oja*_*jas 5 python neural-network theano
谁能解释一下以下python代码的输出:
from theano import tensor as T
from theano import function, shared
a, b = T.dmatrices('a', 'b')
diff = a - b
abs_diff = abs(diff)
diff_squared = diff ** 2
f = function([a, b], [diff, abs_diff, diff_squared])
print f([[1, 1], [1, 1]], [[0, 1], [2, 3]])
Run Code Online (Sandbox Code Playgroud)
print f( [ [1,1],[1,1] ],
[ [0,1],[2,3] ])
Output: [ [[ 1., 0.], [-1., -2.]],
[[ 1., 0.], [ 1., 2.]],
[[ 1., 0.], [ 1., 4.]]]
Run Code Online (Sandbox Code Playgroud)
您实际上是在告诉Theano计算三个不同的函数,其中每个后续函数都取决于先前执行的函数的输出.
在您的示例中,您使用两个输入参数:矩阵A和矩阵B.
A = [[ 1, 1 ],
[ 1, 1 ]]
B = [[ 0, 1 ],
[ 2, 3 ]]
Run Code Online (Sandbox Code Playgroud)
第一个输出行:[[ 1., 0.], [-1., -2.]]通过减去A和B矩阵来计算:
[[1, 1], - [[0, 1], = [[ 1, 0 ],
[1, 1]] [2, 3]] [-1, -2]
Run Code Online (Sandbox Code Playgroud)
第二个输出线[[ 1., 0.], [ 1., x2.]]只是我们刚刚计算出的差值的绝对值:
abs [[ 1, 0 ], = [[ 1, 0],
[-1, -2]] [ 1, 2]]
Run Code Online (Sandbox Code Playgroud)
第三行和最后一行按元素计算平方值.
Theano实际上解释了您的Python代码,并推断出给定变量所依赖的变量(或数学运算).因此,如果你只对有兴趣的diff_squared输出,你并不需要包括对呼叫diff和abs_diff太.
f = function([a, b], [diff_squared])
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
154 次 |
| 最近记录: |