如何在 keras 的条件下更换张量的某些部分？

Question

I would like to perform an operation similar to np.where on keras tensors with tensorflow backend. That means lets say I have two tensors: diff and sum. I divide those vectors as:

rel_dev = diff / sum

For np.arrays I would write:

rel_dev = np.where((diff == 0.0) & (sum == 0.0), 0.0, rel_dev)

rel_dev = np.where((diff != 0.0) & (sum == 0.0), np.sign(diff), rel_dev)

That is, eg if I have zeros in both diff and sum, I wish I will not get an np.Inf, but set the rel_dev to zero. Now in keras with tensors it did not work. I have tried K.switch, K.set_value, etc. As I understand it works for the whole tensor, but not for it separate parts, right? It works without setting these conditions though, but I have no idea what actually happens where. I did not succeed to debug it yet.

Could you please tell me how to write both conditions for rel_dev in Keras?

Answer 1

You can do that in Keras like this:

import keras.backend as K

diff = K.constant([0, 1, 2, -2, 3, 0])
sum = K.constant([2, 4, 1, 0, 5, 0])
rel_dev = diff / sum
d0 = K.equal(diff, 0)
s0 = K.equal(sum, 0)
rel_dev = K.switch(d0 & s0, K.zeros_like(rel_dev), rel_dev)
rel_dev = K.switch(~d0 & s0, K.sign(diff), rel_dev)
print(K.eval(rel_dev))
# [ 0.    0.25  2.   -1.    0.6   0.  ]

EDIT: The above formulation has an insidious problem, which is that, even though the result is right, nan values will propagate back through the gradients (namely because dividing by zero gives inf or nan , and multiplying inf or nan by zero gives nan ). Indeed, if you check the gradients:

gd, gs = K.gradients(rel_dev, (diff, sum))
print(K.eval(gd))
# [0.5  0.25 1.    nan 0.2   nan]
print(K.eval(gs))
# [-0.     -0.0625 -2.         nan -0.12       nan]

The trick you can use to avoid that is to change sum in the division in a way that does not affect the result but prevents the nan values, for example like this:

import keras.backend as K

diff = K.constant([0, 1, 2, -2, 3, 0])
sum = K.constant([2, 4, 1, 0, 5, 0])
d0 = K.equal(diff, 0)
s0 = K.equal(sum, 0)
# sum zeros are replaced by ones on division
rel_dev = diff / K.switch(s0, K.ones_like(sum), sum)
rel_dev = K.switch(d0 & s0, K.zeros_like(rel_dev), rel_dev)
rel_dev = K.switch(~d0 & s0, K.sign(diff), rel_dev)
print(K.eval(rel_dev))
# [ 0.    0.25  2.   -1.    0.6   0.  ]
gd, gs = K.gradients(rel_dev, (diff, sum))
print(K.eval(gd))
# [0.5  0.25 1.   0.   0.2  0.  ]
print(K.eval(gs))
# [-0.     -0.0625 -2.      0.     -0.12    0.    ]

Answer 2

You can use tensorflow's where function to do what you want with tensors.

Answer 3

Yep Tensorflow's where is what you are looking for if you what to convert tensor to nparray perform all the operation on nparray then you can use tensor.numpy() this will return a numpy array of tensor. You get the numpy array back to tensor with the "tf.convert_to_tensor" API.