以下代码使用 beta 作为先验来预测二项分布的 p。不知何故,有时,我得到毫无意义的结果(接受率 = 0)。当我用 pymc3 编写相同的逻辑时,我没有问题。我看不到我在这里缺少什么。
import numpy as np
import tensorflow as tf
import tensorflow_probability as tfp
import edward2 as ed
from pymc3.stats import hpd
import numpy as np
import seaborn
import matplotlib.pyplot as plt
p_true = .15
N = [10, 100, 1000]
successN = np.random.binomial(p=p_true, n=N)
print(N)
print(successN)
def beta_binomial(N):
p = ed.Beta(
concentration1=tf.ones( len(N) ),
concentration0=tf.ones( len(N) ),
name='p'
)
return ed.Binomial(total_count=N, probs=p, name='obs')
log_joint = ed.make_log_joint_fn(beta_binomial)
def target_log_prob_fn(p):
return log_joint(N=N, p=p, obs=successN)
#kernel = tfp.mcmc.HamiltonianMonteCarlo(
# target_log_prob_fn=target_log_prob_fn,
# step_size=0.01,
# num_leapfrog_steps=5)
kernel = tfp.mcmc.NoUTurnSampler(
target_log_prob_fn=target_log_prob_fn,
step_size=.01
)
trace, kernel_results = tfp.mcmc.sample_chain(
num_results=1000,
kernel=kernel,
num_burnin_steps=500,
current_state=[
tf.random.uniform(( len(N) ,))
],
trace_fn=(lambda current_state, kernel_results: kernel_results),
return_final_kernel_results=False)
p, = trace
p = p.numpy()
print(p.shape)
print('acceptance rate ', np.mean(kernel_results.is_accepted))
def printSummary(name, v):
print(name, v.shape)
print(np.mean(v, axis=0))
print(hpd(v))
printSummary('p', p)
for data in p.T:
print(data.shape)
seaborn.distplot(data, kde=False)
plt.savefig('p.png')
图书馆:
pip install -U pip
pip install -e git+https://github.com/google/edward2.git@4a8ed9f5b1dac0190867c48e816168f9f28b5129#egg=edward2
pip install https://storage.googleapis.com/tensorflow/linux/cpu/tensorflow-2.0.0-cp37-cp37m-manylinux2010_x86_64.whl#egg=tensorflow
pip install tensorflow-probability