How to count and sort repeating strings

Question

i have got a string of names

String str = "A. Walker, L. Gordon, C. Riley, L. Gordon";

I need to count name occurancies and sort the occurancies from biggest to lowest.

I have done the countung part, but I also need to sort it.

String[] array = str.split(", ");        

List asList = Arrays.asList(array);
Set<String> mySet = new HashSet<String>(asList);
for(String s: mySet)
    System.out.println(s + " " +Collections.frequency(asList,s));

Output should look like this

L. Gordon 2, A. Walker 1, C. Riley 1

Use LinkedHashSet instead of HashSet to preserve insertion order, and sort asList before you pass it to the set. (Note, use List<String> asList rather than List asList). — Andy Turner
– Andy Turner, Commented Apr 7, 2017 at 15:14
And one way you can sort the list is with java.util.Collections.sort( List<T> list ). If you need a special order, you can provide a Comparator<T> via similar methods in Collections and List. — Andy Thomas
– Andy Thomas, Commented Apr 7, 2017 at 15:14
Note that if you've sorted the list, you don't need to put them into a set... — Andy Turner
– Andy Turner, Commented Apr 7, 2017 at 15:15
If external libraries are allowed, Guava's TreeMultiset is the perfect ADT for this. — Jacob G.
– Jacob G., Commented Apr 7, 2017 at 15:24

yavuzkavus · Accepted Answer · 2017-04-08 14:52:09Z

4

You can do something like this:

public class Test {
    static class NameFreq {
        public NameFreq(String name, int freq) {
            this.name = name;
            this.freq = freq;
        }
        String name;
        int freq;
        @Override
        public String toString() {
            return name + " " + freq;
        }
    }
    public static void main(String[] args) throws Exception {
        String str = "A. Walker, L. Gordon, C. Riley, L. Gordon";
        Map<String, NameFreq> map = new HashMap<>();
        String[] array = str.split("\\s*,\\s*");
        for(String name : array) {
            NameFreq nameFreq = map.get(name);
            if( nameFreq==null )
                map.put(name, new NameFreq(name, 1));
            else
                nameFreq.freq++;
        }

        List<NameFreq> list = new ArrayList<>(map.values());
        Collections.sort(list, new Comparator<NameFreq>() {
            @Override
            public int compare(NameFreq o1, NameFreq o2) {
                return Integer.compare(o2.freq, o1.freq);
            }
        });

        System.out.println(list);
        //output: [L. Gordon 2, A. Walker 1, C. Riley 1]
    }
}

edited Apr 8, 2017 at 14:52

answered Apr 7, 2017 at 15:21

yavuzkavus

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3 Comments

Andreas Over a year ago

new Integer(o2.freq).compareTo(o1.freq) is unnecessarily boxing the int values. Use Integer.compare(o2.freq, o1.freq) instead.

Andreas Over a year ago

Although question doesn't specify a desired ordering of names with equal count, it would be common to require a secondary sort by name. This can be done by either changing the HashMap to a TreeMap, or by enhancing the Comparator with secondary comparison logic. For performance, the second option would be better.

yavuzkavus Over a year ago

@Andreas is right, Integer.compare(i1, i2) method is better than new Integer(i1).compareTo(i2)

Darshan Mehta · Accepted Answer · 2017-04-07 16:04:48Z

3

You can easily do it with stream, e.g.:

String str = "A. Walker, L. Gordon, C. Riley, L. Gordon";
TreeMap<String,Long> data = Arrays.stream(str.split(","))
    .map(s -> s.trim())
    .collect(Collectors.groupingBy(Function.identity(), TreeMap::new, Collectors.counting()));

LinkedHashMap<String,Long> resultMap= data.entrySet().stream()

.sorted(Map.Entry.comparingByValue().reversed()) .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new)); System.out.println(resultMap);

If you want to strip out curly braces from the beginning and end of string, you can use substring, e.g.:

String result = resultMap.toString();
if(result.length > 2){
   result = result.substring(1, result.length() - 1);
}
System.out.println(result);

edited Apr 7, 2017 at 16:04

answered Apr 7, 2017 at 15:20

Darshan Mehta

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3 Comments

Ashutosh Jha Over a year ago

Nice use of groupingBy.

Andreas Over a year ago

OP wants result sorted descending by count: "sort the occurancies from biggest to lowest"

Andreas Over a year ago

@DarshanMehta descendingMap() is descending by key (name), not by value (count). See my answer for sorting descending by count, and for building result string correctly (your current result is A. Walker=1, C. Riley=1, L. Gordon=2, not the requested L. Gordon 2, A. Walker 1, C. Riley 1).

Andreas · Accepted Answer · 2017-04-07 15:33:31Z

2

First, create a Map, keyed by name, with value being a count for that name. Then sort that descending by the value, secondary sort by key (aka name).

It seems you want result as a comma-separated string, so finally combine result that way.

Using Java 8 Streams, it can be done in a single method chain:

String str = "A. Walker, L. Gordon, C. Riley, L. Gordon";
String res = Pattern.compile(", *")
                    .splitAsStream(str)
                    .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()))
                    .entrySet()
                    .stream()
                    .sorted(Comparator.<Entry<String, Long>, Long>comparing(Entry::getValue)
                                      .reversed()
                                      .thenComparing(Entry::getKey))
                    .map(e -> e.getKey() + " " + e.getValue())
                    .collect(Collectors.joining(", "));
System.out.println(res); // prints: L. Gordon 2, A. Walker 1, C. Riley 1

Notice the use of splitAsStream(), so the result of the split doesn't have to be stored in an intermediate array.

answered Apr 7, 2017 at 15:33

Andreas

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1 Comment

user_3380739 Over a year ago

Isn't the code too complicate for this simple question?

Ashutosh Jha · Accepted Answer · 2017-04-07 15:19:27Z

1

Use trie to count frequency, it saves a lot of space. And use heap to sort them. While pushing in trie you can count number of distinct words. Create heap of that size, max heap if you want to sort in ascending order.

answered Apr 7, 2017 at 15:19

Ashutosh Jha

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1 Comment

Ashutosh Jha Over a year ago

If you want code as well, let me know. I will write it and post.

user_3380739 · Accepted Answer · 2023-01-11 16:25:46Z

0

Here is the simple solution by abacus-common

Stream.of(str.split(", "))
      .toMultiset()
      .toMapSortedByOccurrences(Comparators.reverseOrder());

edited Jan 11, 2023 at 16:25

answered Apr 7, 2017 at 21:41

user_3380739

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