自定义Theano Op进行数值积分
问题内容:
我正在尝试编写一个自定义的Theano
Op,它将在两个值之间的函数数字化集成。Op是PyMC3的自定义可能性,其中涉及一些积分的数值评估。我不能简单地使用@as_op装饰器,因为我需要使用HMC来执行MCMC步骤。任何帮助将不胜感激,因为这个问题似乎已经走到了好几次,但一直没有得到解决(如https://stackoverflow.com/questions/36853015/using-
theano-with-numerical-
integration
,Theano:实现一个积分函数)。
显然,一种解决方案是在Theano中编写一个数值积分器,但是当非常好的积分器已经可以使用(例如通过scipy.integrate)时,这似乎是在浪费精力。
为了将其作为一个最小的示例,让我们尝试在Op内部集成介于0和1之间的函数。下面的代码在Op之外集成了Theano函数,并根据我的测试产生了正确的结果。
import theano
import theano.tensor as tt
from scipy.integrate import quad
x = tt.dscalar('x')
y = x**4 # integrand
f = theano.function([x], y)
print f(0)
print f(1)
ans = integrate.quad(f, 0, 1)[0]
print ans
但是,尝试在Op内进行集成似乎要困难得多。我目前的最大努力是:
import numpy as np
import theano
import theano.tensor as tt
from scipy import integrate
class IntOp(theano.Op):
__props__ = ()
def make_node(self, x):
x = tt.as_tensor_variable(x)
return theano.Apply(self, [x], [x.type()])
def perform(self, node, inputs, output_storage):
x = inputs[0]
z = output_storage[0]
f_to_int = theano.function([x], x)
z[0] = tt.as_tensor_variable(integrate.quad(f_to_int, 0, 1)[0])
def infer_shape(self, node, i0_shapes):
return i0_shapes
def grad(self, inputs, output_grads):
ans = integrate.quad(output_grads[0], 0, 1)[0]
return [ans]
intOp = IntOp()
x = tt.dmatrix('x')
y = intOp(x)
f = theano.function([x], y)
inp = np.asarray([[2, 4], [6, 8]], dtype=theano.config.floatX)
out = f(inp)
print inp
print out
出现以下错误:
Traceback (most recent call last):
File "stackoverflow.py", line 35, in <module>
out = f(inp)
File "/usr/local/lib/python2.7/dist-packages/theano/compile/function_module.py", line 871, in __call__
storage_map=getattr(self.fn, 'storage_map', None))
File "/usr/local/lib/python2.7/dist-packages/theano/gof/link.py", line 314, in raise_with_op
reraise(exc_type, exc_value, exc_trace)
File "/usr/local/lib/python2.7/dist-packages/theano/compile/function_module.py", line 859, in __call__
outputs = self.fn()
File "/usr/local/lib/python2.7/dist-packages/theano/gof/op.py", line 912, in rval
r = p(n, [x[0] for x in i], o)
File "stackoverflow.py", line 17, in perform
f_to_int = theano.function([x], x)
File "/usr/local/lib/python2.7/dist-packages/theano/compile/function.py", line 320, in function
output_keys=output_keys)
File "/usr/local/lib/python2.7/dist-packages/theano/compile/pfunc.py", line 390, in pfunc
for p in params]
File "/usr/local/lib/python2.7/dist-packages/theano/compile/pfunc.py", line 489, in _pfunc_param_to_in
raise TypeError('Unknown parameter type: %s' % type(param))
TypeError: Unknown parameter type: <type 'numpy.ndarray'>
Apply node that caused the error: IntOp(x)
Toposort index: 0
Inputs types: [TensorType(float64, matrix)]
Inputs shapes: [(2, 2)]
Inputs strides: [(16, 8)]
Inputs values: [array([[ 2., 4.],
[ 6., 8.]])]
Outputs clients: [['output']]
Backtrace when the node is created(use Theano flag traceback.limit=N to make it longer):
File "stackoverflow.py", line 30, in <module>
y = intOp(x)
File "/usr/local/lib/python2.7/dist-packages/theano/gof/op.py", line 611, in __call__
node = self.make_node(*inputs, **kwargs)
File "stackoverflow.py", line 11, in make_node
return theano.Apply(self, [x], [x.type()])
HINT: Use the Theano flag 'exception_verbosity=high' for a debugprint and storage map footprint of this apply node.
我对此感到惊讶,尤其是TypeError,因为我以为我已将output_storage变量转换为张量,但似乎在这里仍然相信它仍然是ndarray。
问题答案:
我发现了您的问题,因为我试图在PyMC3中构建一个表示一般点过程(Hawkes,Cox,Poisson等)的随机变量,并且似然函数具有一个积分。我真的很想能够使用Hamiltonian
Monte Carlo或NUTS采样器,因此我需要相对于时间的积分是可区分的。
从您的尝试开始,我制作了一个IntegratedOut theano
Op,它似乎可以正确处理我需要的行为。我已经在一些不同的输入上对其进行了测试(尚未在我的统计模型中进行过测试,但是看起来很有希望!)。我总是theano
n00b,所以请原谅任何愚蠢。如果有人有任何反馈,我将不胜感激。不确定它是否正是您要的东西,但这是我的解决方案(示例位于底部和doc字符串中)。*编辑:简化了一些残留的拧干方式。
import theano
import theano.tensor as T
from scipy.integrate import quad
class integrateOut(theano.Op):
"""
Integrate out a variable from an expression, computing
the definite integral w.r.t. the variable specified
!!! Only implemented in this for scalars !!!
Parameters
----------
f : scalar
input 'function' to integrate
t : scalar
the variable to integrate out
t0: float
lower integration limit
tf: float
upper integration limit
Returns
-------
scalar
a new scalar with the 't' integrated out
Notes
-----
usage of this looks like:
x = T.dscalar('x')
y = T.dscalar('y')
t = T.dscalar('t')
z = (x**2 + y**2)*t
# integrate z w.r.t. t as a function of (x,y)
intZ = integrateOut(z,t,0.0,5.0)(x,y)
gradIntZ = T.grad(intZ,[x,y])
funcIntZ = theano.function([x,y],intZ)
funcGradIntZ = theano.function([x,y],gradIntZ)
"""
def __init__(self,f,t,t0,tf,*args,**kwargs):
super(integrateOut,self).__init__()
self.f = f
self.t = t
self.t0 = t0
self.tf = tf
def make_node(self,*inputs):
self.fvars=list(inputs)
# This will fail when taking the gradient... don't be concerned
try:
self.gradF = T.grad(self.f,self.fvars)
except:
self.gradF = None
return theano.Apply(self,self.fvars,[T.dscalar().type()])
def perform(self,node, inputs, output_storage):
# Everything else is an argument to the quad function
args = tuple(inputs)
# create a function to evaluate the integral
f = theano.function([self.t]+self.fvars,self.f)
# actually compute the integral
output_storage[0][0] = quad(f,self.t0,self.tf,args=args)[0]
def grad(self,inputs,grads):
return [integrateOut(g,self.t,self.t0,self.tf)(*inputs)*grads[0] \
for g in self.gradF]
x = T.dscalar('x')
y = T.dscalar('y')
t = T.dscalar('t')
z = (x**2+y**2)*t
intZ = integrateOut(z,t,0,1)(x,y)
gradIntZ = T.grad(intZ,[x,y])
funcIntZ = theano.function([x,y],intZ)
funcGradIntZ = theano.function([x,y],gradIntZ)
print funcIntZ(2,2)
print funcGradIntZ(2,2)