如何加快PyMC马尔可夫模型？

有没有办法加快这个简单的PyMC模型？在20-40个数据点上,需要约5-11秒才能适应.

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)

只有40个数据点,运行需要11秒:

running on 40 points
 [-----------------100%-----------------] 2000 of 2000 complete in 11.5 sec
11.54 secs to run

(80分需要20秒).这是一个玩具的例子.Lambda()确定转换的内部表达式实际上更复杂.这种基本代码结构是灵活的(而使用转换矩阵编码模型的灵活性较低).有没有办法保持类似的代码结构,但获得更好的性能？如有必要,很高兴切换到PyMC3.谢谢.