我正在尝试将Beta分布拟合到具有离散分数(1,2,3,4,5)的调查结果中。 为此，我需要TensorFlow概率中Beta的有效log_prob。但是，在Beta中如何处理批处理存在问题。 这是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]

相同代码似乎可以在正态分布下正常工作...

我在这里做什么错了？

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.,3.]).astype(np.float32)
beta = np.array([1.,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 (1,5)
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（Chris Suter）所建议的那样。这是手工广播解决方案：

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.,numdata)
data *= 5.0
data = np.ceil(data)
data = data[:,None].astype(np.float32)

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

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

bt = tfd.Beta(concentration1=alpha,concentration0=beta)
scbt = tfd.TransformedDistribution(
            distribution=bt,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，以上解决方案不起作用。 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.,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,scale=5.))

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

        return qdist    

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

这最终会显示一条错误消息：

  

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]等级：[无，无]