So is there a better way to perform 1X1 convolution using the conv2d module
? Just curious.
On Monday, October 31, 2016 at 12:53:11 PM UTC-7, Pascal Lamblin wrote:
>
>
> If you run a convolution with a (1, 1, 1, 1) filter, then you will only
> have one output channel. That could explain why you have a shape of (1,
> 128, 1, 2) instead of (1, 128, 128, 2).
>
> These shapes are not compatible, and Join will not explicitly duplicate
> cost along the broadcastable axis (neither will numpy.join).
>
> If duplication of the cost along axis=2 is what you want, then you could
> use
>
> T.concatenate(
> [img_param[:, :, :, :3], T.alloc(cost, cost.shape[0], cost.shape[1],
> img_param.shape[2], 2)],
> axis=3)
>
> On Sun, Oct 30, 2016, Ido wrote:
> > I'm trying to concatenate two tensors along their 3rd dimension.
> >
> > param = T.concatenate(
> > [img_param[:, :, :, :3], cost.repeat(2, axis=3)],
> > axis=3
> > )
> >
> >
> > img_param has the shape (1, 128, 128, 5), cost is of shape (1, 128, 128,
> 1).
> >
> > If I take cost directly from the input, this works perfectly fine.
> > Yet, if I first run T.nnet.conv2d with a constant filter (1, 1, 1, 1) I
> get:
> >
> > ValueError: all the input array dimensions except for the concatenation
> > axis must match exactly
> > Apply node that caused the error: Join(TensorConstant{3}, Subtensor{::,
> ::,
> > ::, :int64:}.0, Reshape{4}.0)
> > Toposort index: 12
> > Inputs types: [TensorType(int8, scalar), TensorType(float32, 4D),
> > TensorType(float32, 4D)]
> > Inputs shapes: [(), (1, 128, 128, 3), (1, 128, 1, 2)]
> >
> > The conv2d call:
> >
> > cost = T.nnet.conv2d(
> > input=input,
> > filters=self.unary_w,
> > input_shape=self.input_shp,
> > filter_shape=self.unary_filter_shp,
> > border_mode='valid'
> > ). dimshuffle(0,2,3,1)
> >
> >
> > (I also made sure all dimensions are shuffled to the correct shape)
> >
> > I tried returning both tensors to validate their shape and both return
> just
> > fine.
> > I've looked a bit into this and it seems to have something to do with a
> > sort of sparse-matrix conversion. Meaning Theano just automatically
> drops
> > dimensions which are close to 0.
> >
> > Any ideas?
> >
> > Thanks,
> >
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> >
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>
> --
> Pascal
>
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