如果我有一个Guava Multimap,如何根据给定键的值数目对条目进行排序?
例如:
Multimap<String, String> multiMap = ArrayListMultimap.create();
multiMap.put("foo", "1");
multiMap.put("bar", "2");
multiMap.put("bar", "3");
multiMap.put("bar", "99");
鉴于此,当遍历multiMap时,我将如何使“ bar”条目优先出现(因为“ bar”具有3个值,而“ foo”只有1个值?]
提取列表中的条目,然后对列表进行排序:
List<Map.Entry<String, String>> entries = new ArrayList<Map.Entry<String, String>>(map.entries());
Collections.sort(entries, new Comparator<Map.Entry<String, String>>() {
@Override
public int compare(Map.Entry<String, String> e1, Map.Entry<String, String> e2) {
return Ints.compare(map.get(e2.getKey()).size(), map.get(e1.getKey()).size());
}
});
然后遍历条目。
编辑:
如果您实际上想要遍历内部映射(Entry<String, Collection<String>>
)的条目,请执行以下操作:
List<Map.Entry<String, Collection<String>>> entries =
new ArrayList<Map.Entry<String, Collection<String>>>(map.asMap().entrySet());
Collections.sort(entries, new Comparator<Map.Entry<String, Collection<String>>>() {
@Override
public int compare(Map.Entry<String, Collection<String>> e1,
Map.Entry<String, Collection<String>> e2) {
return Ints.compare(e2.getValue().size(), e1.getValue().size());
}
});
// and now iterate
for (Map.Entry<String, Collection<String>> entry : entries) {
System.out.println("Key = " + entry.getKey());
for (String value : entry.getValue()) {
System.out.println(" Value = " + value);
}
}
我将使用Multimap的keys Multiset条目,按降序对它们进行排序(一旦issue 356中描述的功能添加到Guava中,这将变得更加容易),并通过迭代排序的键并获取值来构建新的Multimap来自原始的Multimap:
/**
* @return a {@link Multimap} whose entries are sorted by descending frequency
*/
public Multimap<String, String> sortedByDescendingFrequency(Multimap<String, String> multimap) {
// ImmutableMultimap.Builder preserves key/value order
ImmutableMultimap.Builder<String, String> result = ImmutableMultimap.builder();
for (Multiset.Entry<String> entry : DESCENDING_COUNT_ORDERING.sortedCopy(multimap.keys().entrySet())) {
result.putAll(entry.getElement(), multimap.get(entry.getElement()));
}
return result.build();
}
/**
* An {@link Ordering} that orders {@link Multiset.Entry Multiset entries} by ascending count.
*/
private static final Ordering<Multiset.Entry<?>> ASCENDING_COUNT_ORDERING = new Ordering<Multiset.Entry<?>>() {
@Override
public int compare(Multiset.Entry<?> left, Multiset.Entry<?> right) {
return Ints.compare(left.getCount(), right.getCount());
}
};
/**
* An {@link Ordering} that orders {@link Multiset.Entry Multiset entries} by descending count.
*/
private static final Ordering<Multiset.Entry<?>> DESCENDING_COUNT_ORDERING = ASCENDING_COUNT_ORDERING.reverse();
编辑:如果某些条目具有相同的频率,则不会起作用(请参阅我的评论)
另一种方法,使用基于Multimaps的键Multiset和ImmutableMultimap.Builder.orderKeysBy()的排序:
/**
* @return a {@link Multimap} whose entries are sorted by descending frequency
*/
public Multimap<String, String> sortedByDescendingFrequency(Multimap<String, String> multimap) {
return ImmutableMultimap.<String, String>builder()
.orderKeysBy(descendingCountOrdering(multimap.keys()))
.putAll(multimap)
.build();
}
private static Ordering<String> descendingCountOrdering(final Multiset<String> multiset) {
return new Ordering<String>() {
@Override
public int compare(String left, String right) {
return Ints.compare(multiset.count(left), multiset.count(right));
}
};
}
第二种方法更短,但是我不喜欢Ordering具有状态这一事实(它取决于Multimap的键Multiset来比较键)。
使用Java8流:
ListMultimap<String, String> multiMap = ArrayListMultimap.create();
multiMap.put("foo", "f1");
multiMap.put("baz", "z1");
multiMap.put("baz", "z2");
multiMap.put("bar", "b1");
multiMap.put("bar", "b2");
multiMap.put("bar", "b3");
Multimaps.asMap(multiMap).entrySet().stream()
.sorted(Comparator.comparing(e -> -e.getValue().size()))
.forEach(e -> System.out.println(e.getKey() + " -> "
+ Arrays.toString(e.getValue().toArray())));
输出:
bar -> [b1, b2, b3]
baz -> [z1, z2]
foo -> [f1]