I'm sorry if this question was already asked, but i don't fine an answer to my question.
I'm working on HashMap i put two values (7,"value test 1") (7,"value test 2)
According to the specification java API HashMap put the first values is replaced by the second .
My question is when comes the resolution of collision ? why my second value is not store in linkedList or store in another place in the hashMap ? Is it due to the equals or hascode method ??
Best Regards
This has nothing to do with hash collision. Hash collisions (ie., keys with the same hashcode()) are handled correctly by the HashMap. In your example, both keys are equal (ie., 7.equals(7) == true), so the old value is replaced.
In the following example
Map<Integer, String> map = new HashMap<>();
map.put(7, "value 1");
map.put(7, "value 2");
System.out.println(map.get(7));
what would you expect in the last line to happen?
Maybe you are looking for a multimap?
Apache Commons
Guava
Stackoverflow
Collision handling takes place if two different keys resolve to the same bucket in the hashmap. In that case the second entry would be put into the linked list.
In your case you replace the entry for the same key (7) and thus there is no collision.
If you need a map that contains multiple values per key either use a Map<key, Set<value>> (you can also use a List etc. instead of a Set) and handle adding/removing to that set yourself or use Apache Commons' MultiMap or Google Guava's Multimap, e.g. HashMultimap.
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I am new to the collections in Java , I have a HashMap<String, List<String>>. I want to search for a key from the Map when a value is given.
The map stores the data of a state as key and its cities as the list. So it is assumed that there is no duplicate key or value.
Some of the previous answers pointed out the solutions for many:many and one:one relations of key-value in map, but I don't understand how to check for value in the List.
Do I iterate over the whole map and for each key get the list and search in the list? or is there any other way of doing this?
Please suggest some method. Thank you!
You should make inverted map (city -> state map).
public Map<String, String> invertedMap(Map<String, List<String>> map) {
Map<String, String> inverted = new HashMap<>();
for (Entry<String, List<String>> e : map.entrySet())
for (String city : e.getValue())
inverted.put(city, e.getKey() /* state */);
return inverted;
}
A map is meant to allow fast access to a value by using a key. Doing it the other way round would require you to iterate over all values and look for that. Additionally you need to be aware that the same (or an equal) value could be stored for multiple keys.
In order to efficiently search for all the states for a given city name you could employ an inverse map where the city name is the key and the value is a collection of states (assuming there are several cities each in a different state - e.g. there seem to be multiple Springfield in the US).
An easy way to create such an inverted map would be to use Guava's Multimap for the initial map (state -> cities) and then use Multimaps.invertFrom(intialMap);.
Edit: In reference to Brett Walker's comment, Apache Commons Collections' BidiMap seems to follow a similar approach, i.e. the AbstractDualBidiMap implementation internally uses two maps as described above.
I would use the BidiMap from the Apache Common Collection library as a starting point.
For the List<String> part I would look at using the MultiValueMap as it decorates another map, allowing it to have more than one value for a key.
I've used the BidiMap bit not the MultiValueMap. I'm thinking that both would be useful.
I have the following key-value system (HashMap) , where String would be a key like this "2014/12/06".
LinkedHashMap<String, Value>
So, I can retrieve an item knowing the key, but what I'm looking for is a method to retrieve a list of the value which key matches partialy, I mean, how could I retrieve all the values of 2014?.
I would like to avoid solutions like, test every item in the list, brute-force, or similar.
thanks.
Apart from doing the brute-force solution of iterating over all the keys, I can think of two options :
Use a TreeMap, in which the keys are sorted, so you can find the first key that is >= "2014/01/01" (using map.getCeilingEntry("2014/01/01")) and go over all the keys from there.
Use a hierarchy of Maps - i.e. Map<String,Map<String,Value>>. The key in the outer Map would be the year. The key in the inner map would be the full date.
Not possible with LinkedHashMap only. If you can copy the keys to an ordered list you can perform a binary search on that and then do a LinkedHashMap.get(...) with the full key(s).
If you're only ever going to want to retrieve items using the first part of the key, then you want a TreeMap rather than a LinkedHashMap. A LinkedHashMap is sorted according to insertion order, which is no use for this, but a TreeMap is sorted according to natural ordering, or to a Comparator that you supply. This means that you can find the first entry that starts with 2014 efficiently (in log time), and then iterate through until you get to the first one that doesn't match.
If you want to be able to match on any part of the key, then you need a totally different solution, way beyond a simple Map. You'd need to look into full text searching and indexing. You could try something like Lucene.
You could refine a hash function for your values so that values with similar year would hash around similar prefixed hashes. That wouldn't be efficient (probably poor distribution of hashes) nor to the spirit of HashMaps. Use other map implementations such as TreeMaps that keep an order of your choice.
I have this very simple piece of code, and I was just trying to play a bit with different kind of objects inside a Map.
//There's a bit of spanish, sorry about that
//just think 'persona1' as an object with
//a string and an int
Map mapa = new HashMap();
mapa.put('c', 12850);
mapa.put(38.6, 386540);
mapa.put("Andrés", 238761);
mapa.put(14, "Valor de 14");
mapa.put("p1", persona1);
mapa.put("Andrea", 34500);
System.out.println(mapa.toString());
And then I expect from console something like:
{c=12850, 38.6=386540, Andrés=238761, 14=Valor de 14, p1={nombre: Andres Perea, edad: 10}, Andrea=34500}
But susprisingly for me I got same data in different order:
{38.6=386540, Andrés=238761, c=12850, p1={nombre: Andres Perea, edad: 10}, Andrea=34500, 14=Valor de 14}
It doesn't matter if I try other kind of objects, even just Strings or numeric types, it always does the same, it makes a different without-apparently-any-sense order.
Can someone give me a hint why this happens? Or may be something too obvious I'm missing?
I'm using Java 1.7 and Eclipse Juno.
As per Oracle's documentation
The HashMap class is roughly equivalent to Hashtable, except that it is unsynchronized and permits nulls. This class makes no guarantees as to the order of the map; in particular, it does not guarantee that the order will remain constant over time.
Refer to HashMap JavaDocs.
There are 3 class which implements map interface in java.
1. hashMap: Id does not guarantee any order.
2. Linked HashMap:It will store them in insertion order.
3. TreeMap: It will store in ascending order.(ASCII value)
So As per your requirement you can use Linked HashMap instead of HashMap.so instead of writing
Map mapa = new HashMap();
create object of Linked HashMap
Map mapa = new LinkedHashMap();
follow below link for more info.
http://docs.oracle.com/javase/tutorial/collections/interfaces/map.html
HashMap not guaranteed the order of element. If you want to keep order use LinkedHashMap.
See following case
Map<Integer,String> unOrderedMap=new HashMap<>();
unOrderedMap.put(1,"a");
unOrderedMap.put(3,"a");
unOrderedMap.put(2,"a");
System.out.println("HashMap output: "+unOrderedMap.toString());
Map<Integer,String> orderedMap=new LinkedHashMap<>();
orderedMap.put(1,"a");
orderedMap.put(3,"a");
orderedMap.put(2,"a");
System.out.println("LinkedHashMap output: "+orderedMap.toString());
Output:
HashMap output: {1=a, 2=a, 3=a}
LinkedHashMap output: {1=a, 3=a, 2=a}
Maps does not maintain the order the order in which elements were added, List will maintain the order of elements
"The order of a map is defined as the order in which the iterators on the map's collection views return their elements. Some map implementations, like the TreeMap class, make specific guarantees as to their order; others, like the HashMap class, do not."
This is how a hashmap works: (citing from another source)
It has a number of "buckets" which it uses to store key-value pairs in. Each bucket has a unique number - that's what identifies the bucket. When you put a key-value pair into the map, the hashmap will look at the hash code of the key, and store the pair in the bucket of which the identifier is the hash code of the key. For example: The hash code of the key is 235 -> the pair is stored in bucket number 235. (Note that one bucket can store more then one key-value pair).
When you lookup a value in the hashmap, by giving it a key, it will first look at the hash code of the key that you gave. The hashmap will then look into the corresponding bucket, and then it will compare the key that you gave with the keys of all pairs in the bucket, by comparing them with equals().
Now you can see how this is very efficient for looking up key-value pairs in a map: by the hash code of the key the hashmap immediately knows in which bucket to look, so that it only has to test against what's in that bucket.
Looking at the above mechanism, you can also see what requirements are necessary on the hashCode() and equals() methods of keys:
If two keys are the same (equals() returns true when you compare them), their hashCode() method must return the same number. If keys violate this, then keys that are equal might be stored in different buckets, and the hashmap would not be able to find key-value pairs (because it's going to look in the same bucket).
If two keys are different, then it doesn't matter if their hash codes are the same or not. They will be stored in the same bucket if their hash codes are the same, and in this case, the hashmap will use equals() to tell them apart.
Now, when you put all your "key-value" pairs in the hashmap, and print them, it prints them in some random order of the keys which got generated by hashing the value you supplied for keys.
If your requirement is still to maintain the ordering, you can use the LinkedHashMap in Java.
Hope this helps :-)
Edit: Original Post: How does a Java HashMap handle different objects with the same hash code?
I have a text file which looks like this:
code appearance
----------------
j4t8 1
fj89 3
pf6n 1
j4t8 5
And I want to sort by the codes which appear the most. As you can see (and since I want to perform a group by) there are duplicate codes, so using HashMap would be a problem (duplicate keys). Any ideas?
don't know if this is the best solution but you could create a map of a list like this:
Map<String, List<Integer>> map = new HahsMap<String, List<Integer>>();
if(map.contains.(key))
{
map.get(key).add(new_appearance_value);
}
else
{
List<Integer> app = new ArrayList<Integer>();
app.add(new_appearance_value);
map.put(key, app);
}
Where the map key would be the code and the values of appearance would go into the list.
Note: to determine which code has more appearances just check for the size of the list of each code.
You can use
HashMap map = new HashMap<String, List<Integer>>();
The appearances will be stored in a list associated with every code.
Then given a code you just retrieve the list of integers and iterate over it.
You need a Collection of Pair objects. Each pair holds the code and the appearance. You then sort the collection using a Comparator, which only compares the appearance in each Pair object, and disregards the code.
The Commons Collections MultiValueMap can be used to decorate another map, allowing it to have more than one value for a key.
I am have the Hashmap like this,
HashMap<String,String> epnSource = new HashMap<String, String>();
Now I have added the keys/values like this,
epnSource.put("10.3.2.227","EPN1");
epnSource.put("10.3.2.227","EPN2");
epnSource.put("10.3.2.166","EPN3");
epnSource.put("10.3.2.166","EPN4");
epnSource.put("10.3.2.161","EPN5");
I am trying to do every time before adding a value, I want to check number of occurrences of a key present in the HashMap. Suppose if key 10.3.2.227 has more than two occurrences I shouldn't added it and go for new one. Any suggestions will be helpful.
Suppose if value 10.3.2.227 has more than two occurrences ...
It won't. The way that you have implemented it, the "10.3.2.227" is a key of the Map, and a given key cannot appear more than once in a Map.
If you want a given key (e.g. "10.3.2.227") to map to multiple values (e.g. "EPN1" and "EPN1"), you need to use either a Map<String,Set<String>> or a MultiMap class from the Apache or Google/Guava collections libraries.
If the map previously contained a mapping for the key, the old value is replaced.
It is not possible duplicate key in HashMap.