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我正在尝试将 beta 分布拟合到具有离散分数的调查结果中(1, 2, 3, 4, 5)。为此，我需要一个 TensorFlow 概率中 Beta 的工作 log_prob。但是，在 Beta 中如何处理批处理存在问题。这是一个给我一个错误的最小示例：

InvalidArgumentError：a 和 x 的形状不一致：[3] 与 [1000,1] [Op:Betainc]

相同的代码似乎适用于正态分布......

我在这里做错了什么？

import numpy as np
import tensorflow_probability as tfp
tfd = tfp.distributions

#Generate fake data
np.random.seed(2)
data = np.random.beta(2.,2.,1000)*5.0
data = np.ceil(data)
data = data[:,None]

# Create a batch of three Beta distributions.
alpha = np.array([1., 2., 3.]).astype(np.float32)
beta = np.array([1., 2., 3.]).astype(np.float32)

bt = tfd.Beta(alpha, beta)
#bt = tfd.Normal(loc=alpha, scale=beta)


#Scale beta to 0-5
scbt = tfd.TransformedDistribution(
            distribution=bt,
            bijector=tfp.bijectors.AffineScalar(
            shift=0.,
            scale=5.))

# quantize beta to (1,2,3,4,5)
qdist = tfd.QuantizedDistribution(distribution=scbt,low=1,high=5)

#calc log_prob for 3 distributions
print(np.sum(qdist.log_prob(data),axis=0))
print(qdist.log_prob(data).shape)

TensorFlow 2.0.0 tensorflow_probability 0.8.0

编辑：正如克里斯苏特所建议的那样。这是手动广播解决方案：

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

#Generate fake data
numdata = 100
numbeta = 3
np.random.seed(2)
data = np.random.beta(2.,2.,numdata)
data *= 5.0
data = np.ceil(data)
data = data[:,None].astype(np.float32)

#alpha and beta [[1., 2., 3.]]
alpha = np.expand_dims(np.arange(1,4),0).astype(np.float32)
beta =  np.expand_dims(np.arange(1,4),0).astype(np.float32)

#tile to compensate for betainc
alpha = tf.tile(alpha,[numdata,1])
beta = tf.tile(beta,[numdata,1])
data = tf.tile(data,[1,numbeta])

bt = tfd.Beta(concentration1=alpha, concentration0=beta)
scbt = tfd.TransformedDistribution(
            distribution=bt,
            bijector=tfp.bijectors.AffineScalar(
            shift=0.,
            scale=5.))

# quantize beta to (1,2,3,4,5)
qdist = tfd.QuantizedDistribution(distribution=scbt,low=1,high=5)
#calc log_prob for numbeta number of distributions
print(np.sum(qdist.log_prob(data),axis=0))

EDIT2：当我尝试将其应用于 MCMC 采样时，上述解决方案不起作用。新代码如下所示：

import os
os.environ["CUDA_VISIBLE_DEVICES"] = "-1"
from time import time

import tensorflow as tf
import tensorflow_probability as tfp
tfd = tfp.distributions
import numpy as np

#Generate fake data
numdata = 100
np.random.seed(2)
data = np.random.beta(2.,2.,numdata)
data *= 5.0
data = np.ceil(data)
data = data[:,None].astype(np.float32)

@tf.function
def sample_chain():
    #Parameters of MCMC
    num_burnin_steps = 300
    num_results = 200
    num_chains = 50
    step_size = 0.01

    #data tensor
    outcomes =  tf.convert_to_tensor(data, dtype=tf.float32)

    def modeldist(alpha,beta):
        bt = tfd.Beta(concentration1=alpha, concentration0=beta)
        scbt = tfd.TransformedDistribution(
            distribution=bt,
            bijector=tfp.bijectors.AffineScalar(
            shift=0.,
            scale=5.))

        # quantize beta to (1,2,3,4,5)
        qdist = tfd.QuantizedDistribution(distribution=scbt,low=1,high=5)

        return qdist    

    def joint_log_prob(con1,con0):
        #manual broadcast
        tcon1 = tf.tile(con1[None,:],[numdata,1])
        tcon0 = tf.tile(con0[None,:],[numdata,1])
        toutcomes = tf.tile(outcomes,[1,num_chains])

        #model distribution with manual broadcast
        dist = modeldist(tcon1,tcon0)

        #joint log prob
        return tf.reduce_sum(dist.log_prob(toutcomes),axis=0)

    kernel = tfp.mcmc.HamiltonianMonteCarlo(
        target_log_prob_fn=joint_log_prob,
        num_leapfrog_steps=5,
        step_size=step_size)

    kernel = tfp.mcmc.SimpleStepSizeAdaptation(
        inner_kernel=kernel, num_adaptation_steps=int(num_burnin_steps * 0.8))

    init_state = [tf.identity(tf.random.uniform([num_chains])*10.0,name='init_alpha'),
                  tf.identity(tf.random.uniform([num_chains])*10.0,name='init_beta')]

    samples, [step_size, is_accepted] = tfp.mcmc.sample_chain(
        num_results=num_results,
        num_burnin_steps=num_burnin_steps,
        current_state=init_state,
        kernel=kernel,
        trace_fn=lambda _, pkr: [pkr.inner_results.accepted_results.step_size,
                                pkr.inner_results.is_accepted])

    return samples

samples = sample_chain()

这最终会出现一条错误消息：

ValueError：遇到None渐变。fn_arg_list：[tf.Tensor 'init_alpha:0' shape=(50,) dtype=float32, tf.Tensor 'init_beta:0' shape=(50,) dtype=float32] 毕业生：[无，无]

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1 回答 1

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遗憾的是 tf.math.betainc 目前不支持广播，这会导致 QuantizedDistribution 调用的 cdf 计算失败。如果您必须使用 Beta，我能想到的唯一解决方法是通过平铺数据和 Beta 参数来“手动”广播。

或者，您可以使用Kumaraswamy分布，它类似于 Beta，但具有更好的分析属性。

于 2019-11-19T16:17:05.907 回答