I have a TensorFlow vector that only contains 1s and 0s, like a = [0, 0, 0, 1, 0, 1], and conditional on the value of a, I want to draw new random values 0 or 1. If the value of a is 1, I want to draw a new value but if the value of a is 0 I want to leave it alone. So I've tried this:
import tensorflow as tf
import tensorflow_probability as tfp
tfd = tfp.distributions
# random draw of zeros and ones
a = tfd.Binomial(total_count = 1.0, probs = 0.5).sample(6)
which gives me <tf.Tensor: shape=(6,), dtype=float32, numpy=array([0., 0., 0., 1., 0., 1.], dtype=float32)> then if I redraw
# redraw with a different probability if value is 1. in the original draw
b = tf.where(a == 1.0, tfd.Binomial(total_count = 1., probs = 0.5).sample(1), a)
I would expect tf.where to give me a new vector b that has, on average, half of the 1s become 0s but instead it either returns a copy of a or a vector of all 0s. Example output would be one of b = [0, 0, 0, 0, 0, 0], b = [0, 0, 0, 0, 0, 1], b = [0, 0, 0, 1, 0, 0], or b = [0, 0, 0, 1, 0, 1] . I could of course just use b = tfd.Binomial(total_count = 1.0, probs = 0.25).sample(6) but in my particular case the order of the original vector matters.
A more general situation might use a different distribution so that bit-wise operations can't be easily used. For example
# random draw of normals
a = tfd.Normal(loc = 0., scale = 1.).sample(6)
# redraw with a different scale if value is greater than zero in the original draw
b = tf.where(a > 0, tfd.Normal(loc = 0., scale = 2.).sample(1), a)
APPROACH 1:
Not tested, but I think the middle param should be a tensor that matches the original one. E.g. 6 elements:
First, make a second random sequence, of same length:
a2 = tfd.Binomial(total_count = 1.0, probs = 0.5).sample(6)
NOTE: If you need a different probability, you simply use that probability when creating a2.
prob = 0.3
a2 = tfd.Binomial(total_count = 1.0, probs = prob).sample(6)
Then:
b = tf.where(a == 1.0, a2, a)
Explanation:
The values in a2 are irrelevant where a is 0, and are "prob" on average where a is 1.
APPROACH 2:
If that doesn't work, then first param needs to be mapped to a tensor of [true, false, ..]:
def pos(n):
return n > 0
cond = list(map(pos,a)) # I don't have TensorFlow handy; may need to replace `list` with appropriate function to create a Tensor.
b = tf.where(cond, a2, 0.0)
APPROACH 3:
Tested. Doesn't use tf.where.
First, make a second random sequence, of same length:
a2 = tfd.Binomial(total_count = 1.0, probs = prob).sample(6)
Then combine the two, "bitwise-and"ing corresponding elements:
def and2(a, b):
return (a & b)
b = list(map(and2, a, a2))
NOTE: could alternatively use any other function to combine the two corresponding elements.
Example data:
a = [0,0,1,1]
a2 = [0,1,0,1]
Result:
b = [0,0,0,1]
Explanation:
The values in a2 are irrelevant where a is 0, and are "prob" on average where a is 1.