问题
I was using Multiset to have easy access to the freq of elements, but I realize there is Collections#frequency(Collection<?>, Object) that does the same for any collection. What is the point of using Multiset
then? Is performance an issue here?
回答1:
Guava documentation for Multiset#count() has to say:
Note that for an Object.equals(java.lang.Object)-based multiset, this gives the same result as Collections.frequency(java.util.Collection, java.lang.Object) (which would presumably perform more poorly).
So, yes, I suspect that performance is the issue here.
I think Multiset#count
is more efficient because Collections#frequency
iterates through the entire collection. For an object o whose frequency you're checking, it goes through all elements e in the collection and checks (o == null ? e == null : o.equals(e))
.
For Multiset (which is an interface), the exact implementation of count
depends on the class. If it is a HashMultiset
, for example, then it is backed by a HashMap
. For details about how that is more efficient than iterating through the whole collection, take a look at this answer: How does a Java HashMap handle different objects with the same hash code?.
The Guava code is as follows
public int count(@Nullable Object element) {
Count frequency = Maps.safeGet(backingMap, element);
return (frequency == null) ? 0 : frequency.get();
}
Similarly, for a TreeMultiset
, which maintains the ordering of its elements and is backed by an AVL tree, count
can be obtained in O(log(n)) steps instead of O(n), where n is the size of the collection. The Guava code is as follows:
public int count(@Nullable Object element) {
try {
@SuppressWarnings("unchecked")
E e = (E) element;
AvlNode<E> root = rootReference.get();
if (!range.contains(e) || root == null) {
return 0;
}
return root.count(comparator(), e);
} catch (ClassCastException e) {
return 0;
} catch (NullPointerException e) {
return 0;
}
}
来源:https://stackoverflow.com/questions/27374200/using-guava-multisets-or-collections-frequency