Sample without replacement - tensorflow

Question

How to sample without replacement in TensorFlow? Like numpy.random.choice(n, size=k, replace=False) for some very large integer n (e.g. 100k-100M), and smaller k (e.g. 100-10k).
Also, I want it to be efficient and on the GPU, so other solutions like this with tf.py_func are not really an option for me. Anything which would use tf.range(n) or so is also not an option because n could be very large.

Answer 1

This is one way:
n = ...
sample_size = ...
idx = tf.random_shuffle(tf.range(n))[:sample_size]
EDIT:
I had posted the answer below but then read the last line of your post. I don't think there is a good way to do it if you absolutely cannot produce a tensor with size O(n) (numpy.random.choice with replace=False is also implemented as a slice of a permutation). You could resort to a tf.while_loop until you have unique indices:
n = ...
sample_size = ...
idx = tf.zeros(sample_size, dtype=tf.int64)
idx = tf.while_loop(
lambda i: tf.size(idx) == tf.size(tf.unique(idx)),
lambda i: tf.random_uniform(sample_size, maxval=n, dtype=int64))
EDIT 2:
About the average number of iterations in the previous method. If we call n the number of possible values and k the length of the desired vector (with k ≤ n), the probability that an iteration is successful is:
p = product((n - (i - 1) / n) for i in 1 .. k)
Since each iteartion can be considered a Bernoulli trial, the average number of trials unitl first success is 1 / p (proof here). Here is a function that calculates the average numbre of trials in Python for some k and n values:
def avg_iter(k, n):
if k > n or n <= 0 or k < 0:
raise ValueError()
avg_it = 1.0
for p in (float(n) / (n - i) for i in range(k)):
avg_it *= p
return avg_it
And here are some results:
+-------+------+----------+
| n | k | Avg iter |
+-------+------+----------+
| 10 | 5 | 3.3 |
| 100 | 10 | 1.6 |
| 1000 | 10 | 1.1 |
| 1000 | 100 | 167.8 |
| 10000 | 10 | 1.0 |
| 10000 | 100 | 1.6 |
| 10000 | 1000 | 2.9e+22 |
+-------+------+----------+
You can see it varies wildy depending on the parameters.
It is possible, though, to construct a vector in a fixed number of steps, although the only algorithm I can think of is O(k2). In pure Python it goes like this:
import random
def sample_wo_replacement(n, k):
sample = [0] * k
for i in range(k):
sample[i] = random.randint(0, n - 1 - len(sample))
for i, v in reversed(list(enumerate(sample))):
for p in reversed(sample[:i]):
if v >= p:
v += 1
sample[i] = v
return sample
random.seed(100)
print(sample_wo_replacement(10, 5))
# [2, 8, 9, 7, 1]
print(sample_wo_replacement(10, 10))
# [6, 5, 8, 4, 0, 9, 1, 2, 7, 3]
This is a possible way to do it in TensorFlow (not sure if the best one):
import tensorflow as tf
def sample_wo_replacement_tf(n, k):
# First loop
sample = tf.constant([], dtype=tf.int64)
i = 0
sample, _ = tf.while_loop(
lambda sample, i: i < k,
# This is ugly but I did not want to define more functions
lambda sample, i: (tf.concat([sample,
tf.random_uniform([1], maxval=tf.cast(n - tf.shape(sample)[0], tf.int64), dtype=tf.int64)],
axis=0),
i + 1),
[sample, i], shape_invariants=[tf.TensorShape((None,)), tf.TensorShape(())])
# Second loop
def inner_loop(sample, i):
sample_size = tf.shape(sample)[0]
v = sample[i]
j = i - 1
v, _ = tf.while_loop(
lambda v, j: j >= 0,
lambda v, j: (tf.cond(v >= sample[j], lambda: v + 1, lambda: v), j - 1),
[v, j])
return (tf.where(tf.equal(tf.range(sample_size), i), tf.tile([v], (sample_size,)), sample), i - 1)
i = tf.shape(sample)[0] - 1
sample, _ = tf.while_loop(lambda sample, i: i >= 0, inner_loop, [sample, i])
return sample
And an example:
with tf.Graph().as_default(), tf.Session() as sess:
tf.set_random_seed(100)
sample = sample_wo_replacement_tf(10, 5)
for i in range(10):
print(sess.run(sample))
# [3 0 6 8 4]
# [5 4 8 9 3]
# [1 4 0 6 8]
# [8 9 5 6 7]
# [7 5 0 2 4]
# [8 4 5 3 7]
# [0 5 7 4 3]
# [2 0 3 8 6]
# [3 4 8 5 1]
# [5 7 0 2 9]
This is quite intesive on tf.while_loops, though, which are well-known not to be particularly fast in TensorFlow, so I wouldn't know how fast can you really get with this method without some kind of benchmarking.
EDIT 4:
One last possible method. You can divide the range of possible values (0 to n) in "chunks" of size c and pick a random amount of numbers from each chunk, then shuffle everything. The amount of memory that you use is limited by c, and you don't need nested loops. If n is divisible by c, then you should get about a perfect random distribution, otherwise values in the last "short" chunk would receive some extra probability (this may be negligible depending on the case). Here is a NumPy implementation. It is somewhat long to account for different corner cases and pitfalls, but if c ≥ k and n mod c = 0 several parts get simplified.
import numpy as np
def sample_chunked(n, k, chunk=None):
chunk = chunk or n
last_chunk = chunk
parts = n // chunk
# Distribute k among chunks
max_p = min(float(chunk) / k, 1.0)
max_p_last = max_p
if n % chunk != 0:
parts += 1
last_chunk = n % chunk
max_p_last = min(float(last_chunk) / k, 1.0)
p = np.full(parts, 2)
# Iterate until a valid distribution is found
while not np.isclose(np.sum(p), 1) or np.any(p > max_p) or p[-1] > max_p_last:
p = np.random.uniform(size=parts)
p /= np.sum(p)
dist = (k * p).astype(np.int64)
sample_size = np.sum(dist)
# Account for rounding errors
while sample_size < k:
i = np.random.randint(len(dist))
while (dist[i] >= chunk) or (i == parts - 1 and dist[i] >= last_chunk):
i = np.random.randint(len(dist))
dist[i] += 1
sample_size += 1
while sample_size > k:
i = np.random.randint(len(dist))
while dist[i] == 0:
i = np.random.randint(len(dist))
dist[i] -= 1
sample_size -= 1
assert sample_size == k
# Generate sample parts
sample_parts = []
for i, v in enumerate(np.nditer(dist)):
if v <= 0:
continue
c = chunk if i < parts - 1 else last_chunk
base = chunk * i
sample_parts.append(base + np.random.choice(c, v, replace=False))
sample = np.concatenate(sample_parts, axis=0)
np.random.shuffle(sample)
return sample
np.random.seed(100)
print(sample_chunked(15, 5, 4))
# [ 8 9 12 13 3]
A quick benchmark of sample_chunked(100000000, 100000, 100000) takes about 3.1 seconds in my computer, while I haven't been able to run the previous algorithm (sample_wo_replacement function above) to completion with the same parameters. It should be possible to implement it in TensorFlow, maybe using tf.TensorArray, although it would require significant effort to get it exactly right.

Answer 2

use the gumbel-max trick here: https://github.com/tensorflow/tensorflow/issues/9260
z = -tf.log(-tf.log(tf.random_uniform(tf.shape(logits),0,1)))
_, indices = tf.nn.top_k(logits + z,K)
indices are what you want. This tick is so easy~!

Answer 3

The following works fairly fast on the GPU, and I did not encounter memory issues when using n~100M and k~10k (using NVIDIA GeForce GTX 1080 Ti):
def random_choice_without_replacement(n, k):
"""equivalent to 'numpy.random.choice(n, size=k, replace=False)'"""
return tf.math.top_k(tf.random.uniform(shape=[n]), k, sorted=False).indices