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Since I need to repeat over a specific axis, I want to avoid unnecessary memory reallocation as much as possible.

For example, given a numpy array A of shape (3, 4, 5), I want to create a view named B of shape (3, 4, 100, 5) on the original A. The 3rd axis of A is repeated 100 times.

In numpy, this can be achieved like this:

    B=numpy.repeat(A.reshape((3, 4, 1, 5)), repeats=100, axis=2)

or:

    B=numpy.broadcast_to(A.reshape((3, 4, 1, 5)), repeats=100, axis=2)

The former allocates a new memory and then do some copy stuff, while the latter just create a view over A without extra memory reallocation. This can be identified by the method described in the answer Size of numpy strided array/broadcast array in memory? .

In theano, however, the theano.tensor.repeat seems to be the only way, of course it's not preferable.

I wonder if there is a `numpy.broadcast_to' like theano method can do this in an efficient way?

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Tqri
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1 Answers1

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There is a nice method dimshuffle, which makes a theano variable broadcastable over some dimension

At = theano.tensor.tensor3()
Bt = At.dimshuffle(0,1,'x',2)

Now you've got a tensor variable with shape (3,4,'x',5), where 'x' means any dimension you want to add.

Ct=theano.tensor.zeros((Bt.shape[0],Bt.shape[1],100,Bt.shape[3]))+Bt

Example

f=theano.function([At],[Bt,Ct])
A = np.random.random((3,4,5)).astype(np.float32)
B,C=f(A)
print B.shape
print C.shape

(3, 4, 1, 5)

(3, 4, 100, 5)

Unless specified, it's better to work with variable Bt.

Oleg Grinchuk
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  • First of all, thank you for your answer. But in this way, `Ct` still needs extra memory storage for that newly created zero 4d-tensor and its memory efficiency seems to be the same with theano.repeat operation. – Tqri Oct 13 '16 at 14:22