com.google.common.collect.TopKSelector Maven / Gradle / Ivy
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/*
* Copyright (C) 2014 The Guava Authors
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
package com.google.common.collect;
import static com.google.common.base.Preconditions.checkArgument;
import static com.google.common.base.Preconditions.checkNotNull;
import com.google.common.annotations.GwtCompatible;
import com.google.common.math.IntMath;
import java.math.RoundingMode;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Iterator;
import java.util.List;
import org.checkerframework.checker.nullness.compatqual.NullableDecl;
/**
* An accumulator that selects the "top" {@code k} elements added to it, relative to a provided
* comparator. "Top" can mean the greatest or the lowest elements, specified in the factory used to
* create the {@code TopKSelector} instance.
*
* If your input data is available as an {@link Iterable} or {@link Iterator}, prefer {@link
* Ordering#leastOf(Iterable, int)}, which provides the same implementation with an interface
* tailored to that use case.
*
*
This uses the same efficient implementation as {@link Ordering#leastOf(Iterable, int)},
* offering expected O(n + k log k) performance (worst case O(n log k)) for n calls to {@link
* #offer} and a call to {@link #topK}, with O(k) memory. In comparison, quickselect has the same
* asymptotics but requires O(n) memory, and a {@code PriorityQueue} implementation takes O(n log
* k). In benchmarks, this implementation performs at least as well as either implementation, and
* degrades more gracefully for worst-case input.
*
*
The implementation does not necessarily use a stable sorting algorithm; when multiple
* equivalent elements are added to it, it is undefined which will come first in the output.
*
* @author Louis Wasserman
*/
@GwtCompatible
final class TopKSelector {
/**
* Returns a {@code TopKSelector} that collects the lowest {@code k} elements added to it,
* relative to the natural ordering of the elements, and returns them via {@link #topK} in
* ascending order.
*
* @throws IllegalArgumentException if {@code k < 0} or {@code k > Integer.MAX_VALUE / 2}
*/
public static > TopKSelector least(int k) {
return least(k, Ordering.natural());
}
/**
* Returns a {@code TopKSelector} that collects the lowest {@code k} elements added to it,
* relative to the specified comparator, and returns them via {@link #topK} in ascending order.
*
* @throws IllegalArgumentException if {@code k < 0} or {@code k > Integer.MAX_VALUE / 2}
*/
public static TopKSelector least(int k, Comparator comparator) {
return new TopKSelector(comparator, k);
}
/**
* Returns a {@code TopKSelector} that collects the greatest {@code k} elements added to it,
* relative to the natural ordering of the elements, and returns them via {@link #topK} in
* descending order.
*
* @throws IllegalArgumentException if {@code k < 0} or {@code k > Integer.MAX_VALUE / 2}
*/
public static > TopKSelector greatest(int k) {
return greatest(k, Ordering.natural());
}
/**
* Returns a {@code TopKSelector} that collects the greatest {@code k} elements added to it,
* relative to the specified comparator, and returns them via {@link #topK} in descending order.
*
* @throws IllegalArgumentException if {@code k < 0} or {@code k > Integer.MAX_VALUE / 2}
*/
public static TopKSelector greatest(int k, Comparator comparator) {
return new TopKSelector(Ordering.from(comparator).reverse(), k);
}
private final int k;
private final Comparator comparator;
/*
* We are currently considering the elements in buffer in the range [0, bufferSize) as candidates
* for the top k elements. Whenever the buffer is filled, we quickselect the top k elements to the
* range [0, k) and ignore the remaining elements.
*/
private final T[] buffer;
private int bufferSize;
/**
* The largest of the lowest k elements we've seen so far relative to this comparator. If
* bufferSize ≥ k, then we can ignore any elements greater than this value.
*/
@NullableDecl private T threshold;
private TopKSelector(Comparator comparator, int k) {
this.comparator = checkNotNull(comparator, "comparator");
this.k = k;
checkArgument(k >= 0, "k (%s) must be >= 0", k);
checkArgument(k <= Integer.MAX_VALUE / 2, "k (%s) must be <= Integer.MAX_VALUE / 2", k);
this.buffer = (T[]) new Object[IntMath.checkedMultiply(k, 2)];
this.bufferSize = 0;
this.threshold = null;
}
/**
* Adds {@code elem} as a candidate for the top {@code k} elements. This operation takes amortized
* O(1) time.
*/
public void offer(@NullableDecl T elem) {
if (k == 0) {
return;
} else if (bufferSize == 0) {
buffer[0] = elem;
threshold = elem;
bufferSize = 1;
} else if (bufferSize < k) {
buffer[bufferSize++] = elem;
if (comparator.compare(elem, threshold) > 0) {
threshold = elem;
}
} else if (comparator.compare(elem, threshold) < 0) {
// Otherwise, we can ignore elem; we've seen k better elements.
buffer[bufferSize++] = elem;
if (bufferSize == 2 * k) {
trim();
}
}
}
/**
* Quickselects the top k elements from the 2k elements in the buffer. O(k) expected time, O(k log
* k) worst case.
*/
private void trim() {
int left = 0;
int right = 2 * k - 1;
int minThresholdPosition = 0;
// The leftmost position at which the greatest of the k lower elements
// -- the new value of threshold -- might be found.
int iterations = 0;
int maxIterations = IntMath.log2(right - left, RoundingMode.CEILING) * 3;
while (left < right) {
int pivotIndex = (left + right + 1) >>> 1;
int pivotNewIndex = partition(left, right, pivotIndex);
if (pivotNewIndex > k) {
right = pivotNewIndex - 1;
} else if (pivotNewIndex < k) {
left = Math.max(pivotNewIndex, left + 1);
minThresholdPosition = pivotNewIndex;
} else {
break;
}
iterations++;
if (iterations >= maxIterations) {
// We've already taken O(k log k), let's make sure we don't take longer than O(k log k).
Arrays.sort(buffer, left, right, comparator);
break;
}
}
bufferSize = k;
threshold = buffer[minThresholdPosition];
for (int i = minThresholdPosition + 1; i < k; i++) {
if (comparator.compare(buffer[i], threshold) > 0) {
threshold = buffer[i];
}
}
}
/**
* Partitions the contents of buffer in the range [left, right] around the pivot element
* previously stored in buffer[pivotValue]. Returns the new index of the pivot element,
* pivotNewIndex, so that everything in [left, pivotNewIndex] is ≤ pivotValue and everything in
* (pivotNewIndex, right] is greater than pivotValue.
*/
private int partition(int left, int right, int pivotIndex) {
T pivotValue = buffer[pivotIndex];
buffer[pivotIndex] = buffer[right];
int pivotNewIndex = left;
for (int i = left; i < right; i++) {
if (comparator.compare(buffer[i], pivotValue) < 0) {
swap(pivotNewIndex, i);
pivotNewIndex++;
}
}
buffer[right] = buffer[pivotNewIndex];
buffer[pivotNewIndex] = pivotValue;
return pivotNewIndex;
}
private void swap(int i, int j) {
T tmp = buffer[i];
buffer[i] = buffer[j];
buffer[j] = tmp;
}
/**
* Adds each member of {@code elements} as a candidate for the top {@code k} elements. This
* operation takes amortized linear time in the length of {@code elements}.
*
* If all input data to this {@code TopKSelector} is in a single {@code Iterable}, prefer
* {@link Ordering#leastOf(Iterable, int)}, which provides a simpler API for that use case.
*/
public void offerAll(Iterable elements) {
offerAll(elements.iterator());
}
/**
* Adds each member of {@code elements} as a candidate for the top {@code k} elements. This
* operation takes amortized linear time in the length of {@code elements}. The iterator is
* consumed after this operation completes.
*
*
If all input data to this {@code TopKSelector} is in a single {@code Iterator}, prefer
* {@link Ordering#leastOf(Iterator, int)}, which provides a simpler API for that use case.
*/
public void offerAll(Iterator elements) {
while (elements.hasNext()) {
offer(elements.next());
}
}
/**
* Returns the top {@code k} elements offered to this {@code TopKSelector}, or all elements if
* fewer than {@code k} have been offered, in the order specified by the factory used to create
* this {@code TopKSelector}.
*
*
The returned list is an unmodifiable copy and will not be affected by further changes to
* this {@code TopKSelector}. This method returns in O(k log k) time.
*/
public List topK() {
Arrays.sort(buffer, 0, bufferSize, comparator);
if (bufferSize > k) {
Arrays.fill(buffer, k, buffer.length, null);
bufferSize = k;
threshold = buffer[k - 1];
}
// we have to support null elements, so no ImmutableList for us
return Collections.unmodifiableList(Arrays.asList(Arrays.copyOf(buffer, bufferSize)));
}
}