Support for square root operation on Node

[Ramana Sundararaman]

I was trying out this code for profiling cgt and theano for understanding how better the graph optimisations are done in CGT.
x = cgt.scalar(name='x', dtype='float32') y = cgt.scalar(name='y', dtype='float32') func = np.sqrt((x**2)*(y/x)*(x**3/y**3) -(x**2)*(y/x)*(x**3/y**3) + (x**2)*(y/x)*(x**3/y**3))
and then i got this error while compiling the function,
AttributeError: 'Node' object has no attribute 'sqrt'
Would be glad to have square root support
Thanks

[Frédéric Bastien]

If you cann cgt.sqrt (or similar) instead of numpy np.sqrt(), does it work?

On Sun, Mar 20, 2016 at 12:21 PM, Ramana Subramanyam <
[e-mail]> wrote:

I was trying out this code for profiling cgt and theano for understanding
how better the graph optimisations are done in CGT.
x = cgt.scalar(name='x', dtype='float32')
y = cgt.scalar(name='y', dtype='float32')
func = np.sqrt((x2)(y/x)(x3/y3) -(x2)(y/x)(x3/y3) +
(x2)(y/x)(x3/y**3))

and then i got this error while compiling the function,
AttributeError: 'Node' object has no attribute 'sqrt'

Would be glad to have square root support
Thanks

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#62

[John Schulman]

Yes, cgt.sqrt will work.