Question

Traceback (most recent call last):
  File "/home/axoren1/SmokingDataProject/Rotation Test.py", line 40, in <module>
    dJ = T.grad((R(n, t) - R(n, angles)).norm(2), t)
  File "/usr/local/lib/python2.7/dist-packages/theano/gradient.py", line 529, in grad
    handle_disconnected(elem)
  File "/usr/local/lib/python2.7/dist-packages/theano/gradient.py", line 516, in handle_disconnected
    raise DisconnectedInputError(message)
theano.gradient.DisconnectedInputError: grad method was asked to compute the gradient with respect to a variable that is not part of the computational graph of the cost, or is used only by a non-differentiable operator: Theta

这是什么意思？下面是我的代码，解释了为什么我认为这个错误是空洞的。

import numpy as np
import theano
import theano.tensor as T
import theano.tensor.nlinalg as Tn

n = 5

angles = 2 * np.pi * np.random.rand(n, 1)

def R(n, angles):
    sines   = T.sin(angles)
    cosines = T.cos(angles)

    def r(i, j):
        sign = -1 * -1 ** ((i + j) % 2)
        c = cosines[i - 1] * cosines[j]
        s = T.prod(sines[i:j])

        return sign * c * s

    R = T.zeros((n, n))
    for i in range(n):
        for j in range(i, n):
            T.inc_subtensor(R[i:i+1][j:j+1], r(i, j))
    for i in range(0, n - 1):
        T.inc_subtensor(R[i+1:i+2][i:i+1], sines[i])

    return R

guess = np.random.rand(n, 1)

t = T.vector("Theta")
for i in range(100):
    J = (R(n, t) - R(n, angles)).norm(2)
    dJ = T.grad((R(n, t) - R(n, angles)).norm(2), t)
    guess -= dJ.eval({t:guess})
    print J.eval({t:guess}), guess

如您所见，Theta节点由成本函数定义和使用。我根本不知道函数R是如何不连续的。为什么这会破裂？

Answer 1

问题是您需要将inc_subtensor来电的结果分配回R。

而不是

T.inc_subtensor(R[i:i+1][j:j+1], r(i, j))

和

T.inc_subtensor(R[i+1:i+2][i:i+1], sines[i])

使用

R = T.inc_subtensor(R[i:i+1][j:j+1], r(i, j))

和

R = T.inc_subtensor(R[i+1:i+2][i:i+1], sines[i])

inc_subtensor是一个符号操作，它返回一个对象，表示按提供的值递增提供的子指标的符号结果。

在theano中，为什么我明白连接时会得到theano.gradient.DisconnectedInputError？

1 个答案: