Python - Theano scan() 函数 [英] Python - Theano scan() function
问题描述
我无法完全理解 theano.scan() 的行为.
I cannot fully understand the behaviour of theano.scan().
这是一个例子:
import numpy as np
import theano
import theano.tensor as T
def addf(a1,a2):
return a1+a2
i = T.iscalar('i')
x0 = T.ivector('x0')
step= T.iscalar('step')
results, updates = theano.scan(fn=addf,
outputs_info=[{'initial':x0, 'taps':[-2]}],
non_sequences=step,
n_steps=i)
f=theano.function([x0,i,step],results)
print f([1,1],10,2)
上面的代码片段打印了以下序列,这是完全合理的:
The above snippet prints the following sequence, which is perfectly reasonable:
[ 3 3 5 5 7 7 9 9 11 11]
但是,如果我将抽头索引从 -2 切换到 -1,即
However if I switch the tap index from -2 to -1, i.e.
outputs_info=[{'initial':x0, 'taps':[-1]}]
结果变成:
[[ 3 3]
[ 5 5]
[ 7 7]
[ 9 9]
[11 11]
[13 13]
[15 15]
[17 17]
[19 19]
[21 21]]
而不是我认为合理的(只需取向量的最后一个值并添加 2):
instead of what would seem reasonable to me (just take the last value of the vector and add 2):
[ 3 5 7 9 11 13 15 17 19 21]
任何帮助将不胜感激.
谢谢!
推荐答案
当您使用 taps=[-1] 时,假设输出信息中的信息按原样使用.这意味着将使用向量和 non_sequence 作为输入调用 addf 函数.如果将 x0 转换为标量,它将按预期工作:
When you use taps=[-1], scan suppose that the information in the output info is used as is. That mean the addf function will be called with a vector and the non_sequence as inputs. If you convert x0 to a scalar, it will work as you expect:
import numpy as np
import theano
import theano.tensor as T
def addf(a1,a2):
print a1.type
print a2.type
return a1+a2
i = T.iscalar('i')
x0 = T.iscalar('x0')
step= T.iscalar('step')
results, updates = theano.scan(fn=addf,
outputs_info=[{'initial':x0, 'taps':[-1]}],
non_sequences=step,
n_steps=i)
f=theano.function([x0,i,step],results)
print f(1,10,2)
这给出了这个输出:
TensorType(int32, scalar)
TensorType(int32, scalar)
[ 3 5 7 9 11 13 15 17 19 21]
在你的情况下,当它做 addf(vector,scalar) 时,它会广播 elemwise 值.
In your case as it do addf(vector,scalar), it broadcast the elemwise value.
换一种方式解释,如果 taps 为 [-1],x0 将按原样"传递给内部函数.如果 taps 包含其他任何内容,则传递给内部函数的内容将比 x0 小 1 维,因为 x0 必须提供许多初始步长值(-2 和 -1).
Explained in another way, if taps is [-1], x0 will be passed "as is" to the inner function. If taps contain anything else, what is passed to the inner function will have 1 dimension less then x0, as x0 must provide many initial steps value (-2 and -1).
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