Python 我可以配置Theano';x除以0
我使用Theano有点问题。似乎Python 我可以配置Theano';x除以0,python,numpy,theano,Python,Numpy,Theano,我使用Theano有点问题。似乎除以0会产生inf而不是使用例如Numpy这会产生0(至少逆函数的行为是这样的)。看一看: from theano import function, sandbox, Out, shared import theano.tensor as T import numpy as np reservoirSize = 7 _eye = np.eye(reservoirSize) gpu_I = shared( np.asarray(_eye
除以0
会产生inf
而不是使用例如Numpy这会产生0(至少逆函数的行为是这样的)。看一看:
from theano import function, sandbox, Out, shared
import theano.tensor as T
import numpy as np
reservoirSize = 7
_eye = np.eye(reservoirSize)
gpu_I = shared( np.asarray(_eye, np.float32 ) )
simply_inverse = function(
[],
Out(sandbox.cuda.basic_ops.gpu_from_host(
T.inv( gpu_I )
),
borrow=True
)
)
gpu_wOut = simply_inverse()
Wout = np.linalg.inv(_eye)
print "gpu_wOut:\n"
print np.asarray(gpu_wOut)
print "\nWout:\n"
print np.asarray(Wout)
diff_wOut = np.asarray(gpu_wOut) - Wout
diff_wOut = [ diff_wOut[0][i] if diff_wOut[0][i] > epsilon else 0 for i in range(reservoirSize)]
print "\n\nDifference of output weights: (only first row)\n"
print np.asarray(diff_wOut)
结果:
gpu_wOut:
[[ 1. inf inf inf inf inf inf]
[ inf 1. inf inf inf inf inf]
[ inf inf 1. inf inf inf inf]
[ inf inf inf 1. inf inf inf]
[ inf inf inf inf 1. inf inf]
[ inf inf inf inf inf 1. inf]
[ inf inf inf inf inf inf 1.]]
Wout:
[[ 1. 0. 0. 0. 0. 0. 0.]
[ 0. 1. 0. 0. 0. 0. 0.]
[ 0. 0. 1. 0. 0. 0. 0.]
[ 0. 0. 0. 1. 0. 0. 0.]
[ 0. 0. 0. 0. 1. 0. 0.]
[ 0. 0. 0. 0. 0. 1. 0.]
[ 0. 0. 0. 0. 0. 0. 1.]]
Difference of output weights (only first row):
[ 0. inf inf inf inf inf inf]
对于我想在GPU中执行的一些计算来说,这是一个问题,我不想从GPU中获取数据,用
0
替换inf
,以继续我的计算,当然,因为这会大大减慢计算过程。无张量
计算元素逆
np.linalg.inv
计算逆矩阵
从数学上讲,这不是一回事
您可能正在寻找实验性的在代码中没有
numpy
除以0…-你的矩阵不是奇异的,所以行列式是非零的,将所有元素倒置意味着计算(x_ij)^(-1),这意味着元素等于0 1/0。但无论出于什么原因,这种情况正在发生,我都需要一个解决方案,来应对这种情况。可能只是因为我不再研究张量数学(如果我曾经研究过的话),但是有没有理由假设矩阵的张量逆等于矩阵的逆?@StefanR.Falk:你知道theano张量倒数和numpy矩阵逆是完全不同的数学运算,不是吗?你可以阅读文档:,