How to sort a Map in Java by its String keys which are numeric

Question

I have created a map called result.

In the sortByKeys method as my keys are String with Numeric values, I have converted them to Integer key type Map then sorted them.

The sorting is working fine when I am looping and printing individually, but not when I am setting them in another Map.

public class TestDate {
    public static void main (String args[]){

    Map<String, String> result = new HashMap<String, String>();

        result.put("error", "10");
        result.put("1","hii");
        result.put("Update","herii");
        result.put("insert","insert");
        result.put("10","hiiuu");
        result.put("7","hii");
        result.put("21","hii");
        result.put("15","hii"); 

        Map<String, String> sorted = sortByKeys(result);
        //System.out.println(sorted);   
    }

    private static Map<String, String> sortByKeys(Map<String, String> map) {
        Map <Integer,String> unSorted = new  HashMap<Integer, String>();
        Map <String,String> sorted = new  HashMap<String, String>();
        for (Map.Entry<String, String> entry : map.entrySet())
        {
            try{
                int foo = Integer.parseInt(entry.getKey());            
                unSorted.put(foo, entry.getValue());                
            }catch (Exception e){

            }
        }
        Map<Integer, String> newMap = new TreeMap<Integer, String>(unSorted); 
        Set set = newMap.entrySet();
        Iterator iterator = set.iterator();
        while(iterator.hasNext()) {
            Map.Entry me = (Map.Entry)iterator.next();
            System.out.println(me.getKey());
            System.out.println(me.getValue());
            sorted.put(me.getKey().toString(),  me.getValue().toString());

       }
        System.out.println(sorted);

        return null;
    }   
}

Here is the o/p :

1
hii
7
hii
10
hiiuu
15
hii
21
hii
{21=hii, 10=hiiuu, 1=hii, 7=hii, 15=hii}

Marko Topolnik · Accepted Answer · 2014-08-27 09:15:40Z

8

If you don't need the last inch of performance, you can solve this rather directly, without an extra step to sort the map, by using SortedMap:

Map<String,String> result = new TreeMap<>(Comparator.comparingInt(Integer::parseInt));

If you are among the unfortunate bunch who are still being denied access to Java 8, you'll have to implement the Comparator in long-hand:

new TreeMap<>(new Comparator<String,String> { public int compare(String a, String b) {
  return Integer.compare(Integer.parseInt(a), Integer.parseInt(b));
}});

The above approach works only under the assumption that all keys are parseable integers. If that is not the case, then you won't be able to use the SortedMap directly, but transform your original map into it, filtering out the unparseable keys.

edited Aug 27, 2014 at 9:15

answered Aug 27, 2014 at 7:11

Marko Topolnik

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10 Comments

curiousguy Over a year ago

I am not sure how to use it .

Marko Topolnik Over a year ago

You just put your stuff into it, and when you want the sorted keys, just use result.keySet() and it will be in order.

curiousguy Over a year ago

Ok , But i am getting an error here Integer::parseInt , Integer cannot be resolved to a variable .

Marko Topolnik Over a year ago

You are not on Java 8, then. In that case you cannot use the method reference and need to write out the full implementation of Comparator as an anonymous class instance.

chiastic-security Over a year ago

But OP is putting in some non-integer Strings as keys (into result, though not into the second Map he was creating). Surely this will give exceptions in the Comparator when it tries to call Integer.parseInt() on something that isn't parseable?

|

chiastic-security · Accepted Answer · 2014-08-27 07:13:28Z

2

It's because the Map you're putting them into is a HashMap, which isn't sorted. There's no guarantee of the ordering of results you'll get out of the HashMap, even if you put them in in the right order.

(And calling it sorted doesn't change anything :) )

answered Aug 27, 2014 at 7:13

chiastic-security

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Comments

icza · Accepted Answer · 2014-08-27 07:13:04Z

1

You print 2 different maps and not the same: you iterate over and print the entries of newMap map, and at the end you print sorted map.

You see the sorted entries printed because you iterate over your sorted newMap.

Then you print the sorted map which is unsorted (despite by its name). You print a different map instance.

Print this:

System.out.println(newMap); // This is the instance of the sorted "TreeMap"

answered Aug 27, 2014 at 7:13

icza

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3 Comments

curiousguy Over a year ago

Yes that is working . But I want the sorted Map (newMap) in <String,String> format . Which is now <Integer,String> . So I tried to loop through .

icza Over a year ago

But sorted is a HashMap which is not even sorted, but even if you would use TreeMap<String, String> it cannot do what you want because the lexicographical order of Strings is different from the natural order of numbers even if the Strings contain numbers: "2" is greater than "11" even though 2 is less than 11.

curiousguy Over a year ago

Yes I can see the issue.

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