我正在尝试使用最大后验估计来估计泊松过程的速率,其中速率随时间变化。这是一个简化的示例,其速率呈线性变化 (λ = ax+b):
import numpy as np
import pymc
# Observation
a_actual = 1.3
b_actual = 2.0
t = np.arange(10)
obs = np.random.poisson(a_actual * t + b_actual)
# Model
a = pymc.Uniform(name='a', value=1., lower=0, upper=10)
b = pymc.Uniform(name='b', value=1., lower=0, upper=10)
@pymc.deterministic
def linear(a=a, b=b):
return a * t + b
r = pymc.Poisson(mu=linear, name='r', value=obs, observed=True)
model = pymc.Model([a, b, r])
map = pymc.MAP(model)
map.fit()
map.revert_to_max()
print "a :", a._value
print "b :", b._value
这工作正常。但是我的实际泊松过程受到确定性值的限制。由于我无法将观察到的值与确定性函数相关联,因此我正在为我的观察添加一个方差很小的正态随机函数:
import numpy as np
import pymc
# Observation
a_actual = 1.3
b_actual = 2.0
t = np.arange(10)
obs = np.random.poisson(a_actual * t + b_actual).clip(0, 10)
# Model
a = pymc.Uniform(name='a', value=1., lower=0, upper=10)
b = pymc.Uniform(name='b', value=1., lower=0, upper=10)
@pymc.deterministic
def linear(a=a, b=b):
return a * t + b
r = pymc.Poisson(mu=linear, name='r')
@pymc.deterministic
def clip(r=r):
return r.clip(0, 10)
rc = pymc.Normal(mu=r, tau=0.001, name='rc', value=obs, observed=True)
model = pymc.Model([a, b, r, rc])
map = pymc.MAP(model)
map.fit()
map.revert_to_max()
print "a :", a._value
print "b :", b._value
此代码产生以下错误:
Traceback (most recent call last):
File "pymc-bug-2.py", line 59, in <module>
map.revert_to_max()
File "pymc/NormalApproximation.py", line 486, in revert_to_max
self._set_stochastics([self.mu[s] for s in self.stochastics])
File "pymc/NormalApproximation.py", line 58, in __getitem__
tot_len += self.owner.stochastic_len[p]
KeyError: 0
知道我在做什么错吗?