有没有办法加速这个简单的PyMC模型？在20-40个数据点上，拟合需要大约5-11秒。
import pymc
import time
import numpy as np
from collections import OrderedDict
# prior probability of rain
p_rain = 0.5
variables = OrderedDict()
# rain observations
data = [True, True, True, True, True,
False, False, False, False, False]*4
num_steps = len(data)
p_rain_given_rain = 0.9
p_rain_given_norain = 0.2
p_umbrella_given_rain = 0.8
p_umbrella_given_norain = 0.3
for n in range(num_steps):
if n == 0:
# Rain node at time t = 0
rain = pymc.Bernoulli("rain_%d" %(n), p_rain)
else:
rain_trans = \
pymc.Lambda("rain_trans",
lambda prev_rain=variables["rain_%d" %(n-1)]: \
prev_rain*p_rain_given_rain + (1-prev_rain)*p_rain_given_norain)
rain = pymc.Bernoulli("rain_%d" %(n), p=rain_trans)
umbrella_obs = \
pymc.Lambda("umbrella_obs",
lambda rain=rain: \
rain*p_umbrella_given_rain + (1-rain)*p_umbrella_given_norain)
umbrella = pymc.Bernoulli("umbrella_%d" %(n), p=umbrella_obs,
observed=True,
value=data[n])
variables["rain_%d" %(n)] = rain
variables["umbrella_%d" %(n)] = umbrella
print "running on %d points" %(len(data))
all_vars = variables.values()
t_start = time.time()
model = pymc.Model(all_vars)
m = pymc.MCMC(model)
m.sample(iter=2000)
t_end = time.time()
print "\n%.2f secs to run" %(t_end - t_start)
仅有40个数据点，运行需要11秒钟：
running on 40 points
[-----------------100%-----------------] 2000 of 2000 complete in 11.5 sec
11.54 secs to run
(使用80个点需要20秒). 这只是一个玩具示例。在实践中，确定转换的Lambda()内部表达式更加复杂。这种基本代码结构是灵活的（而用转换矩阵对模型进行编码则不太灵活）。有没有办法保持类似的代码结构但获得更好的性能？如果必要，可以切换到PyMC3。谢谢。