Use scipy.integrate.quad with Tensorflow

问题

I am trying to use scipy.integrate.quad with Tensorflow as following. time and Lambda are two Tensors with shape (None, 1).

def f_t(self, time, Lambda):
    h = Lambda * self.shape * time ** (self.shape - 1)
    S = tf.exp(-1 * Lambda * time ** self.shape)
    return h * S


def left_censoring(self, time, Lambda):
    return tf.map_fn(lambda x: integrate.quad(self.f_t,
                                              0.0,
                                              x[0], # it is not a float before evaluation
                                              args=(x[1],)),
                     tf.concat([time, Lambda], 1))

However, I get an error as below:

File "J:\Workspace\Distributions.py", line 30, in <lambda>
    args=(x[1],)),
  File "I:\Anaconda3\envs\tensorflow\lib\site-packages\scipy\integrate\quadpack.py", line 323, in quad
    points)
  File "I:\Anaconda3\envs\tensorflow\lib\site-packages\scipy\integrate\quadpack.py", line 388, in _quad
    return _quadpack._qagse(func,a,b,args,full_output,epsabs,epsrel,limit)
TypeError: a float is required

X[0] is a Tensor with shape=(). It is not a float value before evaluation. Is it possible to solve the problem? How should I calculate integration in Tensorflow?

回答1:

If you have at least TensorFlow 1.8.0, you're probably best off using tf.contrib.integrate.odeint_fixed() like this code (tested):

from __future__ import print_function
import tensorflow as tf

assert tf.VERSION >= "1.8.0", "This code only works with TensorFlow 1.8.0 or later."

def f( y, a ):
    return a * a

x = tf.constant( [ 0.0, 1.0, 2, 3, 4 ], dtype = tf.float32 )

i = tf.contrib.integrate.odeint_fixed( f, 0.0, x, method = "rk4" )

with tf.Session() as sess:
    res = sess.run( i )
    print( res )

will output:

[ 0. 0.33333334 2.6666667 9. 21.333334 ]

properly integrating x² over the intervals of [ 0, 0 ], [ 0, 1 ], [ 0, 2 ], [ 0, 3 ], and [ 0, 4 ] as per x = [ 0, 1, 2, 3, 4 ] above. (The primitive function of x² is ⅓ x³, so for example 4³ / 3 = 64/3 = 21 ⅓.)

Otherwise, for earlier TensorFlow versions, here's how to fix your code.

So the main issue is that you have to use tf.py_func() to map a Python function (scipy.integrate.quad() in this case) on a tensor. tf.map_fn() will map other TensorFlow operations and passes and expects tensors as operands. Therefore x[ 0 ] will never be a simple float, it will be a scalar tensor and scipy.integrate.quad() will not know what to do with that.

You can't completely get rid of tf.map_fn() either, unless you want to manually loop over numpy arrays.

Furthermore, scipy.integrate.quad() returns a double (float64), whereas your tensors are float32.

I've simplified your code a lot, because I don't have access to the rest of it and it looks too complicated compared to the core of this question. The following code (tested):

from __future__ import print_function
import tensorflow as tf
from scipy import integrate

def f( a ):
    return a * a

def integrated( f, x ):
    return tf.map_fn( lambda y: tf.py_func( 
               lambda z: integrate.quad( f, 0.0, z )[ 0 ], [ y ], tf.float64 ),
                      x )

x = tf.constant( [ 1.0, 2, 3, 4 ], dtype = tf.float64 )

i = integrated( f, x )

with tf.Session() as sess:
    res = sess.run( i )
    print( res )

will also output:

[ 0.33333333 2.66666667 9. 21.33333333]

来源：https://stackoverflow.com/questions/50356467/use-scipy-integrate-quad-with-tensorflow

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scipy