java 按元素频率的顺序迭代 Multiset 的最简单方法？
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Simplest way to iterate through a Multiset in the order of element frequency?
提问by Jonik
Consider this example which prints out some device type stats. ("DeviceType" is an enum with a dozenish values.)
考虑这个打印出一些设备类型统计信息的例子。（“DeviceType”是一个有十几个值的枚举。）
Multiset<DeviceType> histogram = getDeviceStats();
for (DeviceType type : histogram.elementSet()) {
System.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)?
按频率顺序打印不同元素的最简单、最优雅的方法是什么（最常见的类型在前）？
With a quick look at the Multisetinterface, there's no ready-made method for this, and none of Guava's Multisetimplementations (HashMultiset, TreeMultiset, etc) seem to automatically keep elements frequency-ordered either.
随着快速浏览一下Multiset界面，有一个为这个没有现成的方法，并没有番石榴的的Multiset实现（HashMultiset，TreeMultiset，等）似乎自动保持要素频率有序无论是。
回答by Louis Wasserman
I just added this feature to Guava, see herefor the Javadoc.
我刚刚将此功能添加到 Guava，请参阅此处获取 Javadoc。
Edit: usage example of Multisets.copyHighestCountFirst()as per the original question:
编辑：Multisets.copyHighestCountFirst()根据原始问题的使用示例：
Multiset<DeviceType> histogram = getDeviceStats();
for (DeviceType type : Multisets.copyHighestCountFirst(histogram).elementSet()) {
System.out.println(type + ": " + histogram.count(type));
}
回答by Sean Patrick Floyd
Here's a method that returns a Listof 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 ):
这是一个返回List条目的方法，按频率排序（更新：使用一个标志来切换升序/降序并使用番石榴最喜欢的玩具：the Enum Singleton Pattern，如Effective Java，第 3 项中所述）：
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:
测试代码：
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:
输出：
ascending:
One (1)
Two (2)
Three (3)
Four (4)
descending:
Four (4)
Three (3)
Two (2)
One (1)
升序：
一 (1)
二 (2)
三 (3)
四 (4)
降序：
四 (4)
三 (3)
二 (2)
一 (1)
回答by Emil
An Implementation using ForwardingMultiSet:
使用ForwardingMultiSet的实现：
(EntryCompfrom seanizer'sanswer)
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());
}*/
}
回答by Bozho
Since it is not yet implemented, I guess you can create a Mapwith key=type and value=count. Then sort that map - see here
由于它还没有实现，我猜你可以Map用 key=type 和 value=count创建一个。然后对该地图进行排序 - 请参见此处
