Why is grouped summation slower with sorted groups than unsorted groups?

Question

I have 2 columns of tab delimited integers, the first of which is a random integer, the second an integer identifying the group, which can be generated by this program. (generate_groups.cc)

#include <cstdlib>
#include <iostream>
#include <ctime>

int main(int argc, char* argv[]) {
  int num_values = atoi(argv[1]);
  int num_groups = atoi(argv[2]);

  int group_size = num_values / num_groups;
  int group = -1;

  std::srand(42);

  for (int i = 0; i < num_values; ++i) {
    if (i % group_size == 0) {
      ++group;
    }
    std::cout << std::rand() << '\t' << group << '\n';
  }

  return 0;
}

I then use a second program (sum_groups.cc) to calculate the sums per group.

#include <iostream>
#include <chrono>
#include <vector>

// This is the function whose performance I am interested in
void grouped_sum(int* p_x, int *p_g, int n, int* p_out) {
  for (size_t i = 0; i < n; ++i) {
    p_out[p_g[i]] += p_x[i];
  }
}

int main() {
  std::vector<int> values;
  std::vector<int> groups;
  std::vector<int> sums;

  int n_groups = 0;

  // Read in the values and calculate the max number of groups
  while(std::cin) {
    int value, group;
    std::cin >> value >> group;
    values.push_back(value);
    groups.push_back(group);
    if (group > n_groups) {
      n_groups = group;
    }
  }
  sums.resize(n_groups);

  // Time grouped sums
  std::chrono::system_clock::time_point start = std::chrono::system_clock::now();
  for (int i = 0; i < 10; ++i) {
    grouped_sum(values.data(), groups.data(), values.size(), sums.data());
  }
  std::chrono::system_clock::time_point end = std::chrono::system_clock::now();

  std::cout << (end - start).count() << std::endl;

  return 0;
}

If I then run these programs on a dataset of given size, and then shuffle the order of the rows of the same dataset the shuffled data computes the sums ~2x or more faster than the ordered data.

g++ -O3 generate_groups.cc -o generate_groups
g++ -O3 sum_groups.cc -o sum_groups
generate_groups 1000000 100 > groups
shuf groups > groups2
sum_groups < groups
sum_groups < groups2
sum_groups < groups2
sum_groups < groups
20784
8854
8220
21006

I would have expected the original data which is sorted by group to have better data locality and be faster, but I observe the opposite behavior. I was wondering if anyone can hypothesize the reason?

Answer 1

设置/使其变慢

首先，该程序将在大约相同的时间运行，无论：

sumspeed$ time ./sum_groups < groups_shuffled 
11558358

real    0m0.705s
user    0m0.692s
sys 0m0.013s

sumspeed$ time ./sum_groups < groups_sorted
24986825

real    0m0.722s
user    0m0.711s
sys 0m0.012s

大部分时间都花在输入循环中。但是，由于我们对感兴趣，因此请grouped_sum()忽略它。

将基准循环从10次迭代更改为1000次迭代，grouped_sum()开始控制运行时间：

sumspeed$ time ./sum_groups < groups_shuffled 
1131838420

real    0m1.828s
user    0m1.811s
sys 0m0.016s

sumspeed$ time ./sum_groups < groups_sorted
2494032110

real    0m3.189s
user    0m3.169s
sys 0m0.016s

性能差异

现在，我们可以使用它perf来找到程序中最热门的地方。

sumspeed$ perf record ./sum_groups < groups_shuffled
1166805982
[ perf record: Woken up 1 times to write data ]
[kernel.kallsyms] with build id 3a2171019937a2070663f3b6419330223bd64e96 not found, continuing without symbols
Warning:
Processed 4636 samples and lost 6.95% samples!

[ perf record: Captured and wrote 0.176 MB perf.data (4314 samples) ]

sumspeed$ perf record ./sum_groups < groups_sorted
2571547832
[ perf record: Woken up 2 times to write data ]
[kernel.kallsyms] with build id 3a2171019937a2070663f3b6419330223bd64e96 not found, continuing without symbols
[ perf record: Captured and wrote 0.420 MB perf.data (10775 samples) ]

它们之间的区别：

sumspeed$ perf diff
[...]
# Event 'cycles:uppp'
#
# Baseline  Delta Abs  Shared Object        Symbol                                                                  
# ........  .........  ...................  ........................................................................
#
    57.99%    +26.33%  sum_groups           [.] main
    12.10%     -7.41%  libc-2.23.so         [.] _IO_getc
     9.82%     -6.40%  libstdc++.so.6.0.21  [.] std::num_get<char, std::istreambuf_iterator<char, std::char_traits<c
     6.45%     -4.00%  libc-2.23.so         [.] _IO_ungetc
     2.40%     -1.32%  libc-2.23.so         [.] _IO_sputbackc
     1.65%     -1.21%  libstdc++.so.6.0.21  [.] 0x00000000000dc4a4
     1.57%     -1.20%  libc-2.23.so         [.] _IO_fflush
     1.71%     -1.07%  libstdc++.so.6.0.21  [.] std::istream::sentry::sentry
     1.22%     -0.77%  libstdc++.so.6.0.21  [.] std::istream::operator>>
     0.79%     -0.47%  libstdc++.so.6.0.21  [.] __gnu_cxx::stdio_sync_filebuf<char, std::char_traits<char> >::uflow
[...]

中有更多时间main()，这可能是grouped_sum()内联的。太好了，非常感谢。

性能注释

有没有在时间都花在差别里面 main()？

随机播放：

sumspeed$ perf annotate -i perf.data.old
[...]
       ?     // This is the function whose performance I am interested in
       ?     void grouped_sum(int* p_x, int *p_g, int n, int* p_out) {
       ?       for (size_t i = 0; i < n; ++i) {
       ?180:   xor    %eax,%eax
       ?       test   %rdi,%rdi
       ?     ? je     1a4
       ?       nop
       ?         p_out[p_g[i]] += p_x[i];
  6,88 ?190:   movslq (%r9,%rax,4),%rdx
 58,54 ?       mov    (%r8,%rax,4),%esi
       ?     #include <chrono>
       ?     #include <vector>
       ?
       ?     // This is the function whose performance I am interested in
       ?     void grouped_sum(int* p_x, int *p_g, int n, int* p_out) {
       ?       for (size_t i = 0; i < n; ++i) {
  3,86 ?       add    $0x1,%rax
       ?         p_out[p_g[i]] += p_x[i];
 29,61 ?       add    %esi,(%rcx,%rdx,4)
[...]

排序：

sumspeed$ perf annotate -i perf.data
[...]
       ?     // This is the function whose performance I am interested in
       ?     void grouped_sum(int* p_x, int *p_g, int n, int* p_out) {
       ?       for (size_t i = 0; i < n; ++i) {
       ?180:   xor    %eax,%eax
       ?       test   %rdi,%rdi
       ?     ? je     1a4
       ?       nop
       ?         p_out[p_g[i]] += p_x[i];
  1,00 ?190:   movslq (%r9,%rax,4),%rdx
 55,12 ?       mov    (%r8,%rax,4),%esi
       ?     #include <chrono>
       ?     #include <vector>
       ?
       ?     // This is the function whose performance I am interested in
       ?     void grouped_sum(int* p_x, int *p_g, int n, int* p_out) {
       ?       for (size_t i = 0; i < n; ++i) {
  0,07 ?       add    $0x1,%rax
       ?         p_out[p_g[i]] += p_x[i];
 43,28 ?       add    %esi,(%rcx,%rdx,4)
[...]

不，这是两个相同的指令。因此，在两种情况下它们都需要花费很长时间，但对数据进行排序时甚至更糟。

性能统计

好的。但是我们应该将它们运行相同的次数，因此由于某种原因，每条指令必须变慢。让我们看看怎么perf stat说。

sumspeed$ perf stat ./sum_groups < groups_shuffled 
1138880176

 Performance counter stats for './sum_groups':

       1826,232278      task-clock (msec)         #    0,999 CPUs utilized          
                72      context-switches          #    0,039 K/sec                  
                 1      cpu-migrations            #    0,001 K/sec                  
             4 076      page-faults               #    0,002 M/sec                  
     5 403 949 695      cycles                    #    2,959 GHz                    
       930 473 671      stalled-cycles-frontend   #   17,22% frontend cycles idle   
     9 827 685 690      instructions              #    1,82  insn per cycle         
                                                  #    0,09  stalled cycles per insn
     2 086 725 079      branches                  # 1142,639 M/sec                  
         2 069 655      branch-misses             #    0,10% of all branches        

       1,828334373 seconds time elapsed

sumspeed$ perf stat ./sum_groups < groups_sorted
2496546045

 Performance counter stats for './sum_groups':

       3186,100661      task-clock (msec)         #    1,000 CPUs utilized          
                 5      context-switches          #    0,002 K/sec                  
                 0      cpu-migrations            #    0,000 K/sec                  
             4 079      page-faults               #    0,001 M/sec                  
     9 424 565 623      cycles                    #    2,958 GHz                    
     4 955 937 177      stalled-cycles-frontend   #   52,59% frontend cycles idle   
     9 829 009 511      instructions              #    1,04  insn per cycle         
                                                  #    0,50  stalled cycles per insn
     2 086 942 109      branches                  #  655,014 M/sec                  
         2 078 204      branch-misses             #    0,10% of all branches        

       3,186768174 seconds time elapsed

只有一件事很突出：stalled-cycles-frontend。

好的，指令流水线正在停滞。在前端。确切的说，这可能在微体系结构之间有所不同。

我有一个猜测。如果您很慷慨，您甚至可以称其为假设。

假设

通过对输入进行排序，可以增加写入的局部性。实际上，它们将非常本地化；您所做的几乎所有添加操作都将写入与上一个相同的位置。

这对缓存很有用，但对管道却没有用。您正在引入数据依赖关系，从而阻止下一条加法指令继续执行，直到前一条加法完成（或使结果可用于后续指令）为止。

那是你的问题。

我认为。

修复它

多个和向量

实际上，让我们尝试一下。如果我们使用多个和向量，在每次加法之间切换它们，然后在最后求和，该怎么办？它花费了我们一些局部性，但是应该删除数据依赖项。

（代码不是很漂亮；不要判断我，互联网！）

#include <iostream>
#include <chrono>
#include <vector>

#ifndef NSUMS
#define NSUMS (4) // must be power of 2 (for masking to work)
#endif

// This is the function whose performance I am interested in
void grouped_sum(int* p_x, int *p_g, int n, int** p_out) {
  for (size_t i = 0; i < n; ++i) {
    p_out[i & (NSUMS-1)][p_g[i]] += p_x[i];
  }
}

int main() {
  std::vector<int> values;
  std::vector<int> groups;
  std::vector<int> sums[NSUMS];

  int n_groups = 0;

  // Read in the values and calculate the max number of groups
  while(std::cin) {
    int value, group;
    std::cin >> value >> group;
    values.push_back(value);
    groups.push_back(group);
    if (group >= n_groups) {
      n_groups = group+1;
    }
  }
  for (int i=0; i<NSUMS; ++i) {
    sums[i].resize(n_groups);
  }

  // Time grouped sums
  std::chrono::system_clock::time_point start = std::chrono::system_clock::now();
  int* sumdata[NSUMS];
  for (int i = 0; i < NSUMS; ++i) {
    sumdata[i] = sums[i].data();
  }
  for (int i = 0; i < 1000; ++i) {
    grouped_sum(values.data(), groups.data(), values.size(), sumdata);
  }
  for (int i = 1; i < NSUMS; ++i) {
    for (int j = 0; j < n_groups; ++j) {
      sumdata[0][j] += sumdata[i][j];
    }
  }
  std::chrono::system_clock::time_point end = std::chrono::system_clock::now();

  std::cout << (end - start).count() << " with NSUMS=" << NSUMS << std::endl;

  return 0;
}

（哦，我还修复了n_groups的计算；它被减一了。）

结果

配置我的makefile以将-DNSUMS=...arg赋予编译器后，我可以这样做：

sumspeed$ for n in 1 2 4 8 128; do make -s clean && make -s NSUMS=$n && (perf stat ./sum_groups < groups_shuffled && perf stat ./sum_groups < groups_sorted)  2>&1 | egrep '^[0-9]|frontend'; done
1134557008 with NSUMS=1
       924 611 882      stalled-cycles-frontend   #   17,13% frontend cycles idle   
2513696351 with NSUMS=1
     4 998 203 130      stalled-cycles-frontend   #   52,79% frontend cycles idle   
1116188582 with NSUMS=2
       899 339 154      stalled-cycles-frontend   #   16,83% frontend cycles idle   
1365673326 with NSUMS=2
     1 845 914 269      stalled-cycles-frontend   #   29,97% frontend cycles idle   
1127172852 with NSUMS=4
       902 964 410      stalled-cycles-frontend   #   16,79% frontend cycles idle   
1171849032 with NSUMS=4
     1 007 807 580      stalled-cycles-frontend   #   18,29% frontend cycles idle   
1118732934 with NSUMS=8
       881 371 176      stalled-cycles-frontend   #   16,46% frontend cycles idle   
1129842892 with NSUMS=8
       905 473 182      stalled-cycles-frontend   #   16,80% frontend cycles idle   
1497803734 with NSUMS=128
     1 982 652 954      stalled-cycles-frontend   #   30,63% frontend cycles idle   
1180742299 with NSUMS=128
     1 075 507 514      stalled-cycles-frontend   #   19,39% frontend cycles idle   

和向量的最佳数量可能取决于您CPU的流水线深度。我7岁的超极本CPU可能可以用比新型花式台式机CPU所需的更少的向量最大化处理流程。

显然，更多并不一定更好。当我疯狂使用128个和向量时，我们开始遭受缓存未命中的更多痛苦-改组后的输入变得比排序慢，就像您最初预期的那样。我们来了整整一圈！:)

寄存器中的每组总和

（这是在编辑中添加的）

啊，书呆子了！如果您知道输入将被排序并且正在寻找更高的性能，那么至少在我的计算机上，函数的以下重写（没有额外的和数组）甚至更快。

// This is the function whose performance I am interested in
void grouped_sum(int* p_x, int *p_g, int n, int* p_out) {
  int i = n-1;
  while (i >= 0) {
    int g = p_g[i];
    int gsum = 0;
    do {
      gsum += p_x[i--];
    } while (i >= 0 && p_g[i] == g);
    p_out[g] += gsum;
  }
}

这其中的技巧是，它允许编译器将gsum变量（组的总和）保留在寄存器中。我猜测（但可能是非常错误的），这样做速度更快，因为此处的管道中的反馈循环可能更短，并且/或者更少的内存访问。一个好的分支预测器会使对组相等性的额外检查便宜。

结果

混音输入太糟糕了...

sumspeed$ time ./sum_groups < groups_shuffled
2236354315

real    0m2.932s
user    0m2.923s
sys 0m0.009s

...但是比排序输入的“许多总和”解决方案快40％。

sumspeed$ time ./sum_groups < groups_sorted
809694018

real    0m1.501s
user    0m1.496s
sys 0m0.005s

很多小组的速度要比几个小组的慢，因此这是否是更快的实现实际上取决于您的数据。而且，与以往一样，在您的CPU型号上。

多个和向量，具有偏移量而不是位掩码

Sopel建议了四个展开的扩展，以替代我的位掩码方法。我已经实施了他们建议的通用版本，可以处理不同的建议NSUMS。我指望编译器为我们展开内部循环（至少这样做NSUMS=4）。

#include <iostream>
#include <chrono>
#include <vector>

#ifndef NSUMS
#define NSUMS (4) // must be power of 2 (for masking to work)
#endif

#ifndef INNER
#define INNER (0)
#endif
#if INNER
// This is the function whose performance I am interested in
void grouped_sum(int* p_x, int *p_g, int n, int** p_out) {
  size_t i = 0;
  int quadend = n & ~(NSUMS-1);
  for (; i < quadend; i += NSUMS) {
    for (int k=0; k<NSUMS; ++k) {
      p_out[k][p_g[i+k]] += p_x[i+k];
    }
  }
  for (; i < n; ++i) {
    p_out[0][p_g[i]] += p_x[i];
  }
}
#else
// This is the function whose performance I am interested in
void grouped_sum(int* p_x, int *p_g, int n, int** p_out) {
  for (size_t i = 0; i < n; ++i) {
    p_out[i & (NSUMS-1)][p_g[i]] += p_x[i];
  }
}
#endif


int main() {
  std::vector<int> values;
  std::vector<int> groups;
  std::vector<int> sums[NSUMS];

  int n_groups = 0;

  // Read in the values and calculate the max number of groups
  while(std::cin) {
    int value, group;
    std::cin >> value >> group;
    values.push_back(value);
    groups.push_back(group);
    if (group >= n_groups) {
      n_groups = group+1;
    }
  }
  for (int i=0; i<NSUMS; ++i) {
    sums[i].resize(n_groups);
  }

  // Time grouped sums
  std::chrono::system_clock::time_point start = std::chrono::system_clock::now();
  int* sumdata[NSUMS];
  for (int i = 0; i < NSUMS; ++i) {
    sumdata[i] = sums[i].data();
  }
  for (int i = 0; i < 1000; ++i) {
    grouped_sum(values.data(), groups.data(), values.size(), sumdata);
  }
  for (int i = 1; i < NSUMS; ++i) {
    for (int j = 0; j < n_groups; ++j) {
      sumdata[0][j] += sumdata[i][j];
    }
  }
  std::chrono::system_clock::time_point end = std::chrono::system_clock::now();

  std::cout << (end - start).count() << " with NSUMS=" << NSUMS << ", INNER=" << INNER << std::endl;

  return 0;
}

结果

该测量了。请注意，由于昨天我在/ tmp中工作，因此我没有完全相同的输入数据。因此，这些结果不能直接与之前的结果进行比较（但可能足够接近）。

sumspeed$ for n in 2 4 8 16; do for inner in 0 1; do make -s clean && make -s NSUMS=$n INNER=$inner && (perf stat ./sum_groups < groups_shuffled && perf stat ./sum_groups < groups_sorted)  2>&1 | egrep '^[0-9]|frontend'; done; done1130558787 with NSUMS=2, INNER=0
       915 158 411      stalled-cycles-frontend   #   16,96% frontend cycles idle   
1351420957 with NSUMS=2, INNER=0
     1 589 408 901      stalled-cycles-frontend   #   26,21% frontend cycles idle   
840071512 with NSUMS=2, INNER=1
     1 053 982 259      stalled-cycles-frontend   #   23,26% frontend cycles idle   
1391591981 with NSUMS=2, INNER=1
     2 830 348 854      stalled-cycles-frontend   #   45,35% frontend cycles idle   
1110302654 with NSUMS=4, INNER=0
       890 869 892      stalled-cycles-frontend   #   16,68% frontend cycles idle   
1145175062 with NSUMS=4, INNER=0
       948 879 882      stalled-cycles-frontend   #   17,40% frontend cycles idle   
822954895 with NSUMS=4, INNER=1
     1 253 110 503      stalled-cycles-frontend   #   28,01% frontend cycles idle   
929548505 with NSUMS=4, INNER=1
     1 422 753 793      stalled-cycles-frontend   #   30,32% frontend cycles idle   
1128735412 with NSUMS=8, INNER=0
       921 158 397      stalled-cycles-frontend   #   17,13% frontend cycles idle   
1120606464 with NSUMS=8, INNER=0
       891 960 711      stalled-cycles-frontend   #   16,59% frontend cycles idle   
800789776 with NSUMS=8, INNER=1
     1 204 516 303      stalled-cycles-frontend   #   27,25% frontend cycles idle   
805223528 with NSUMS=8, INNER=1
     1 222 383 317      stalled-cycles-frontend   #   27,52% frontend cycles idle   
1121644613 with NSUMS=16, INNER=0
       886 781 824      stalled-cycles-frontend   #   16,54% frontend cycles idle   
1108977946 with NSUMS=16, INNER=0
       860 600 975      stalled-cycles-frontend   #   16,13% frontend cycles idle   
911365998 with NSUMS=16, INNER=1
     1 494 671 476      stalled-cycles-frontend   #   31,54% frontend cycles idle   
898729229 with NSUMS=16, INNER=1
     1 474 745 548      stalled-cycles-frontend   #   31,24% frontend cycles idle   

NSUMS=8是的，内部循环是我计算机上最快的。与我的“本地gsum”方法相比，它还具有不为混洗输入带来可怕影响的额外好处。

有趣的是：NSUMS=16变得比差NSUMS=8。这可能是因为我们开始看到更多的高速缓存未命中，或者是因为我们没有足够的寄存器来正确展开内部循环。