Is there a tensorflow equivalent to np.empty?

Question

Numpy has this helper function, np.empty, which will:

Return a new array of given shape and type, without initializing entries.

I find it pretty useful when I want to create a tensor using tf.concat since:

The number of dimensions of the input tensors must match, and all dimensions except axis must be equal.

So it comes in handy to start with an empty tensor of an expected shape. Is there any way to achieve this in tensorflow?

[edit]

A simplified example of why I want this

    netInput = np.empty([0, 4])
    netTarget = np.empty([0, 4])
    inputWidth = 2

    for step in range(data.shape.as_list()[-2]-frames_width-1):
        netInput = tf.concat([netInput, data[0, step:step + frames_width, :]], -2)
        target = tf.concat([target, data[0, step + frames_width + 1:step + frames_width + 2, :]], -2)

In this example, if netInput or netTarget are initialized, I'll be concatenating an extra example with that initialization. And to initialize them with the first value, I need to hack the loop. Nothing mayor, I just wondered if there is a 'tensorflow' way to solve this.

Answer 1

If you're creating an empty tensor, tf.zeros will do

>>> a = tf.zeros([0, 4])
>>> tf.concat([a, [[1, 2, 3, 4], [5, 6, 7, 8]]], axis=0)
<tf.Tensor: shape=(2, 4), dtype=float32, numpy=
array([[1., 2., 3., 4.],
       [5., 6., 7., 8.]], dtype=float32)>

Answer 2

In TF 2,

tensor = tf.reshape(tf.convert_to_tensor(()), (0, n))

worked for me.