sigmoid的Jax - autograd总是返回nan

Question

我正在尝试微分一个函数，该函数近似包含在2个限制(截断高斯)内的高斯的分数，给定一个偏移的平均值。jnp.grad不让我区分加起来的布尔过滤器(注释行)，所以我不得不临时使用一个sigmoid。

然而，现在当截断边界很高时，梯度总是nan，我不明白为什么。

在下面的例子中，我计算了一个均值为0，std=1为0的高斯曲线的梯度，然后用x对其进行平移。

如果我减小边界，那么函数就会按预期运行。但这不是一个解决方案。当边界较高时，belows始终变为1。但是如果是这种情况，x对下面没有影响，那么它对梯度的贡献应该是0而不是nan。但是如果我返回belows[0][0]而不是jnp.mean(filt, axis=0)，我仍然会得到nan。

有什么想法吗？提前感谢( github上也有一个公开的问题)

import os

from tqdm import tqdm

os.environ["XLA_FLAGS"] = '--xla_force_host_platform_device_count=4' # Use 8 CPU devices
import numpy as np
from jax.config import config
config.update("jax_enable_x64", True)
import jax
import jax.numpy as jnp
from jax import vmap

from functools import reduce

def sigmoid(x, scale=100):
    return 1 / (1 + jnp.exp(-x*scale))

def above_lower(x, l, scale=100):
    return sigmoid(x - l, scale)

def below_upper(x, u, scale=100):
    return 1 - sigmoid(x - u, scale)

def combine_soft_filters(a):
    return jnp.prod(jnp.stack(a), axis=0)


def fraction_not_truncated(mu, v, limits, stdnorm_samples):
    L = jnp.linalg.cholesky(v)
    y = vmap(lambda x: jnp.dot(L, x))(stdnorm_samples) + mu
    # filt = reduce(jnp.logical_and, [(y[..., i] > l) & (y[..., i] < u) for i, (l, u) in enumerate(limits)])
    aboves = [above_lower(y[..., i], l) for i, (l, u) in enumerate(limits)]
    belows = [below_upper(y[..., i], u) for i, (l, u) in enumerate(limits)]
    filt = combine_soft_filters(aboves+belows)
    return jnp.mean(filt, axis=0)

limits = np.array([
        [0.,1000],
])

stdnorm_samples = np.random.multivariate_normal([0], np.eye(1), size=1000)

def func(x):
    return fraction_not_truncated(jnp.zeros(1)+x, jnp.eye(1), limits, stdnorm_samples)

_x = np.linspace(-2, 2, 500)
gradfunc = jax.grad(func)
vals = [func(x) for x in tqdm(_x)]
grads = [gradfunc(x) for x in tqdm(_x)]
print(vals)
print(grads)
import matplotlib.pyplot as plt
plt.plot(_x, np.asarray(vals))
plt.ylabel('f(x)')
plt.twinx()
plt.plot(_x, np.asarray(grads), c='r')
plt.ylabel("f(x)'")
plt.title('Fraction not truncated')
plt.axhline(0, color='k', alpha=0.2)
plt.xlabel('shift')
plt.tight_layout()
plt.show()

​

​

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Answer 1

问题是，您的sigmoid函数的实现方式使得自动确定的梯度对于较大的x负值不稳定

import jax.numpy as jnp
import jax

def sigmoid(x, scale=100):
    return 1 / (1 + jnp.exp(-x*scale))

print(jax.grad(sigmoid)(-1000.0))
# nan

您可以使用jax.make_jaxpr函数来内省由自动确定的渐变生成的操作(注释是我的注释)，从而了解发生这种情况的原因：

>>> jax.make_jaxpr(jax.grad(sigmoid))(-1000.0)
{ lambda  ; a.                    # a = -1000
  let b = neg a                   # b = 1000
      c = mul b 100.0             # c = 100,000
      d = exp c                   # d = inf
      e = add d 1.0
      _ = div 1.0 e
      f = integer_pow[ y=-2 ] e   # f = 0
      g = mul 1.0 f               # g = 0
      h = mul g 1.0               # h = 0
      i = neg h                   # i = 0
      j = mul i d                 # j = 0 * inf = NaN
      k = mul j 100.0             # k = NaN
      l = neg k                   # l = NaN
  in (l,) }                       # return NaN

这是64位浮点算术失败的情况之一:它没有处理exp(100000)这样的数字的范围。

那么你能做什么呢？一个重要的选项是使用custom derivative rule来告诉autodiff如何以更稳定的方式处理sigmoid函数。但是，在这种情况下，一个更简单的选择是用在autodiff转换下表现更好的东西来重新表示sigmoid函数。一种选择是：

def sigmoid(x, scale=100):
    return 0.5 * (jnp.tanh(x * scale / 2) + 1)

在脚本中使用此版本可以解决此问题。