Sampling without replacement from a given non-uniform distribution in TensorFlow

Question

I'm looking for something similar to numpy.random.choice(range(3),replacement=False,size=2,p=[0.1,0.2,0.7])
in TensorFlow.

The closest Op to it seems to be tf.multinomial(tf.log(p)) which takes logits as input but it can't sample without replacement. Is there any other way to do sampling from a non-uniform distribution in TensorFlow?

Thanks.

Answer 1

Yes, there is. See here and here for some background information. The solution is:

z = -tf.log(-tf.log(tf.random_uniform(tf.shape(p),0,1))) 
_, indices = tf.nn.top_k(tf.log(p) + z, size)

Answer 2

You could just use tf.py_func to wrap numpy.random.choice and make it available as a TensorFlow op:

a = tf.placeholder(tf.float32)
size = tf.placeholder(tf.int32)
replace = tf.placeholder(tf.bool)
p = tf.placeholder(tf.float32)

y = tf.py_func(np.random.choice, [a, size, replace, p], tf.float32)

with tf.Session() as sess:
    print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))

You can specify the numpy seed as usual:

np.random.seed(1)
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))
np.random.seed(1)
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))
np.random.seed(1)
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))

would print:

[ 2.  0.]
[ 2.  1.]
[ 0.  1.]
[ 2.  0.]
[ 2.  1.]
[ 0.  1.]
[ 2.  0.]