2

这是一个简单的例子,它对两个高斯 pdf 的乘积进行数值积分。其中一个高斯是固定的,均值始终为 0。另一个高斯的均值不同:

import time

import jax.numpy as np
from jax import jit
from jax.scipy.stats.norm import pdf

# set up evaluation points for numerical integration
integr_resolution = 6400
lower_bound = -100
upper_bound = 100
integr_grid = np.linspace(lower_bound, upper_bound, integr_resolution)
proba = pdf(integr_grid)
integration_weight = (upper_bound - lower_bound) / integr_resolution


# integrate with new mean
def integrate(mu_new):
    x_new = integr_grid - mu_new

    proba_new = pdf(x_new)
    total_proba = sum(proba * proba_new * integration_weight)

    return total_proba


print('starting jit')
start = time.perf_counter()
integrate = jit(integrate)
integrate(1)
stop = time.perf_counter()
print('took: ', stop - start)

该函数看起来很简单,但它根本无法扩展。以下列表包含成对的(integr_resolution 的值,运行代码所用的时间):

  • 100 | 0.107s
  • 200 | 0.23s
  • 400 | 0.537s
  • 800 | 1.52s
  • 1600 | 5.2s
  • 3200 | 19 岁
  • 6400 | 134s

作为参考,应用到的 unjitted 函数integr_resolution=6400需要 0.02s。

我认为这可能与函数正在访问全局变量这一事实有关。但是移动代码以在函数内部设置积分点对时序没有显着影响。以下代码需要 5.36 秒才能运行。它对应于先前花费 5.2 秒的 1600 表条目:

# integrate with new mean
def integrate(mu_new):
    # set up evaluation points for numerical integration
    integr_resolution = 1600
    lower_bound = -100
    upper_bound = 100
    integr_grid = np.linspace(lower_bound, upper_bound, integr_resolution)
    proba = pdf(integr_grid)
    integration_weight = (upper_bound - lower_bound) / integr_resolution

    x_new = integr_grid - mu_new

    proba_new = pdf(x_new)
    total_proba = sum(proba * proba_new * integration_weight)

    return total_proba

这里发生了什么?

4

1 回答 1

5

我也在https://github.com/google/jax/issues/1776回答了这个问题,但也在这里添加了答案。

这是因为代码使用sum了它应该使用的地方np.sum

sum是一个 Python 内置程序,它提取序列的每个元素并使用+运算符将​​它们一一相加。这具有构建大型、展开的添加链的效果,XLA 需要很长时间才能编译。

如果使用np.sum,那么 JAX 会构建一个 XLA 归约运算符,它的编译速度要快得多。

并且只是为了说明我是如何解决这个问题的:我使用了jax.make_jaxpr,它转储了 JAX 的函数的内部跟踪表示。在这里,它显示:

In [3]: import jax

In [4]: jax.make_jaxpr(integrate)(1)
Out[4]:
{ lambda b c ;  ; a.
  let d = convert_element_type[ new_dtype=float32
                                old_dtype=int32 ] a
      e = sub c d
      f = sub e 0.0
      g = pow f 2.0
      h = div g 1.0
      i = add 1.8378770351409912 h
      j = neg i
      k = div j 2.0
      l = exp k
      m = mul b l
      n = mul m 2.0
      o = slice[ start_indices=(0,)
                 limit_indices=(1,)
                 strides=(1,)
                 operand_shape=(100,) ] n
      p = reshape[ new_sizes=()
                   dimensions=None
                   old_sizes=(1,) ] o
      q = add p 0.0
      r = slice[ start_indices=(1,)
                 limit_indices=(2,)
                 strides=(1,)
                 operand_shape=(100,) ] n
      s = reshape[ new_sizes=()
                   dimensions=None
                   old_sizes=(1,) ] r
      t = add q s
      u = slice[ start_indices=(2,)
                 limit_indices=(3,)
                 strides=(1,)
                 operand_shape=(100,) ] n
      v = reshape[ new_sizes=()
                   dimensions=None
                   old_sizes=(1,) ] u
      w = add t v
      x = slice[ start_indices=(3,)
                 limit_indices=(4,)
                 strides=(1,)
                 operand_shape=(100,) ] n
      y = reshape[ new_sizes=()
                   dimensions=None
                   old_sizes=(1,) ] x
      z = add w y
... similarly ...

然后很明显为什么这很慢:程序非常大。

对比np.sum版本:

In [5]: def integrate(mu_new):
   ...:     x_new = integr_grid - mu_new
   ...:
   ...:     proba_new = pdf(x_new)
   ...:     total_proba = np.sum(proba * proba_new * integration_weight)
   ...:
   ...:     return total_proba
   ...:

In [6]: jax.make_jaxpr(integrate)(1)
Out[6]:
{ lambda b c ;  ; a.
  let d = convert_element_type[ new_dtype=float32
                                old_dtype=int32 ] a
      e = sub c d
      f = sub e 0.0
      g = pow f 2.0
      h = div g 1.0
      i = add 1.8378770351409912 h
      j = neg i
      k = div j 2.0
      l = exp k
      m = mul b l
      n = mul m 2.0
      o = reduce_sum[ axes=(0,)
                      input_shape=(100,) ] n
  in [o] }

希望有帮助!

于 2019-11-27T16:56:47.020 回答