


Why does the performance of moving_avg_concurrent2 not improve with the increase of concurrent execution?
moving_avg_concurrent2 splits the list into smaller pieces and uses a single goroutine to handle each piece. For some reason (it's not clear why), this function using one goroutine is faster than moving_avg_serial4, but using multiple goroutines starts to perform worse than moving_avg_serial4.
Why moving_avg_concurrent3 is much slower than moving_avg_serial4?
The performance of moving_avg_concurrent3 is worse than moving_avg_serial4 when using a goroutine. Although increasing num_goroutines can improve performance, it is still worse than moving_avg_serial4.
Even though goroutines are lightweight, they are not completely free, is it possible that the overhead incurred is so large that it is even slower than moving_avg_serial4?
Yes, although goroutines are lightweight, they are not free. When using multiple goroutines, the overhead of launching, managing, and scheduling them may outweigh the benefits of increased parallelism.
Code
Function:
// 返回包含输入移动平均值的列表(已提供,即未优化) func moving_avg_serial(input []float64, window_size int) []float64 { first_time := true var output = make([]float64, len(input)) if len(input) > 0 { var buffer = make([]float64, window_size) // 初始化缓冲区为 NaN for i := range buffer { buffer[i] = math.NaN() } for i, val := range input { old_val := buffer[int((math.Mod(float64(i), float64(window_size))))] buffer[int((math.Mod(float64(i), float64(window_size))))] = val if !NaN_in_slice(buffer) && first_time { sum := 0.0 for _, entry := range buffer { sum += entry } output[i] = sum / float64(window_size) first_time = false } else if i > 0 && !math.IsNaN(output[i-1]) && !NaN_in_slice(buffer) { output[i] = output[i-1] + (val-old_val)/float64(window_size) // 无循环的解决方案 } else { output[i] = math.NaN() } } } else { // 空输入 fmt.Println("moving_avg is panicking!") panic(fmt.Sprintf("%v", input)) } return output } // 返回包含输入移动平均值的列表 // 重新排列控制结构以利用短路求值 func moving_avg_serial4(input []float64, window_size int) []float64 { first_time := true var output = make([]float64, len(input)) if len(input) > 0 { var buffer = make([]float64, window_size) // 初始化缓冲区为 NaN for i := range buffer { buffer[i] = math.NaN() } for i := range input { // fmt.Printf("in mvg_avg4: i=%v\n", i) old_val := buffer[int((math.Mod(float64(i), float64(window_size))))] buffer[int((math.Mod(float64(i), float64(window_size))))] = input[i] if first_time && !NaN_in_slice(buffer) { sum := 0.0 for j := range buffer { sum += buffer[j] } output[i] = sum / float64(window_size) first_time = false } else if i > 0 && !math.IsNaN(output[i-1]) /* && !NaN_in_slice(buffer)*/ { output[i] = output[i-1] + (input[i]-old_val)/float64(window_size) // 无循环的解决方案 } else { output[i] = math.NaN() } } } else { // 空输入 fmt.Println("moving_avg is panicking!") panic(fmt.Sprintf("%v", input)) } return output } // 返回包含输入移动平均值的列表 // 将列表拆分为较小的片段以使用 goroutine,但不使用串行版本,即我们仅在开头具有 NaN,因此希望减少一些开销 // 仍然不能扩展(随着大小和 num_goroutines 的增加,性能下降) func moving_avg_concurrent2(input []float64, window_size, num_goroutines int) []float64 { var output = make([]float64, window_size-1, len(input)) for i := 0; i 0 { num_items := len(input) - (window_size - 1) var barrier_wg sync.WaitGroup n := num_items / num_goroutines go_avg := make([][]float64, num_goroutines) for i := 0; i 0 { num_windows := len(input) - (window_size - 1) var output = make([]float64, len(input)) for i := 0; i
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