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

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.]