Bri*_*007 4 c sorting algorithm performance
我uint64_t使用来自.RAW文件的RGB数据排序超过1000万s,花费了79%的C程序时间qsort.我正在寻找这种特定数据类型的更快排序.
作为RAW图形数据,数字是非常随机的,并且~80%是唯一的.不需要对排序数据进行部分排序或运行.4 uint16_tS的内部的uint64_t是R,G,B和零(可能是一个小计数<=〜20).
我有最简单的比较函数,我可以想到使用unsigned long longs(你不能只是减去它们):
qsort(hpidx, num_pix, sizeof(uint64_t), comp_uint64);
...
int comp_uint64(const void *a, const void *b) {
if(*((uint64_t *)a) > *((uint64_t *)b)) return(+1);
if(*((uint64_t *)a) < *((uint64_t *)b)) return(-1);
return(0);
} // End Comp_uint64().
StackExchange上有一个非常有趣的"编程拼图和代码高尔夫",但他们使用了floats.然后有QSort,RecQuick,堆,stooge,树,基数......
swenson/sort看起来很有趣,但对我的数据类型没有(明显的)支持uint64_t.而"快速排序"时间是最好的.有些消息称,系统qsort可以是任何东西,不一定是"快速排序".
C++排序绕过了void指针的通用转换,并实现了对C的性能的巨大改进.必须有一种优化的方法,以经线速度通过64位处理器猛击U8.
系统/编译器信息:
我目前正在使用GCC和Strawberry Perl
gcc version 4.9.2 (x86_64-posix-sjlj, built by strawberryperl.com
Intel 2700K Sandy Bridge CPU, 32GB DDR3
windows 7/64 pro
gcc -D__USE_MINGW_ANSI_STDIO -O4 -ffast-math -m64 -Ofast -march=corei7-avx -mtune=corei7 -Ic:/bin/xxHash-master -Lc:/bin/xxHash-master c:/bin/stddev.c -o c:/bin/stddev.g6.exe
qsort,QSORT()!试图使用迈克尔托卡列夫的内联 qsort.
"可以用了"？从qsort.h文档
-----------------------------
* Several ready-to-use examples:
*
* Sorting array of integers:
* void int_qsort(int *arr, unsigned n) {
* #define int_lt(a,b) ((*a)<(*b))
* QSORT(int, arr, n, int_lt);
--------------------------------
Change from type "int" to "uint64_t"
compile error on TYPE???
c:/bin/bpbfct.c:586:8: error: expected expression before 'uint64_t'
QSORT(uint64_t, hpidx, num_pix, islt);
我找不到一个真实的,编译的,有效的示例程序,只是用"一般概念"来评论
#define QSORT_TYPE uint64_t
#define islt(a,b) ((*a)<(*b))
uint64_t *QSORT_BASE;
int QSORT_NELT;
hpidx=(uint64_t *) calloc(num_pix+2, sizeof(uint64_t)); // Hash . PIDX
QSORT_BASE = hpidx;
QSORT_NELT = num_pix; // QSORT_LT is function QSORT_LT()
QSORT(uint64_t, hpidx, num_pix, islt);
//QSORT(uint64_t *, hpidx, num_pix, QSORT_LT); // QSORT_LT mal-defined?
//qsort(hpidx, num_pix, sizeof(uint64_t), comp_uint64); // << WORKS
"准备使用的"示例使用类型的int,char *和struct elt.不是uint64_t一个类型？尝试long long
QSORT(long long, hpidx, num_pix, islt);
c:/bin/bpbfct.c:586:8: error: expected expression before 'long'
QSORT(long long, hpidx, num_pix, islt);
RADIXSORT:结果:RADIX_SORT是根本的!
I:\br3\pf.249465>grep "Event" bb12.log | grep -i Sort
<< 1.40 sec average
4) Time=1.411 sec = 49.61%, Event RADIX_SORT , hits=1
4) Time=1.396 sec = 49.13%, Event RADIX_SORT , hits=1
4) Time=1.392 sec = 49.15%, Event RADIX_SORT , hits=1
16) Time=1.414 sec = 49.12%, Event RADIX_SORT , hits=1
I:\br3\pf.249465>grep "Event" bb11.log | grep -i Sort
<< 5.525 sec average = 3.95 time slower
4) Time=5.538 sec = 86.34%, Event QSort , hits=1
4) Time=5.519 sec = 79.41%, Event QSort , hits=1
4) Time=5.519 sec = 79.02%, Event QSort , hits=1
4) Time=5.563 sec = 79.49%, Event QSort , hits=1
4) Time=5.684 sec = 79.83%, Event QSort , hits=1
4) Time=5.509 sec = 79.30%, Event QSort , hits=1
比qsort开箱即用的任何类型快3.94倍!
而且,更重要的是,有一些实际的,有效的代码,而不仅仅是一些Guru所需要的80%,他们假设你知道他们所知道的一切,并且可以填写其他20%.
很棒的解决方案!谢谢Louis Ricci!
我会使用Radix Sort和8bit基数.对于64位值,优化良好的基数排序将不得不在列表上迭代9次(一次用于预先计算计数和偏移量,8次用于64位/ 8位).9*N时间和2*N空间(使用阴影阵列).
这是优化的基数排序的样子.
typedef union {
struct {
uint32_t c8[256];
uint32_t c7[256];
uint32_t c6[256];
uint32_t c5[256];
uint32_t c4[256];
uint32_t c3[256];
uint32_t c2[256];
uint32_t c1[256];
};
uint32_t counts[256 * 8];
} rscounts_t;
uint64_t * radixSort(uint64_t * array, uint32_t size) {
rscounts_t counts;
memset(&counts, 0, 256 * 8 * sizeof(uint32_t));
uint64_t * cpy = (uint64_t *)malloc(size * sizeof(uint64_t));
uint32_t o8=0, o7=0, o6=0, o5=0, o4=0, o3=0, o2=0, o1=0;
uint32_t t8, t7, t6, t5, t4, t3, t2, t1;
uint32_t x;
// calculate counts
for(x = 0; x < size; x++) {
t8 = array[x] & 0xff;
t7 = (array[x] >> 8) & 0xff;
t6 = (array[x] >> 16) & 0xff;
t5 = (array[x] >> 24) & 0xff;
t4 = (array[x] >> 32) & 0xff;
t3 = (array[x] >> 40) & 0xff;
t2 = (array[x] >> 48) & 0xff;
t1 = (array[x] >> 56) & 0xff;
counts.c8[t8]++;
counts.c7[t7]++;
counts.c6[t6]++;
counts.c5[t5]++;
counts.c4[t4]++;
counts.c3[t3]++;
counts.c2[t2]++;
counts.c1[t1]++;
}
// convert counts to offsets
for(x = 0; x < 256; x++) {
t8 = o8 + counts.c8[x];
t7 = o7 + counts.c7[x];
t6 = o6 + counts.c6[x];
t5 = o5 + counts.c5[x];
t4 = o4 + counts.c4[x];
t3 = o3 + counts.c3[x];
t2 = o2 + counts.c2[x];
t1 = o1 + counts.c1[x];
counts.c8[x] = o8;
counts.c7[x] = o7;
counts.c6[x] = o6;
counts.c5[x] = o5;
counts.c4[x] = o4;
counts.c3[x] = o3;
counts.c2[x] = o2;
counts.c1[x] = o1;
o8 = t8;
o7 = t7;
o6 = t6;
o5 = t5;
o4 = t4;
o3 = t3;
o2 = t2;
o1 = t1;
}
// radix
for(x = 0; x < size; x++) {
t8 = array[x] & 0xff;
cpy[counts.c8[t8]] = array[x];
counts.c8[t8]++;
}
for(x = 0; x < size; x++) {
t7 = (cpy[x] >> 8) & 0xff;
array[counts.c7[t7]] = cpy[x];
counts.c7[t7]++;
}
for(x = 0; x < size; x++) {
t6 = (array[x] >> 16) & 0xff;
cpy[counts.c6[t6]] = array[x];
counts.c6[t6]++;
}
for(x = 0; x < size; x++) {
t5 = (cpy[x] >> 24) & 0xff;
array[counts.c5[t5]] = cpy[x];
counts.c5[t5]++;
}
for(x = 0; x < size; x++) {
t4 = (array[x] >> 32) & 0xff;
cpy[counts.c4[t4]] = array[x];
counts.c4[t4]++;
}
for(x = 0; x < size; x++) {
t3 = (cpy[x] >> 40) & 0xff;
array[counts.c3[t3]] = cpy[x];
counts.c3[t3]++;
}
for(x = 0; x < size; x++) {
t2 = (array[x] >> 48) & 0xff;
cpy[counts.c2[t2]] = array[x];
counts.c2[t2]++;
}
for(x = 0; x < size; x++) {
t1 = (cpy[x] >> 56) & 0xff;
array[counts.c1[t1]] = cpy[x];
counts.c1[t1]++;
}
free(cpy);
return array;
}
编辑这个实现是基于JavaScript版本最快的方式来在JavaScript中对32位有符号整数数组进行排序？
这是C基数排序的IDEONE http://ideone.com/JHI0d9