为TensorFlow 2中的许多伪数据实现优化功能

时间：2019-04-26 10:40:54

标签： python tensorflow tensorflow2.0 tensorflow-probability

我的最终目标是模拟似然比测试统计数据，但是，我的核心问题是我不了解如何让TensorFlow 2对不同的数据输入执行许多优化。这是我的尝试，希望它可以使您了解我正在尝试的事情：

import tensorflow as tf
import tensorflow_probability as tfp
from tensorflow_probability import distributions as tfd
import numpy as np

# Bunch of independent Poisson distributions that we want to combine
poises0 = [tfp.distributions.Poisson(rate = 10) for i in range(5)]

# Construct joint distributions
joint0 = tfd.JointDistributionSequential(poises0)

# Generate samples
N = int(1e3)
samples0 = joint0.sample(N)

# Now we need the same distributions but with floating parameters,
# and need to define the function to be minimised
mus = [tf.Variable(np.random.randn(), name='mu{0}'.format(i)) for i in range(5)]

#@tf.function
def loss():
    poises_free = [tfp.distributions.Poisson(rate = mus[i]) for i in range(5)]
    joint_free = tfd.JointDistributionSequential(poises_free)
    # Construct (half of) test statistic
    return -2*(joint_free.log_prob(samples0))

# Minimise (for all samples? Apparently not?)
opt = tf.optimizers.SGD(0.1).minimize(loss,var_list=mus)

print(mus)
print(loss())
print(opt)
quit()

输出：

[<tf.Variable 'mu0:0' shape=() dtype=float32, numpy=53387.016>, <tf.Variable 'mu1:0' shape=() dtype=float32, numpy=2540.568>, <tf.Variable 'mu2:0' shape=() dtype=float32, numpy=-5136.6226>, <tf.Variable 'mu3:0' shape=() dtype=float32, numpy=-3714.5227>, <tf.Variable 'mu4:0' shape=() dtype=float32, numpy=1062.9396>]
tf.Tensor(
[nan nan nan nan ... nan nan nan], shape=(1000,), dtype=float32)
<tf.Variable 'UnreadVariable' shape=() dtype=int64, numpy=1>

最后我要计算测试统计量

q = -2*joint0.log_prob(samples0) - loss()

并表明它具有5个自由度的卡方分布。

我是TensorFlow的新手，所以也许我这样做完全错了，但是希望您对我想要的东西有所了解。

编辑：

所以我玩了更多，并且我想TensorFlow根本不会像我想象的那样对输入张量进行优化。也许可以，但是我需要进行不同的设置，即一次为所有最小化提供一个张量输入参数和一个巨大的联合损失函数？

我也尝试通过一个简单的循环来做事，只是为了看看会发生什么。正如预料的那样，它速度缓慢，但我什至没有得到正确的答案：

poises0 = [tfp.distributions.Poisson(rate = 10) for i in range(5)]
joint0 = tfd.JointDistributionSequential(poises0)

N = int(5e2)
samples0 = joint0.sample(N)

mus = [tf.Variable(10., name='mu{0}'.format(i)) for i in range(5)]

#@tf.function
def loss(xi):
    def loss_inner():
        poises_free = [tfp.distributions.Poisson(rate = mus[i]) for i in range(5)]
        joint_free = tfd.JointDistributionSequential(poises_free)
        # Construct (half of) test statistic
        return -2*(joint_free.log_prob(xi))
    return loss_inner

# Minimise
# I think I have to loop over the samples... bit lame. Can perhaps parallelise though.
q = []
for i in range(N):
   xi = [x[i] for x in samples0]
   opt = tf.optimizers.SGD(0.1).minimize(loss=loss(xi),var_list=mus)
   q += [-2*joint0.log_prob(xi) - loss(xi)()]

fig = plt.figure()
ax = fig.add_subplot(111)
sns.distplot(q, kde=False, ax=ax, norm_hist=True)
qx = np.linspace(np.min(q),np.max(q),1000)
qy = np.exp(tfd.Chi2(df=5).log_prob(qx))
sns.lineplot(qx,qy)
plt.show()

输出不是DOF = 5的卡方分布。确实，检验统计量经常具有负值，这意味着优化结果通常比零假设的拟合度差，这应该是不可能的。

Not a chi-squared distribution with DOF=5

编辑2：

这里是“怪物”解决方案的尝试，在该解决方案中，我一次将每个伪数据实现的不同输入变量的巨型网络最小化了。这感觉更像TensorFlow可能擅长的事情，尽管我觉得一旦我进入大量的伪数据集，我将耗尽RAM。不过，我可能可以遍历一批伪数据。

poises0 = [tfp.distributions.Poisson(rate = 10) for i in range(5)]
joint0 = tfd.JointDistributionSequential(poises0)

N = int(5e3)
samples0 = joint0.sample(N)

mus = [tf.Variable(10*np.ones(N, dtype='float32'), name='mu{0}'.format(i)) for i in range(5)]

poises_free = [tfp.distributions.Poisson(rate = mus[i]) for i in range(5)]
joint_free = tfd.JointDistributionSequential(poises_free)
qM = -2*(joint_free.log_prob(samples0))

@tf.function
def loss():
    return tf.math.reduce_sum(qM,axis=0)

# Minimise
opt = tf.optimizers.SGD(0.1).minimize(loss,var_list=mus)
print("parameters:", mus)
print("loss:", loss())
q0 =-2*joint0.log_prob(samples0)
print("q0:", q0)
print("qM:", qM)
q = q0 - qM

fig = plt.figure()
ax = fig.add_subplot(111)
sns.distplot(q, kde=False, ax=ax, norm_hist=True)
qx = np.linspace(np.min(q),np.max(q),1000)
qy = np.exp(tfd.Chi2(df=5).log_prob(qx))
sns.lineplot(qx,qy)
plt.show()

不幸的是，我现在得到了错误：

Traceback (most recent call last):
  File "testing3.py", line 35, in <module>
    opt = tf.optimizers.SGD(0.1).minimize(loss,var_list=mus)   
  File "/home/farmer/anaconda3/envs/general/lib/python3.6/site-packages/tensorflow/python/keras/optimizer_v2/optimizer_v2.py", line 298, in minimize
    return self.apply_gradients(grads_and_vars, name=name)
  File "/home/farmer/anaconda3/envs/general/lib/python3.6/site-packages/tensorflow/python/keras/optimizer_v2/optimizer_v2.py", line 396, in apply_gradients
    grads_and_vars = _filter_grads(grads_and_vars)
  File "/home/farmer/anaconda3/envs/general/lib/python3.6/site-packages/tensorflow/python/keras/optimizer_v2/optimizer_v2.py", line 924, in _filter_grads
    ([v.name for _, v in grads_and_vars],))
ValueError: No gradients provided for any variable: ['mu0:0', 'mu1:0', 'mu2:0', 'mu3:0', 'mu4:0'].
我认为

是一种基本的错误。我想我只是不明白TensorFlow如何跟踪需要计算的导数。如果我在损失函数内部而不是外部定义变量，似乎一切正常，但是我需要在外部使用它们以便以后访问它们的值。所以我想我在这里听不懂。

1 个答案:

答案 0 :(得分：0)

好的，这就是我的想法。我缺少的关键是：

  1. 将输入变量定义为大张量，以便所有最小化都可以立即发生。
  2. 一次构造所有损失最小的单一组合损失函数
  3. 在损失函数定义内构造用于损失计算的中间变量，以便TensorFlow可以跟踪梯度（我认为minimize函数将损失函数包装在梯度带或类似的磁带中）。
  4. 将损失函数定义为类的一部分，以便可以存储中间变量。
  5. minimize仅执行最小化的一个步骤，因此我们需要对其进行多次遍历，直到根据某些准则收敛为止。
  6. 由于Poisson分布的均值小于零的无效性，我遇到了一些NaN。因此，我需要为输入变量添加约束。

有了这个，我现在可以在笔记本电脑上像10秒钟那样完成一百万次最小化操作，这真是太好了！

import tensorflow as tf
import tensorflow_probability as tfp
from tensorflow_probability import distributions as tfd
import seaborn as sns
import numpy as np
import matplotlib.pyplot as plt

# Bunch of independent Poisson distributions that we want to combine
poises0 = [tfd.Poisson(rate = 10) for i in range(5)]

# Construct joint distributions
joint0 = tfd.JointDistributionSequential(poises0)

N = int(1e6)
samples0 = joint0.sample(N)

class Model(object):
  def __init__(self):
     self.mus = [tf.Variable(10*np.ones(N, dtype='float32'), name='mu{0}'.format(i),
                    constraint=lambda x: tf.clip_by_value(x, 0.000001, np.infty)) for i in range(5)]

  def loss(self):
     poises_free = [tfd.Poisson(rate = self.mus[i]) for i in range(5)]
     joint_free = tfd.JointDistributionSequential(poises_free)
     # Construct (half of) test statistic
     self.qM = -2*(joint_free.log_prob(samples0))
     self.last_loss = tf.math.reduce_sum(self.qM,axis=0)
     return self.last_loss

model = Model()

# Minimise
tol = 0.01 * N
delta_loss = 1e99
prev_loss = 1e99
i = 0
print("tol:", tol)
while delta_loss > tol:
    opt = tf.optimizers.SGD(0.1).minimize(model.loss,var_list=model.mus)
    delta_loss = np.abs(prev_loss - model.last_loss)
    print("i:", i," delta_loss:", delta_loss)
    i+=1
    prev_loss = model.last_loss

q0 =-2*joint0.log_prob(samples0)
q = q0 - model.qM

print("parameters:", model.mus)
print("loss:", model.last_loss)
print("q0:", q0)
print("qM:", model.qM)

fig = plt.figure()
ax = fig.add_subplot(111)
sns.distplot(q, kde=False, ax=ax, norm_hist=True)
qx = np.linspace(np.min(q),np.max(q),1000)
qy = np.exp(tfd.Chi2(df=5).log_prob(qx))
sns.lineplot(qx,qy)
plt.show()

输出：

tol: 10000.0
i: 0  delta_loss: inf
i: 1  delta_loss: 197840.0
i: 2  delta_loss: 189366.0
i: 3  delta_loss: 181456.0
i: 4  delta_loss: 174040.0
i: 5  delta_loss: 167042.0
i: 6  delta_loss: 160448.0
i: 7  delta_loss: 154216.0
i: 8  delta_loss: 148310.0
i: 9  delta_loss: 142696.0
i: 10  delta_loss: 137352.0
i: 11  delta_loss: 132268.0
i: 12  delta_loss: 127404.0
...
i: 69  delta_loss: 11894.0
i: 70  delta_loss: 11344.0
i: 71  delta_loss: 10824.0
i: 72  delta_loss: 10318.0
i: 73  delta_loss: 9860.0
parameters: [<tf.Variable 'mu0:0' shape=(1000000,) dtype=float32, numpy=
array([ 6.5849004, 14.81182  ,  7.506216 , ..., 10.       , 11.491933 ,
       10.760278 ], dtype=float32)>, <tf.Variable 'mu1:0' shape=(1000000,) dtype=float32, numpy=
array([12.881036,  7.506216, 12.881036, ...,  7.506216, 14.186232,
       10.760278], dtype=float32)>, <tf.Variable 'mu2:0' shape=(1000000,) dtype=float32, numpy=
array([16.01586  ,  8.378036 , 12.198007 , ...,  6.5849004, 12.198007 ,
        8.378036 ], dtype=float32)>, <tf.Variable 'mu3:0' shape=(1000000,) dtype=float32, numpy=
array([10.      ,  7.506216, 12.198007, ...,  9.207426, 10.760278,
       11.491933], dtype=float32)>, <tf.Variable 'mu4:0' shape=(1000000,) dtype=float32, numpy=
array([ 8.378036 , 14.81182  , 10.       , ...,  6.5849004, 12.198007 ,
       10.760278 ], dtype=float32)>]
loss: tf.Tensor(20760090.0, shape=(), dtype=float32)
q0: tf.Tensor([31.144037 31.440613 25.355555 ... 24.183338 27.195362 22.123463], shape=(1000000,), dtype=float32)
qM: tf.Tensor([21.74377  21.64162  21.526024 ... 19.488544 22.40428  21.08519 ], shape=(1000000,), dtype=float32)

结果现在是卡方DOF = 5！或至少很接近。 enter image description here

相关问题
最新问题