I have an array of elements (in the example, these are simply integers), which are compared using some custom comparator. In this example, I simulate this comparator by defining i SMALLER j if and only if scores[i] <= scores[j].
I have two approaches:
I update the upper two structures in the following way:
PriorityQueue.poll and PriorityQueue.offer,top of the worst among top k candidates in the array of candidates is stored. If a newly seen example is better than the element at the index top, the latter is replaced by the former and top is updated by iterating through all k elements of the array.However, when I have tested, which of the approaches is faster, I found out that this is the second. The questions are:
- Is my use of
PriorityQueuesuboptimal?- What is the fastest way to compute k smallest elements?
I am interested in the case, when the number of examples can be large, but the number of neighbours is relatively small (between 10 and 20).
Here is the code:
public static void main(String[] args) {
long kopica, navadno, sortiranje;
int numTries = 10000;
int numExamples = 1000;
int numNeighbours = 10;
navadno = testSimple(numExamples, numNeighbours, numTries);
kopica = testHeap(numExamples, numNeighbours, numTries);
sortiranje = testSort(numExamples, numNeighbours, numTries, false);
System.out.println(String.format("tries: %d examples: %d neighbours: %d\n time heap[ms]: %d\n time simple[ms]: %d", numTries, numExamples, numNeighbours, kopica, navadno));
}
public static long testHeap(int numberExamples, int numberNeighbours, int numberTries){
Random rnd = new Random(123);
long startTime = System.currentTimeMillis();
for(int iteration = 0; iteration < numberTries; iteration++){
final double[] scores = new double[numberExamples];
for(int i = 0; i < numberExamples; i++){
scores[i] = rnd.nextDouble();
}
PriorityQueue<Integer> myHeap = new PriorityQueue(numberNeighbours, new Comparator<Integer>(){
@Override
public int compare(Integer o1, Integer o2) {
return -Double.compare(scores[o1], scores[o2]);
}
});
int top;
for(int i = 0; i < numberExamples; i++){
if(i < numberNeighbours){
myHeap.offer(i);
} else{
top = myHeap.peek();
if(scores[top] > scores[i]){
myHeap.poll();
myHeap.offer(i);
}
}
}
}
long endTime = System.currentTimeMillis();
return endTime - startTime;
}
public static long testSimple(int numberExamples, int numberNeighbours, int numberTries){
Random rnd = new Random(123);
long startTime = System.currentTimeMillis();
for(int iteration = 0; iteration < numberTries; iteration++){
final double[] scores = new double[numberExamples];
for(int i = 0; i < numberExamples; i++){
scores[i] = rnd.nextDouble();
}
int[] candidates = new int[numberNeighbours];
int top = 0;
for(int i = 0; i < numberExamples; i++){
if(i < numberNeighbours){
candidates[i] = i;
if(scores[candidates[top]] < scores[candidates[i]]) top = i;
} else{
if(scores[candidates[top]] > scores[i]){
candidates[top] = i;
top = 0;
for(int j = 1; j < numberNeighbours; j++){
if(scores[candidates[top]] < scores[candidates[j]]) top = j;
}
}
}
}
}
long endTime = System.currentTimeMillis();
return endTime - startTime;
}
This produces the following result:
tries: 10000 examples: 1000 neighbours: 10
time heap[ms]: 393
time simple[ms]: 388
First of all, your benchmarking method is incorrect. You are measuring input data creation along with an algorithm performance, and you aren't warming up the JVM before measuring. Results for your code, when tested through the JMH:
Benchmark Mode Cnt Score Error Units
CounterBenchmark.testHeap thrpt 2 18103,296 ops/s
CounterBenchmark.testSimple thrpt 2 59490,384 ops/s
Modified benchmark pastebin.
Regarding 3x times difference between two provided solutions. In the terms of big-O notation your first algorithm may seem better, but in fact big-O notation only tells your how good the algorithm is in the terms of scaling, it never tells you how fast it performs (see this question also). And in your case scaling is not the issue, as your numNeighbours is limited to 20. In other words big-O notation describes how many ticks of algorithm is necessary for it to complete, but it doesn't limit the duration of a tick, it just says that the tick duration doesn't change when inputs change. And in terms of tick complexity your second algorithm surely wins.
What is the fastest way to compute k smallest elements?
I've came up with the next solution which I do believe allows branch prediction to do its job:
@Benchmark
public void testModified(Blackhole bh) {
final double[] scores = sampleData;
int[] candidates = new int[numberNeighbours];
for (int i = 0; i < numberNeighbours; i++) {
candidates[i] = i;
}
// sorting candidates so scores[candidates[0]] is the largest
for (int i = 0; i < numberNeighbours; i++) {
for (int j = i+1; j < numberNeighbours; j++) {
if (scores[candidates[i]] < scores[candidates[j]]) {
int temp = candidates[i];
candidates[i] = candidates[j];
candidates[j] = temp;
}
}
}
// processing other scores, while keeping candidates array sorted in the descending order
for (int i = numberNeighbours; i < numberExamples; i++) {
if (scores[i] > scores[candidates[0]]) {
continue;
}
// moving all larger candidates to the left, to keep the array sorted
int j; // here the branch prediction should kick-in
for (j = 1; j < numberNeighbours && scores[i] < scores[candidates[j]]; j++) {
candidates[j - 1] = candidates[j];
}
// inserting the new item
candidates[j - 1] = i;
}
bh.consume(candidates);
}
Benchmark results (2x times faster than your current solution):
(10 neighbours) CounterBenchmark.testModified thrpt 2 136492,151 ops/s
(20 neighbours) CounterBenchmark.testModified thrpt 2 118395,598 ops/s
Others mentioned quickselect, but as one may expect, the complexity of that algorithm neglects its strong sides in your case:
@Benchmark
public void testQuickSelect(Blackhole bh) {
final int[] candidates = new int[sampleData.length];
for (int i = 0; i < candidates.length; i++) {
candidates[i] = i;
}
final int[] resultIndices = new int[numberNeighbours];
int neighboursToAdd = numberNeighbours;
int left = 0;
int right = candidates.length - 1;
while (neighboursToAdd > 0) {
int partitionIndex = partition(candidates, left, right);
int smallerItemsPartitioned = partitionIndex - left;
if (smallerItemsPartitioned <= neighboursToAdd) {
while (left < partitionIndex) {
resultIndices[numberNeighbours - neighboursToAdd--] = candidates[left++];
}
} else {
right = partitionIndex - 1;
}
}
bh.consume(resultIndices);
}
private int partition(int[] locations, int left, int right) {
final int pivotIndex = ThreadLocalRandom.current().nextInt(left, right + 1);
final double pivotValue = sampleData[locations[pivotIndex]];
int storeIndex = left;
for (int i = left; i <= right; i++) {
if (sampleData[locations[i]] <= pivotValue) {
final int temp = locations[storeIndex];
locations[storeIndex] = locations[i];
locations[i] = temp;
storeIndex++;
}
}
return storeIndex;
}
Benchmark results are pretty upsetting in this case:
CounterBenchmark.testQuickSelect thrpt 2 11586,761 ops/s
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