap(entries, count));
}
return list;
}
<K, V> Map<K, V> subMap(Iterator<Entry<K, V>> iterator, int count) {
Map<K, V> map = new LinkedHashMap<K, V>();
for (int i = 0; i < count && iterator.hasNext(); i++) {
Entry<K, V> entry = iterator.next();
map.put(entry.getKey(), entry.getValue());
}
return map;
}
Then just call splitMap(yourMap, 5) in order to get the collection (actually a list) of maps containing 5 entries each (or less for the last one if there are not enough entries in the initial map).

Finding the first value greater than in a SortedMap

I'd like to know what is there a better way to find the first value greater than an inputted value in a large SortedMap instead of looping through all values in my example below. Or if SortedMap is a the best structure to use for this.
Could this be achieved using google-collections?
Thanks in advance
public class mapTest {
public static void main(String[] args) {
SortedMap<Double, Object> sortedMap = new TreeMap<Double, Object>();
sortedMap.put(30d, "lala");
sortedMap.put(10d, "foo");
sortedMap.put(25d, "bar");
System.out.println("result: " + findFirstValueGreaterThan(sortedMap, 28d));
}
public static Object findFirstValueGreaterThan(SortedMap<Double, Object> sortedMap, Double value) {
for (Entry<Double, Object> entry : sortedMap.entrySet()) {
if (entry.getKey() > value) {
// return first value with a key greater than the inputted value
return entry.getValue();
}
}
return null;
}
}
It's all in the docs:
ceilingKey(K key)
Returns the least key greater than or equal to the given key, or null if there is no such key.
So,
findFirstValueGreaterThan(sortedMap, 28d)
should be
sortedMap.ceilingKey(28d)
Pay attention at difference between "greater than" and "greater than or equal to", though.
This solution only requires SortedMap. Please note that tailMap typically doesn't create a new map, so it's fast.
public static <K extends Comparable<K>, V> V
findFirstValueGreaterThan(SortedMap<K, V> map, K value) {
Iterator<Entry<K, V>> it = map.tailMap(value).entrySet().iterator();
if (it.hasNext()) {
Entry<K, V> e = it.next();
if (e.getKey().compareTo(value) > 0) {
return e.getValue();
} else if (it.hasNext()) {
return it.next().getValue();
}
}
return null;
}

Checking contents of a map in another map in java

I need to write a small snippet of code where I need to check contents of a map (key value) if it exists in another map , remove it from the map
E.g
Map1:
1=>obj1
2=>obj21
3=>obj3
4=>obj4
Other map
Map2:
10=>obj10
20=>obj20
2=>obj2
30=>obj30
3=>obj3
The result of fun (Map1, Map2)
after it executes it has the following ouput
Map2:
10=>obj10
2=>obj2
20=>obj20
30=>obj30
Is iterating over the smaller map and checking contents (key, value) is iterating over the smaller map and checking the key and contents in the bigger map the most efficient way to go about it.
m1.entrySet().removeAll(m2.entrySet());
where m1 is the Map to be modified, and m2 is the map with the mappings that need to be removed from m1.
private static <K, V> void fun(Map<K, V> a, Map<K, V> b) {
Map<K, V> shortestMap = a.size() < b.size() ? a : b;
Map<K, V> longestMap = a.size() > b.size() ? a : b;
Set<Entry<K, V>> shortestMapEntries = shortestMap.entrySet();
Set<Entry<K, V>> longestMapEntries = longestMap.entrySet();
longestMapEntries.removeAll(shortestMapEntries);
}
See java.util.Collection
boolean removeAll(Collection<?> c)
private static <K, V> removeDuplicates(Map<K, V> map1, Map<K, V> map2) {
for (K key : map1.keySet()) {
V val1 = map1.get(key);
V val2 = map2.get(key);
if (val2 != null && val2.equals(val1)
map2.remove(key);
}
}

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