Consider this example which prints out some device type stats. ("DeviceType" is an enum with a dozenish values.)
Multiset<DeviceType> histogram = getDeviceStats();
for (DeviceType type : histogram.elementSet()) {
System.开发者_运维知识库out.println(type + ": " + histogram.count(type));
}
What's the simplest, most elegant way to print the distinct elements in the order of their frequency (most common type first)?
With a quick look at the Multiset
interface, there's no ready-made method for this, and none of Guava's Multiset
implementations (HashMultiset
, TreeMultiset
, etc) seem to automatically keep elements frequency-ordered either.
I just added this feature to Guava, see here for the Javadoc.
Edit: usage example of Multisets.copyHighestCountFirst()
as per the original question:
Multiset<DeviceType> histogram = getDeviceStats();
for (DeviceType type : Multisets.copyHighestCountFirst(histogram).elementSet()) {
System.out.println(type + ": " + histogram.count(type));
}
Here's a method that returns a List
of entries, sorted by frequency (UPDATE: used a flag to toggle ascending / descending order and used Guava's favorite toy: the Enum Singleton Pattern
, as found in Effective Java, Item 3 ):
private enum EntryComp implements Comparator<Multiset.Entry<?>>{
DESCENDING{
@Override
public int compare(final Entry<?> a, final Entry<?> b){
return Ints.compare(b.getCount(), a.getCount());
}
},
ASCENDING{
@Override
public int compare(final Entry<?> a, final Entry<?> b){
return Ints.compare(a.getCount(), b.getCount());
}
},
}
public static <E> List<Entry<E>> getEntriesSortedByFrequency(
final Multiset<E> ms, final boolean ascending){
final List<Entry<E>> entryList = Lists.newArrayList(ms.entrySet());
Collections.sort(entryList, ascending
? EntryComp.ASCENDING
: EntryComp.DESCENDING);
return entryList;
}
Test code:
final Multiset<String> ms =
HashMultiset.create(Arrays.asList(
"One",
"Two", "Two",
"Three", "Three", "Three",
"Four", "Four", "Four", "Four"
));
System.out.println("ascending:");
for(final Entry<String> entry : getEntriesSortedByFrequency(ms, true)){
System.out.println(MessageFormat.format("{0} ({1})",
entry.getElement(), entry.getCount()));
}
System.out.println("descending:");
for(final Entry<String> entry : getEntriesSortedByFrequency(ms, false)){
System.out.println(MessageFormat.format("{0} ({1})",
entry.getElement(), entry.getCount()));
}
Output:
ascending:
One (1)
Two (2)
Three (3)
Four (4)
descending:
Four (4)
Three (3)
Two (2)
One (1)
An Implementation using ForwardingMultiSet :
(EntryComp from seanizer's answer)
enum EntryComp implements Comparator<Multiset.Entry<?>> {
DESCENDING {
@Override
public int compare(final Entry<?> a, final Entry<?> b) {
return Ints.compare(b.getCount(), a.getCount());
}
},
ASCENDING {
@Override
public int compare(final Entry<?> a, final Entry<?> b) {
return Ints.compare(a.getCount(), b.getCount());
}
},
}
public class FreqSortMultiSet<E> extends ForwardingMultiset<E> {
Multiset<E> delegate;
EntryComp comp;
public FreqSortMultiSet(Multiset<E> delegate, boolean ascending) {
this.delegate = delegate;
if (ascending)
this.comp = EntryComp.ASCENDING;
else
this.comp = EntryComp.DESCENDING;
}
@Override
protected Multiset<E> delegate() {
return delegate;
}
@Override
public Set<Entry<E>> entrySet() {
TreeSet<Entry<E>> sortedEntrySet = new TreeSet<Entry<E>>(comp);
sortedEntrySet.addAll(delegate.entrySet());
return sortedEntrySet;
}
@Override
public Set<E> elementSet() {
Set<E> sortedEntrySet = new LinkedHashSet<E>();
for (Entry<E> en : entrySet())
sortedEntrySet.add(en.getElement());
return sortedEntrySet;
}
public static <E> FreqSortMultiSet<E> create(boolean ascending) {
return new FreqSortMultiSet<E>(HashMultiset.<E> create(), ascending);
}
/*
* For Testing
* public static void main(String[] args) {
Multiset<String> s = FreqSortMultiSet.create(false);
s.add("Hello");
s.add("Hello");
s.setCount("World", 3);
s.setCount("Bye", 5);
System.out.println(s.entrySet());
}*/
}
Since it is not yet implemented, I guess you can create a Map
with key=type and value=count. Then sort that map - see here
精彩评论