tensorflow自定义op：work_shard

强行解释 work_shard

在学习 tensorflow 自定义 op 的时候碰到的，google 了一下，也没有找到详细的介绍，难道是姿势不对？？

通过看 了一些示例，这里打算强行解释一波。

概览

如果想用 work shard，首先 代码能够并行化计算。work shard 是一个代码并行化工具。不用自己头疼的写多线程代码了。

什么样的代码能够并行化计算 —> 每一个输出数据都能表示成相互无关的

work_shard 的最后一个参数就是要 shard 的 work， 这个 work 的签名为 void shard(int64 start, int64 limit)，work_shard 就是将 (start, limit) 给划分成多块，然后 块给 一个线程来计算。

# 如何使用 work_shard# 1. 包含头文件# 2. 该用的地方用就行了# 3. 链接的时候 g++ 会自动找到实现去链接的，不用操心。

代码

work_shard声明代码 地址

// work_sharder.h#ifndef TENSORFLOW_UTIL_WORK_SHARDER_H_#define TENSORFLOW_UTIL_WORK_SHARDER_H_#include <functional>#include "tensorflow/core/lib/core/threadpool.h"#include "tensorflow/core/platform/types.h"namespace tensorflow {// Shards the "total" unit of work assuming each unit of work having// roughly "cost_per_unit". Each unit of work is indexed 0, 1, ...,// total - 1. Each shard contains 1 or more units of work and the// total cost of each shard is roughly the same. The calling thread and the// "workers" are used to compute each shard (calling work(start,// limit). A common configuration is that "workers" is a thread pool// with at least "max_parallelism" threads.//// "cost_per_unit" is an estimate of the number of CPU cycles (or nanoseconds// if not CPU-bound) to complete a unit of work. Overestimating creates too// many shards and CPU time will be dominated by per-shard overhead, such as// Context creation. Underestimating may not fully make use of the specified// parallelism.//// "work" should be a callable taking (int64, int64) arguments.// work(start, limit) computes the work units from [start,// limit), i.e., [start, limit) is a shard.//// REQUIRES: max_parallelism >= 0// REQUIRES: workers != nullptr// REQUIRES: total >= 0// REQUIRES: cost_per_unit >= 0void Shard(int max_parallelism, thread::ThreadPool* workers, int64 total,           int64 cost_per_unit, std::function<void(int64, int64)> work);}  // end namespace tensorflow#endif  // TENSORFLOW_UTIL_WORK_SHARDER_H_

用到 Sharder 的地方（见代码片段最后） 完整代码地址

auto shard = [pooled_height, pooled_width, spatial_scale,num_rois, batch_size, data_height, data_width, num_channels,&bottom_data_flat, &bottom_rois_flat, &output, &argmax](int64 start, int64 limit) {for (int64 b = start; b < limit; ++b){  // (n, ph, pw, c) is an element in the pooled output  int n = b;  int c = n % num_channels;  n /= num_channels;  int pw = n % pooled_width;  n /= pooled_width;  int ph = n % pooled_height;  n /= pooled_height;  const float* bottom_rois = bottom_rois_flat.data() + n * 5;  int roi_batch_ind = bottom_rois[0];  int roi_start_w = round(bottom_rois[1] * spatial_scale);  int roi_start_h = round(bottom_rois[2] * spatial_scale);  int roi_end_w = round(bottom_rois[3] * spatial_scale);  int roi_end_h = round(bottom_rois[4] * spatial_scale);  // Force malformed ROIs to be 1x1  int roi_width = std::max(roi_end_w - roi_start_w + 1, 1);  int roi_height = std::max(roi_end_h - roi_start_h + 1, 1);  const T bin_size_h = static_cast<T>(roi_height)  / static_cast<T>(pooled_height);  const T bin_size_w = static_cast<T>(roi_width)  / static_cast<T>(pooled_width);  int hstart = static_cast<int>(floor(ph * bin_size_h));  int wstart = static_cast<int>(floor(pw * bin_size_w));  int hend = static_cast<int>(ceil((ph + 1) * bin_size_h));  int wend = static_cast<int>(ceil((pw + 1) * bin_size_w));  // Add roi offsets and clip to input boundaries  hstart = std::min(std::max(hstart + roi_start_h, 0), data_height);  hend = std::min(std::max(hend + roi_start_h, 0), data_height);  wstart = std::min(std::max(wstart + roi_start_w, 0), data_width);  wend = std::min(std::max(wend + roi_start_w, 0), data_width);  bool is_empty = (hend <= hstart) || (wend <= wstart);  // Define an empty pooling region to be zero  float maxval = is_empty ? 0 : -FLT_MAX;  // If nothing is pooled, argmax = -1 causes nothing to be backprop'd  int maxidx = -1;  const float* bottom_data = bottom_data_flat.data() + roi_batch_ind * num_channels * data_height * data_width;  for (int h = hstart; h < hend; ++h) {    for (int w = wstart; w < wend; ++w) {      int bottom_index = (h * data_width + w) * num_channels + c;      if (bottom_data[bottom_index] > maxval) {      maxval = bottom_data[bottom_index];      maxidx = bottom_index;      }    }  }  output(b) = maxval;  argmax(b) = maxidx;  }};const DeviceBase::CpuWorkerThreads& worker_threads =*(context->device()->tensorflow_cpu_worker_threads());const int64 shard_cost =num_rois * num_channels * pooled_height * pooled_width * spatial_scale;// 用到 shard 的地方Shard(worker_threads.num_threads, worker_threads.workers, output.size(), shard_cost, shard);

通过调用方法来分析Shard 声明中各参数的意义：

// 声明void Shard(int max_parallelism, thread::ThreadPool* workers, int64 total,           int64 cost_per_unit, std::function<void(int64, int64)> work);// 调用Shard(worker_threads.num_threads, worker_threads.workers, output.size(), shard_cost, shard);// max_parallelism: 最大并行个数，通过调用的形式来看，一般是使用 本机的线程数。// workers: 从声明来看，是代表的线程池。// total: 从调用来看，像是 work 中 unit 的数量，即最外层 for 循环的数量。// cost_per_unit: 对每个 unit 的 cpu 循环的一个估计。// work: 一个可调用对象，work的调用应该是这样的 work(int64, int64)

Shard 的实现源码：地址 地址如果失效，就去 tensorflow/core/util/work_sharder.cc 找

将work 分块执行，[0, limit) 变成 [0,block_size), [block_size, 2*block_size) 这么一块一块。

num_shards = total * cost_per_unit / 10000 为了理解 cost_per_unit 可以只关心这一部分

从这个部分可以看出，如果cost_per_unit 的运算量很大的话，tensorflow 会多分几块，那么问题来了，分成多少是合适的呢？

block_size = (total + num_shards - 1) / num_shards 。num_shards 要分成几块。

#include "tensorflow/core/util/work_sharder.h"#include "tensorflow/core/lib/core/blocking_counter.h"#include "tensorflow/core/platform/logging.h"namespace tensorflow {void Shard(int max_parallelism, thread::ThreadPool* workers, int64 total,           int64 cost_per_unit, std::function<void(int64, int64)> work) {  CHECK_GE(total, 0);  if (total == 0) {    return;  }  if (max_parallelism <= 1) {    // Just inline the whole work since we only have 1 thread (core).    work(0, total);    return;  }  if (max_parallelism >= workers->NumThreads()) {    workers->ParallelFor(total, cost_per_unit, work);    return;  }  cost_per_unit = std::max(1LL, cost_per_unit);  // We shard [0, total) into "num_shards" shards.  //   1 <= num_shards <= num worker threads  //  // If total * cost_per_unit is small, it is not worth shard too  // much. Let us assume each cost unit is 1ns, kMinCostPerShard=10000  // is 10us.  static const int64 kMinCostPerShard = 10000;  const int num_shards =      std::max<int>(1, std::min(static_cast<int64>(max_parallelism),                                total * cost_per_unit / kMinCostPerShard));  // Each shard contains up to "block_size" units. [0, total) is sharded  // into:  //   [0, block_size), [block_size, 2*block_size), ...  // The 1st shard is done by the caller thread and the other shards  // are dispatched to the worker threads. The last shard may be smaller than  // block_size.  const int64 block_size = (total + num_shards - 1) / num_shards;  CHECK_GT(block_size, 0);  // total > 0 guarantees this.  if (block_size >= total) {    work(0, total);    return;  }  const int num_shards_used = (total + block_size - 1) / block_size;  BlockingCounter counter(num_shards_used - 1);  for (int64 start = block_size; start < total; start += block_size) {    auto limit = std::min(start + block_size, total);    workers->Schedule([&work, &counter, start, limit]() {      work(start, limit);        // Compute the shard.      counter.DecrementCount();  // The shard is done.    });  }  // Inline execute the 1st shard.  work(0, std::min(block_size, total));  counter.Wait();}}  // end namespace tensorflow