How to update a value, given a key in a hashmap?

JavaKeyHashmap

Java Problem Overview


Suppose we have a HashMap<String, Integer> in Java.

How do I update (increment) the integer-value of the string-key for each existence of the string I find?

One could remove and reenter the pair, but overhead would be a concern.
Another way would be to just put the new pair and the old one would be replaced.

In the latter case, what happens if there is a hashcode collision with a new key I am trying to insert? The correct behavior for a hashtable would be to assign a different place for it, or make a list out of it in the current bucket.

Java Solutions


Solution 1 - Java

map.put(key, map.get(key) + 1);

should be fine. It will update the value for the existing mapping. Note that this uses auto-boxing. With the help of map.get(key) we get the value of corresponding key, then you can update with your requirement. Here I am updating to increment value by 1.

Solution 2 - Java

Java 8 way:

You can use computeIfPresent method and supply it a mapping function, which will be called to compute a new value based on existing one.

For example,

Map<String, Integer> words = new HashMap<>();
words.put("hello", 3);
words.put("world", 4);
words.computeIfPresent("hello", (k, v) -> v + 1);
System.out.println(words.get("hello"));

Alternatevely, you could use merge method, where 1 is the default value and function increments existing value by 1:

words.merge("hello", 1, Integer::sum);

In addition, there is a bunch of other useful methods, such as putIfAbsent, getOrDefault, forEach, etc.

Solution 3 - Java

The simplified Java 8 way:

map.put(key, map.getOrDefault(key, 0) + 1);

This uses the method of HashMap that retrieves the value for a key, but if the key can't be retrieved it returns the specified default value (in this case a '0').

This is supported within core Java: [HashMap<K,V> getOrDefault(Object key, V defaultValue)][1]

[1]: https://docs.oracle.com/javase/8/docs/api/java/util/HashMap.html#getOrDefault-java.lang.Object-V- "HashMap<K,V> getOrDefault(Object key, V defaultValue)"

Solution 4 - Java

hashmap.put(key, hashmap.get(key) + 1);

The method put will replace the value of an existing key and will create it if doesn't exist.

Solution 5 - Java

Replace Integer by AtomicInteger and call one of the incrementAndGet/getAndIncrement methods on it.

An alternative is to wrap an int in your own MutableInteger class which has an increment() method, you only have a threadsafety concern to solve yet.

Solution 6 - Java

One line solution:

map.put(key, map.containsKey(key) ? map.get(key) + 1 : 1);

Solution 7 - Java

@Matthew's solution is the simplest and will perform well enough in most cases.

If you need high performance, AtomicInteger is a better solution ala @BalusC.

However, a faster solution (provided thread safety is not an issue) is to use TObjectIntHashMap which provides a increment(key) method and uses primitives and less objects than creating AtomicIntegers. e.g.

TObjectIntHashMap<String> map = new TObjectIntHashMap<String>()
map.increment("aaa");

Solution 8 - Java

You can increment like below but you need to check for existence so that a NullPointerException is not thrown

if(!map.containsKey(key)) {
 p.put(key,1);
}
else {
 p.put(key, map.getKey()+1);
}

Solution 9 - Java

Does the hash exist (with 0 as the value) or is it "put" to the map on the first increment? If it is "put" on the first increment, the code should look like:

if (hashmap.containsKey(key)) {
    hashmap.put(key, hashmap.get(key)+1);
} else { 
    hashmap.put(key,1);
}

Solution 10 - Java

It may be little late but here are my two cents.

If you are using Java 8 then you can make use of computeIfPresent method. If the value for the specified key is present and non-null then it attempts to compute a new mapping given the key and its current mapped value.

final Map<String,Integer> map1 = new HashMap<>();
map1.put("A",0);
map1.put("B",0);
map1.computeIfPresent("B",(k,v)->v+1);  //[A=0, B=1]

We can also make use of another method putIfAbsent to put a key. If the specified key is not already associated with a value (or is mapped to null) then this method associates it with the given value and returns null, else returns the current value.

In case the map is shared across threads then we can make use of ConcurrentHashMap and AtomicInteger. From the doc:

> An AtomicInteger is an int value that may be updated atomically. An > AtomicInteger is used in applications such as atomically incremented > counters, and cannot be used as a replacement for an Integer. However, > this class does extend Number to allow uniform access by tools and > utilities that deal with numerically-based classes.

We can use them as shown:

final Map<String,AtomicInteger> map2 = new ConcurrentHashMap<>();
map2.putIfAbsent("A",new AtomicInteger(0));
map2.putIfAbsent("B",new AtomicInteger(0)); //[A=0, B=0]
map2.get("B").incrementAndGet();    //[A=0, B=1]

One point to observe is we are invoking get to get the value for key B and then invoking incrementAndGet() on its value which is of course AtomicInteger. We can optimize it as the method putIfAbsent returns the value for the key if already present:

map2.putIfAbsent("B",new AtomicInteger(0)).incrementAndGet();//[A=0, B=2]

On a side note if we plan to use AtomicLong then as per documentation under high contention expected throughput of LongAdder is significantly higher, at the expense of higher space consumption. Also check this question.

Solution 11 - Java

The cleaner solution without NullPointerException is:

map.replace(key, map.get(key) + 1);

Solution 12 - Java

Since I can't comment to a few answers due to less reputation, I will post a solution which I applied.

for(String key : someArray)
{
   if(hashMap.containsKey(key)//will check if a particular key exist or not 
   {
      hashMap.put(hashMap.get(key),value+1);// increment the value by 1 to an already existing key
   }
   else
   {
      hashMap.put(key,value);// make a new entry into the hashmap
   }
}

Solution 13 - Java

Use a for loop to increment the index:

for (int i =0; i<5; i++){
    HashMap<String, Integer> map = new HashMap<String, Integer>();
    map.put("beer", 100);

    int beer = map.get("beer")+i;
    System.out.println("beer " + beer);
    System.out ....

}

Solution 14 - Java

There are misleading answers to this question here that imply Hashtable put method will replace the existing value if the key exists, this is not true for Hashtable but rather for HashMap. See Javadoc for HashMap http://docs.oracle.com/javase/7/docs/api/java/util/HashMap.html#put%28K,%20V%29

Solution 15 - Java

Integer i = map.get(key);
if(i == null)
   i = (aValue)
map.put(key, i + 1);

or

Integer i = map.get(key);
map.put(key, i == null ? newValue : i + 1);

Integer is Primitive data types http://cs.fit.edu/~ryan/java/language/java-data.html, so you need to take it out, make some process, then put it back. if you have a value which is not Primitive data types, you only need to take it out, process it, no need to put it back into the hashmap.

Solution 16 - Java

Use Java8 built in fuction 'computeIfPresent'

Example:

public class ExampleToUpdateMapValue {

    public static void main(String[] args) {
        Map<String,String> bookAuthors = new TreeMap<>();
        bookAuthors.put("Genesis","Moses");
        bookAuthors.put("Joshua","Joshua");
        bookAuthors.put("Judges","Samuel");

        System.out.println("---------------------Before----------------------");
        bookAuthors.entrySet().stream().forEach(System.out::println);
        // To update the existing value using Java 8
        bookAuthors.computeIfPresent("Judges", (k,v) -> v = "Samuel/Nathan/Gad");

        System.out.println("---------------------After----------------------");
        bookAuthors.entrySet().stream().forEach(System.out::println);
    }
}

Solution 17 - Java

Try:

HashMap hm=new HashMap<String ,Double >();

NOTE:

String->give the new value; //THIS IS THE KEY
else
Double->pass new value; //THIS IS THE VALUE

You can change either the key or the value in your hashmap, but you can't change both at the same time.

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