thanks a lot, Fred. 

I define a theano function as below :
x = T.fmatrix() 
y =  T.fmatrix()
z = x / y 
f = theano.function([x,y],z) 


and give 
a = array([[ 0.,  1.,  2.], 

                [ 3., -4.,  5.]], dtype=float32) 



*b = array([[ 0.,  1.,  0.],                  [ 1.,  0.,  1.]], 
dtype=float32)*
f (a,b) = array([[ nan,   1.,  inf], 

              [ -0., -inf,   5.]], dtype=float32)


as can we see from denominator b ,   there are 0  in b. 

could we modify "z= x/y " to process all the cases??  




在 2016年7月30日星期六 UTC+9上午2:40:37,nouiz写道:
>
> Just to be sure, oou want to replace inf/nan with a numerical value in 
> Theano? This can be done like this:
>
> import numpy as np
>
> a = np.array([0,0,1,1,2], dtype='float')
> b = np.array([0,1,0,1,3], dtype='float')
> with np.errstate(divide='ignore', invalid='ignore'):
>     c = np.true_divide(a,b)
>
> import theano.tensor as T
> import theano
> c=theano.tensor.as_tensor_variable(c)
> inf_mask = T.isinf(c)
> nan_mask = T.isnan(c)
> inf_idx = inf_mask.nonzero()
> nan_idx = nan_mask.nonzero()
>
> c_without_inf = theano.tensor.set_subtensor(c[inf_idx], -999)
> c_without_inf_nan = theano.tensor.set_subtensor(c_without_inf[nan_idx], 0)
> c_without_inf_nan.eval()
> array([  0.00000000e+00,   0.00000000e+00,  -9.99000000e+02,
>          1.00000000e+00,   6.66666667e-01])
>
> I don't know if it make a difference between inf and -inf.
>
> Fred
>
>
>
> On Thu, Jul 28, 2016 at 11:52 PM Varglur <[email protected] 
> <javascript:>> wrote:
>
>> dear all,
>>
>> similar to problem in numpy here 
>> <http://stackoverflow.com/questions/26248654/numpy-return-0-with-divide-by-zero>
>>
>> if element wise divide in theano ,  are there corresponding function in 
>> theano as this answer in numpy  ?  
>>
>> numpy answer:
>>
>> import numpy as np
>>
>> a = np.array([0,0,1,1,2], dtype='float')
>> b = np.array([0,1,0,1,3], dtype='float')
>> with np.errstate(divide='ignore', invalid='ignore'):
>>     c = np.true_divide(a,b)
>>     c[c == np.inf] = 0
>>     c = np.nan_to_num(c)
>> print('c: {0}'.format(c))
>>
>> Output:
>>
>> c: [ 0.          0.          0.          1.          0.66666667]
>>
>>
>>
>> how to process inf and nan problem when 0/0 or 1/0 problem in theano??
>>
>> thanks
>>
>>
>>
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>>
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