Tensorflow 2.2, tf.nn.conv1d in Lambda layer

Question

I'd like to perform a convolution in a Lambda layer, but I can't get it to work any way.

    kernel = [1.0,2.0,1.0]  # weighted moving average
    x = [   # history_size=5, num_features=10
      [1.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0],
      [2.0,2.0,2.0,2.0,2.0,2.0,2.0,2.0,2.0,2.0],
      [3.0,3.0,3.0,3.0,3.0,3.0,3.0,3.0,3.0,3.0],
      [4.0,4.0,4.0,4.0,4.0,4.0,4.0,4.0,4.0,4.0],
      [5.0,5.0,5.0,5.0,5.0,5.0,5.0,5.0,5.0,5.0],
    ]
    k = tf.constant(kernel, dtype=tf.float32)
    y = tf.nn.conv1d(x, k, stride=1, padding='SAME')

I realize dimensions are not correct in the above example, but that's my data's actual format. The training samples have a shape of (history_size, num_features) and the kernel has to convolve along history_size, each feature separately. Any help would be appreciated. I cannot find an example on how to perform tf.nn.conv1d manually.

Answer 1

You could use numpy.convolve() for this.

import numpy as np

kernel = [1.0,2.0,1.0]  # weighted moving average
x = [   # history_size=5, num_features=10
  [1.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0,1.0],
  [2.0,2.0,2.0,2.0,2.0,2.0,2.0,2.0,2.0,2.0],
  [3.0,3.0,3.0,3.0,3.0,3.0,3.0,3.0,3.0,3.0],
  [4.0,4.0,4.0,4.0,4.0,4.0,4.0,4.0,4.0,4.0],
  [5.0,5.0,5.0,5.0,5.0,5.0,5.0,5.0,5.0,5.0],
]

output = []
for i in range(len(x)):
    output.append(list(np.convolve(x[i], kernel, mode = 'same')))
output

'''
[[3.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 4.0, 3.0],
 [6.0, 8.0, 8.0, 8.0, 8.0, 8.0, 8.0, 8.0, 8.0, 6.0],
 [9.0, 12.0, 12.0, 12.0, 12.0, 12.0, 12.0, 12.0, 12.0, 9.0],
 [12.0, 16.0, 16.0, 16.0, 16.0, 16.0, 16.0, 16.0, 16.0, 12.0],
 [15.0, 20.0, 20.0, 20.0, 20.0, 20.0, 20.0, 20.0, 20.0, 15.0]]
'''

You could try changing the mode whichever fits best to you according to the documentation .