Counting occurrences of numbers in a CUDA array
I have an array of unsigned integers stored on the GPU with CUDA (typically 1000000
elements). I would like 开发者_如何学Cto count the occurrence of every number in the array. There are only a few distinct numbers (about 10
), but these numbers can span from 1 to 1000000
. About 9/10
th of the numbers are 0
, I don't need the count of them. The result looks something like this:
58458 -> 1000 occurrences
15 -> 412 occurrences
I have an implementation using atomicAdd
s, but it is too slow (a lot of threads write to the same address). Does someone know of a fast/efficient method?
You can implement a histogram by first sorting the numbers, and then doing a keyed reduction.
The most straightforward method would be to use thrust::sort
and then thrust::reduce_by_key
. It's also often much faster than ad hoc binning based on atomics. Here's an example.
I suppose you can find help in the CUDA examples, specifically the histogram examples. They are part of the GPU computing SDK. You can find it here http://developer.nvidia.com/cuda-cc-sdk-code-samples#histogram. They even have a whitepaper explaining the algorithms.
I'm comparing two approaches suggested at the duplicate question thrust count occurence, namely,
- Using
thrust::counting_iterator
andthrust::upper_bound
, following the histogram Thrust example; - Using
thrust::unique_copy
andthrust::upper_bound
.
Below, please find a fully worked example.
#include <time.h> // --- time
#include <stdlib.h> // --- srand, rand
#include <iostream>
#include <thrust\host_vector.h>
#include <thrust\device_vector.h>
#include <thrust\sort.h>
#include <thrust\iterator\zip_iterator.h>
#include <thrust\unique.h>
#include <thrust/binary_search.h>
#include <thrust\adjacent_difference.h>
#include "Utilities.cuh"
#include "TimingGPU.cuh"
//#define VERBOSE
#define NO_HISTOGRAM
/********/
/* MAIN */
/********/
int main() {
const int N = 1048576;
//const int N = 20;
//const int N = 128;
TimingGPU timerGPU;
// --- Initialize random seed
srand(time(NULL));
thrust::host_vector<int> h_code(N);
for (int k = 0; k < N; k++) {
// --- Generate random numbers between 0 and 9
h_code[k] = (rand() % 10);
}
thrust::device_vector<int> d_code(h_code);
//thrust::device_vector<unsigned int> d_counting(N);
thrust::sort(d_code.begin(), d_code.end());
h_code = d_code;
timerGPU.StartCounter();
#ifdef NO_HISTOGRAM
// --- The number of d_cumsum bins is equal to the maximum value plus one
int num_bins = d_code.back() + 1;
thrust::device_vector<int> d_code_unique(num_bins);
thrust::unique_copy(d_code.begin(), d_code.end(), d_code_unique.begin());
thrust::device_vector<int> d_counting(num_bins);
thrust::upper_bound(d_code.begin(), d_code.end(), d_code_unique.begin(), d_code_unique.end(), d_counting.begin());
#else
thrust::device_vector<int> d_cumsum;
// --- The number of d_cumsum bins is equal to the maximum value plus one
int num_bins = d_code.back() + 1;
// --- Resize d_cumsum storage
d_cumsum.resize(num_bins);
// --- Find the end of each bin of values - Cumulative d_cumsum
thrust::counting_iterator<int> search_begin(0);
thrust::upper_bound(d_code.begin(), d_code.end(), search_begin, search_begin + num_bins, d_cumsum.begin());
// --- Compute the histogram by taking differences of the cumulative d_cumsum
//thrust::device_vector<int> d_counting(num_bins);
//thrust::adjacent_difference(d_cumsum.begin(), d_cumsum.end(), d_counting.begin());
#endif
printf("Timing GPU = %f\n", timerGPU.GetCounter());
#ifdef VERBOSE
thrust::host_vector<int> h_counting(d_counting);
printf("After\n");
for (int k = 0; k < N; k++) printf("code = %i\n", h_code[k]);
#ifndef NO_HISTOGRAM
thrust::host_vector<int> h_cumsum(d_cumsum);
printf("\nCounting\n");
for (int k = 0; k < num_bins; k++) printf("element = %i; counting = %i; cumsum = %i\n", k, h_counting[k], h_cumsum[k]);
#else
thrust::host_vector<int> h_code_unique(d_code_unique);
printf("\nCounting\n");
for (int k = 0; k < N; k++) printf("element = %i; counting = %i\n", h_code_unique[k], h_counting[k]);
#endif
#endif
}
The first approach has shown to be the fastest. On an NVIDIA GTX 960 card, I have had the following timings for a number of N = 1048576
array elements:
First approach: 2.35ms
First approach without thrust::adjacent_difference: 1.52
Second approach: 4.67ms
Please, note that there is no strict need to calculate the adjacent difference explicitly, since this operation can be manually done during a kernel processing, if needed.
As others have said, you can use the sort & reduce_by_key
approach to count frequencies. In my case, I needed to get mode of an array (maximum frequency/occurrence) so here is my solution:
1 - First, we create two new arrays, one containing a copy of input data and another filled with ones to later reduce it (sum):
// Input: [1 3 3 3 2 2 3]
// *(Temp) dev_keys: [1 3 3 3 2 2 3]
// *(Temp) dev_ones: [1 1 1 1 1 1 1]
// Copy input data
thrust::device_vector<int> dev_keys(myptr, myptr+size);
// Fill an array with ones
thrust::fill(dev_ones.begin(), dev_ones.end(), 1);
2 - Then, we sort the keys since the reduce_by_key
function needs the array to be sorted.
// Sort keys (see below why)
thrust::sort(dev_keys.begin(), dev_keys.end());
3 - Later, we create two output vectors, for the (unique) keys and their frequencies:
thrust::device_vector<int> output_keys(N);
thrust::device_vector<int> output_freqs(N);
4 - Finally, we perform the reduction by key:
// Reduce contiguous keys: [1 3 3 3 2 2 3] => [1 3 2 1] Vs. [1 3 3 3 3 2 2] => [1 4 2]
thrust::pair<thrust::device_vector<int>::iterator, thrust::device_vector<int>::iterator> new_end;
new_end = thrust::reduce_by_key(dev_keys.begin(), dev_keys.end(), dev_ones.begin(), output_keys.begin(), output_freqs.begin());
5 - ...and if we want, we can get the most frequent element
// Get most frequent element
// Get index of the maximum frequency
int num_keys = new_end.first - output_keys.begin();
thrust::device_vector<int>::iterator iter = thrust::max_element(output_freqs.begin(), output_freqs.begin() + num_keys);
unsigned int index = iter - output_freqs.begin();
int most_frequent_key = output_keys[index];
int most_frequent_val = output_freqs[index]; // Frequencies
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