Question

我正在完成theano文档/教程，第一个例子是：

>>> import numpy
>>> import theano.tensor as T
>>> from theano import function
>>> x = T.dscalar('x')
>>> y = T.dscalar('y')
>>> z = x + y
>>> f = function([x, y], z)

这看起来很简单，所以我编写了自己的程序，扩展了它：

import numpy as np
import theano.tensor as T
from theano import function

x = T.dscalar('x')
y = T.dscalar('y')

z = x + y
f = function([x, y], z)


print f(2, 3)
print np.allclose(f(16.3, 12.1), 28.4) 
print ""

r  = (2, 3), (2, 2), (2, 1), (2, 0)

for i in r:
    print i
    print f(i)

由于某种原因，它不会迭代：

5.0
True

(2, 3)
Traceback (most recent call last):
  File "TheanoBase2.py", line 20, in <module>
    print f(i)
  File "/usr/local/lib/python2.7/dist-packages/theano/compile/function_module.py", line 786, in __call__
    allow_downcast=s.allow_downcast)
  File "/usr/local/lib/python2.7/dist-packages/theano/tensor/type.py", line 177, in filter
    data.shape))
TypeError: ('Bad input argument to theano function with name "TheanoBase2.py:9"  at index 0(0-based)', 'Wrong number of dimensions: expected 0, got 1 with shape (2,).')

为什么print f(2, 3)有效且print f(i)不起作用，而它们是完全相同的表达式。我尝试用方括号替换/括起括号，结果是一样的。

Answer 1

function f将两个标量作为输入并返回它们的总和，r的每个元素，即（x，y）是tuple而不是标量。这应该有效：

import numpy as np
import theano.tensor as T
from theano import function

x = T.dscalar('x')
y = T.dscalar('y')

z = x + y
f = function([x, y], z)

print f(2, 3)
print np.allclose(f(16.3, 12.1), 28.4) 
print ""

r  = (2, 3), (2, 2), (2, 1), (2, 0)

for i in r:
    print i
    print f(i[0], i[1])

Theano函数不适用于简单的4值数组

1 个答案: