Is there a way to reshape a tensor and pad any overflow with zeros? I know ndarray.reshape does this, but as I understand it, converting a Tensor to an ndarray would require flip-flopping between the GPU and CPU.
Tensorflow's reshape() documentation says the TensorShapes need to have the same number of elements, so perhaps the best way would be a pad() and then reshape()?
I'm trying to achieve:
a = tf.Tensor([[1,2],[3,4]])
tf.reshape(a, [2,3])
a => [[1, 2, 3],
[4, 0 ,0]]
Method Used: tf. pad: This method accepts input tensor and padding tensor with other optional arguments and returns a Tensor with added padding and same type as input Tensor. Padding tensor is a Tensor with shape(n, 2).
To flatten the tensor, we're going to use the TensorFlow reshape operation. So tf. reshape, we pass in our tensor currently represented by tf_initial_tensor_constant, and then the shape that we're going to give it is a -1 inside of a Python list.
As far as I know, there's no built-in operator that does this (tf.reshape()
will give you an error if the shapes do not match). However, you can achieve the same result with a few different operators:
a = tf.constant([[1, 2], [3, 4]])
# Reshape `a` as a vector. -1 means "set this dimension automatically".
a_as_vector = tf.reshape(a, [-1])
# Create another vector containing zeroes to pad `a` to (2 * 3) elements.
zero_padding = tf.zeros([2 * 3] - tf.shape(a_as_vector), dtype=a.dtype)
# Concatenate `a_as_vector` with the padding.
a_padded = tf.concat([a_as_vector, zero_padding], 0)
# Reshape the padded vector to the desired shape.
result = tf.reshape(a_padded, [2, 3])
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