Choosing random elements from a set of items is a problem that appears regularly. If you are already given the array of items, this can be implemented trivially:
import random
def random_element( a, N ):
return a[ int( random.random() * N ) ]
Pick a number between 0 to N - 1, and return the element with that index.
Before we go on, here are “Some Minor Disclaimers”:
I’m aware that Python’s random module has some of these implemented. You are more than encouraged to use those, as the first thing you should do is if someone else did it already! The
Ncan easily be substituted withlen(a). It is there to make the parameter explicit. This article is meant for teaching and the coding style aims to make the code as clear (and language-neutral) as possible, and intentionally does not take advantage of a certain Python conciseness!
The Perl Cookbook has an optimization that is subtle, but amazingly concise:
srand;
rand($.) < 1 && ($line = $_) while <>;
# $line is the random line
Rewritten in Python, with an iterator instead of a file:
import random
def random_element_iterator( iterator ):
N = 0
for item in iterator:
N += 1
# 1/N chance
if random.random() * N < 1:
element = item
return element
For the N-th element of the iterator, there is a 1/N probability of keeping that line. When I first came upon it, it worked like magic. It doesn’t have to stay that way — you can prove this with induction.
The base case N = 1 is true: the first element has 100% chance of being picked. The second element has 50% chance of being picked (and in this case, the first element is kept with 50% probability as well).
Suppose for the first N items, the probability of choosing any of the N elements is equal — 1/N. When you process the item N+1, the probability of it being chosen is 1/(N+1), which is by definition above. The probability of choosing any previous item was 1/N, and with N/(N+1) probability of staying that way - (1/N) * (N /(N+1)) = 1/(N+1). Each of the N+1 items now has 1/(N+1) of being chosen.
This was my first exposure to an online algorithm — you don’t need to know how many elements we need to account for. This is particularly practical, as in the Perl Cookbook example, it means you do not have to keep all the data in the memory to retrieve a random item — you can only need enough memory to store one item.
It turns out that this was covered in Knuth’s “The Art of Computer Programming”, which I read a few years later:
Can we pick a random element from a set of items whose cardinality we do not know?
Let’s revisit the naive algorithm. It’s as simple as returning one element based on a randomly generated index.
Now let’s take a look at the Fisher-Yates algorithm:
import random
def shuffle( a, N ):
for i in range( N - 1, 0, -1 ):
j = int( random.random() * i )
a[i], a[j] = a[j], a[i]
return a
This is the canonical way to create an unbiased shuffled array, meaning each of the possible permutation is equally likely. This is O( N ) time, which is as good as it gets, with O( N ) space. On every step of the iteration, you’re getting one random item from the remaining items. Did you make the connection? The naive algorithm can be viewed as one step of this iteration, returning the first element.
Disclaimer: It is very easy to implement Fisher-Yates incorrectly, including Sattolo’s algorithm. Double check your implementation!
Now we’re at the general problem - picking K distinct elements randomly out of a set of N items. The keyword here is distinct — if you want to choose the same item multiple times, just run the above algorithm K times. That is essentially picking K elements out of a set of N elements, with replacement. What we want is an algorithm for picking K elements out of a set of N elements, without replacement.
A simple, naive solution for choosing distinct elements is to keep track of what elements we’ve already picked:
# Assumes K <= N, or this will never terminate
import random
def random_subset_track( a, N, K ):
used = [ False ] * N
result = []
for i in range( 0, K ):
# get a random element between 0 and N.
j = int( random.random() * N )
while ( used[j] ):
j = int( random.random() * N )
result.append( a[j] )
used[j] = True
return result
This might scream inefficient to you, as it should. Linear in the base case, it probably will take longer. A better solution, with some foreshadowing, is to simply use the Fisher-Yates shuffle! Get one of the N! permutations, and take the first K elements:
import random
# Use the slice a[:K]
def shuffle( a, N ):
for i in range( N - 1, 0, -1 ):
j = int( random.random() * i )
a[i], a[j] = a[j], a[i]
return a
Can we do better? Yes! Wait, didn’t I say this was O( N ), and thus optimal? How can you do better? Well, it turns out it is O( N ), and thus optimal in terms of N, but you can do better if you analyze its running time in terms of K, its output — for this problem, K can be much smaller than N, and thus we can do this problem in O(K) time with a small tweak — as we briefly mentioned before, each iteration of the Fisher-Yates algorithm will net you one item that is unused. Just stop early after K iterations:
import random
def random_subset_fy( a, N, K ):
for i in range( N - 1, N - K - 1, -1 ):
j = int( random.random() * i )
a[i], a[j] = a[j], a[i]
return a[ N - K : N ]
(since we update the end of array, we want the last K items)
This is a simple example of an output-sensitive algorithm.
Don’t have the memory to store all the items at the same time? If you know the number of items, then the probability of choosing an element, assuming K more element is needed out of N elements, is simply K/N:
import random
def random_subset_p( a, N, K ):
result = []
for i in range( N ):
# We want to keep this element with probability K/N
if ( random.random() * ( N - i ) < K ):
K -= 1
result.append( a[i] )
return result
We can prove this, once again, with induction.
However, this still requires us to know how big the set is. To paraphrase Knuth:
Can we pick K random elements from a set of items whose cardinality we do not know?
Yes! Similar to the previous Perl solution, for the N-th item, keep it with K / N probability. If we keep it, replace one of K-th entries at random (with 1/K probability):
import random
def random_subset( iterator, K ):
result = []
N = 0
for item in iterator:
N += 1
if len( result ) < K:
result.append( item )
else:
s = int(random.random() * N)
if s < K:
result[ s ] = item
return result
We will prove this with induction.
Base case:
For the first K items, you have 100% chance of keeping the current items.
For the K+1-th item, this item has a K/(K+1) probability of being chosen.
Each previously chosen item has the probability of 1/(K+1) probability of the new item being ignored, plus a K/(K+1) probability that the new item will be chosen, with a (K-1)/K probability that another previously chosen item will be replaced (instead of this one). This is:
Now suppose for the first N items, the algorithm is correct, that is, each previous item had a K/N chance of being kept.
When you consider the item N+1, the probability of it being kept is K/(N+1), which is by definition.
The probability of keeping the previous items was K/N. With (N-K+1)/(N+1) probability that the current item is not kept, plus the probability that the current item is kept (K/(N+1)), but other items were chosen instead ((K-1)/K).
For the N+1-th item, it has K/(N+1) probability of being kept.
This is now still O( N ) time, but with a reasonable O( K ) space. Great for when you can’t just put everything in memory!
There is a difference between this and the Fisher-Yates version — while both return K items, this algorithm does not return the K uniformly in a random order, e.g., the first element is always found in the first position, if it is chosen in the K items. If the order of the K items matter, then simply perform a Fisher-Yates shuffle on the result. (Extra Credit: There is also an O( N log K ) algorithm that preserves this property, without the memory overhead of the Fisher-Yates version, using only O( K ) space.)
And here we go, from start to finish, the best algorithms (and a wrong one) for choosing K distinct elements out of N elements, even if we don’t necessarily know what N is in the beginning!
