Sorting two List (String, Integer) in java based on Integer value

Question

I am working on a program where i am reading a file and extracting keywords and their count. Later i need to pick up word with topmost frequency and match them with a keyword.

I have stored all the keywords i have found in the file in String list. I wish to sort these on basis of frequency. So if at index 17 i have a word "stack" with value at index 17 in other integer list to be maximum, i wish to take them to position 1.

I can sort these using collections.sort but it does not take care of other list.

Here is my code :

while(m.find()) 
    {
        if(keyword.contains(m.group()))
            {
            keywordcount.set(keyword.indexOf(m.group()),keywordcount.get(keyword.indexOf(m.group()))+1);
            //System.out.println("*"+m.group()+":"+keywordcount.get(keyword.indexOf(m.group())));
            }
        else
            {
            keyword.add(m.group());
            int var=keyword.indexOf(m.group());
            //System.out.println(m.group()+":"+var);
            keywordcount.add(var, 1);
            }
        //System.out.println(keyword.size()+"#"+keywordcount.size());                       
    }
    for(int i=0;i<keyword.size();i++)
    {
        System.out.print(keyword.get(i)+ ":" +keywordcount.get(i)+" ");
    }

daveb · Accepted Answer · 2012-04-19 12:12:20Z

5

Typically, one would put both the String and the Integer into a single class, and sort a list of instances of that class.

E.g.

class StringCount implements Comparable<StringCount> {
    private final String string;
    private final int count;

    public StringCount(String string, int count) {
        this.string = string;
        this.count = count;
    }

    @Override
    public int compareTo(StringCount right) {
        return this.count < right.count ? -1
             : this.count > right.count ? 1
             : 0;
    }

    // implement equals and hashCode too
    // if a.compareTo(b) == 0, then a.equals(b) should return true.

}

Then, you can create a List<StringCount> and call Collections.sort(stringCountList).

Note, that this will put the StringCount instances with the lowest values first, so they come out in ascending order.

edited Apr 19, 2012 at 12:12

answered Apr 19, 2012 at 12:05

daveb

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2 Comments

Jakub Zaverka Over a year ago

The compareTo method's body can be reduced to return count - right.count;

daveb Over a year ago

No, that might overflow. -2 billion - 2 billion for example

Aurelien Ribon · Accepted Answer · 2012-04-19 12:22:13Z

1

final List<String> words = new ArrayList<>();
final Map<String, Integer> frequencies = new HashMap<>();

while (m.find())  {
    String word = ...extract the word from m...;

    if (!words.contains(word)) words.add(word);

    if (!frequencies.contains(word)) frequencies.put(word, 1);
    else frequencies.put(word, frequencies.get(word) + 1);
}

Collections.sort(words, new Comparator<String>() {
    @Override public int compare(String s1, String s2) {
        int f1 = frequencies.get(s1);
        int f2 = frequencies.get(s2);
        if (f1 < f2) return 1;
        if (f1 > f2) return -1;
        return 0;
    }
});

edited Apr 19, 2012 at 12:22

answered Apr 19, 2012 at 12:14

Aurelien Ribon

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10 Comments

CodeMonkey Over a year ago

i was using hashmaps earlier.. but did the sort u suggesting keeps the unique elements ?

daveb Over a year ago

This implementation will probably fit your use case better - however, I would suggest using MutableInt from apache commons lang, because this will avoid a lot of boxing & unboxing in the counting - because you can increment the count rather than putting a new value in for the count.

Aurelien Ribon Over a year ago

What do you mean by "the unique elements"? Every time a keyword is found, it is saved in the list if not already present, and its frequency is increased. Finally, the list is sorted in reverse order based on the frequencies. The list "words" will only contains unique keywords, and will be sorted based on their frequency, in descending order.

Aurelien Ribon Over a year ago

By the way, I agree with @daveb. If you don't want to use the whole apache commons lang library for this, just quickly create a MutableInt class, it's trivial for this use case.

CodeMonkey Over a year ago

@AurélienRibon i tried using your code... i think if (f1<f2) i need to swap the list ? because my code is giving me almost same results even now

|

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

This might be the ideal moment to check into multisets.

A collection that supports order-independent equality, like Set, but may have duplicate elements. A multiset is also sometimes called a bag.

Elements of a multiset that are equal to one another are referred to as occurrences of the same single element. The total number of occurrences of an element in a multiset is called the count of that element (the terms "frequency" and "multiplicity" are equivalent, but not used in this API). Since the count of an element is represented as an int, a multiset may never contain more than Integer.MAX_VALUE occurrences of any one element.

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Apr 19, 2012 at 12:44

Qkyrie

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