[SOLVED] How to shift a tensor like pandas.shift in tensorflow / keras? (Without shift the last row to first row, like tf.roll)

How to shift a tensor like pandas.shift in tensorflow / keras? (Without shift the last row to first row, like tf.roll)

I want to shift a tensor in a given axis. It's easy to do this in pandas or numpy. Like this:

import numpy as np
import pandas as pd

data = np.arange(0, 6).reshape(-1, 2)
pd.DataFrame(data).shift(1).fillna(0).values

Output is:

array([[0., 0.],
[0., 1.],
[2., 3.]])

But in tensorflow, the closest solution I found is tf.roll. But it shift the last row to the first row. (I don't want that). So I have to use something like

tf.roll + tf.slice(remove the last row) + tf.concat(add tf.zeros to the first row).

It's really ugly.

Is there a better way to handle shift in tensorflow or keras?

Thanks.

Solution

I think I find a better way for this problem.

We could use tf.roll, then apply tf.math.multiply to set the first row to zeros.

Sample code is as follows:

Original tensor:

A = tf.cast(tf.reshape(tf.range(27), (-1, 3, 3)), dtype=tf.float32)
A

Output:

<tf.Tensor: id=117, shape=(3, 3, 3), dtype=float32, numpy=
array([[[ 0.,  1.,  2.],
        [ 3.,  4.,  5.],
        [ 6.,  7.,  8.]],

       [[ 9., 10., 11.],
        [12., 13., 14.],
        [15., 16., 17.]],

       [[18., 19., 20.],
        [21., 22., 23.],
        [24., 25., 26.]]], dtype=float32)>

Shift (like pd.shift):

B = tf.concat((tf.zeros((1, 3)), tf.ones((2, 3))), axis=0)
C = tf.expand_dims(B, axis=0)
tf.math.multiply(tf.roll(A, 1, axis=1), C)

Output:

<tf.Tensor: id=128, shape=(3, 3, 3), dtype=float32, numpy=
array([[[ 0.,  0.,  0.],
        [ 0.,  1.,  2.],
        [ 3.,  4.,  5.]],

       [[ 0.,  0.,  0.],
        [ 9., 10., 11.],
        [12., 13., 14.]],

       [[ 0.,  0.,  0.],
        [18., 19., 20.],
        [21., 22., 23.]]], dtype=float32)>