# Reservoir sampling

Reservoir sampling is a family of randomized algorithms for choosing a simple random sample, without replacement, of k items from a population of unknown size n in a single pass over the items. The size of the population n is not known to the algorithm and is typically too large for all n items to fit into main memory. The population is revealed to the algorithm over time, and the algorithm cannot look back at previous items. At any point, the current state of the algorithm must permit extraction of a simple random sample without replacement of size k over the part of the population seen so far.

## Motivation

Suppose we see a sequence of items, one at a time. We want to keep ten items in memory, and we want them to be selected at random from the sequence. If we know the total number of items n and can access the items arbitrarily, then the solution is easy: select 10 distinct indices i between 1 and n with equal probability, and keep the i-th elements. The problem is that we do not always know the exact n in advance.

## Simple algorithm

A simple and popular but slow algorithm, commonly known as Algorithm R, is due to Alan Waterman.[1]

The algorithm works by maintaining a reservoir of size k, which initially contains the first k items of the input. It then iterates over the remaining items until the input is exhausted. Using one-based array indexing, let ${\displaystyle i>k}$ be the index of the item currently under consideration. The algorithm then generates a random number j between (and including) 1 and i. If j is at most k, then the item is selected and replaces whichever item currently occupies the j-th position in the reservoir. Otherwise, the item is discarded. In effect, for all i, the ith element of the input is chosen to be included in the reservoir with probability ${\displaystyle k/i}$. Similarly, at each iteration the jth current element of the reservoir array is chosen to be replaced by the new ith element with probability ${\displaystyle 1/k\times k/i=1/i}$. When the algorithm has finished executing, each item in the input population has equal probability (i.e., ${\displaystyle k/n}$) of being chosen for the reservoir:

${\displaystyle {\frac {k}{j}}\times {\Bigl (}1-{\frac {1}{j+1}}{\Bigr )}\times {\Bigl (}1-{\frac {1}{j+2}}{\Bigr )}\times {\Bigl (}1-{\frac {1}{j+3}}{\Bigr )}\times \cdots \times {\Bigl (}1-{\frac {1}{n}}{\Bigr )}={\frac {k}{j}}\prod _{i=j+1}^{n}{\Bigl (}1-{\frac {1}{i}}{\Bigr )}={\frac {k}{n}}.}$

Here, ${\displaystyle {\frac {k}{j}}}$ is the probability of selecting the jth element, while ${\displaystyle 1-{\frac {k}{i}}\times {\frac {1}{k}}=1-{\frac {1}{i}}}$ is the complementary probability of both selecting the ith element afterwards (probability ${\displaystyle {\frac {k}{i}}}$) and replacing the previously selected jth element with it (probability ${\displaystyle {\frac {1}{k}}}$) for any future index ${\displaystyle i>j}$.

(* S has items to sample, R will contain the result *)
ReservoirSample(S[1..n], R[1..k])
// fill the reservoir array
for i := 1 to k
R[i] := S[i]

// replace elements with gradually decreasing probability
for i := k+1 to n
(* randomInteger(a, b) generates a uniform integer from the inclusive range {a, ..., b} *)
j := randomInteger(1, i)
if j <= k
R[j] := S[i]


While conceptually simple and easy to understand, this algorithm needs to generate a random number for each item of the input, including the items that are discarded. Its asymptotic running time is thus ${\displaystyle O(n)}$. This causes the algorithm to be unnecessarily slow if the input population is large.

## An optimal algorithm

Algorithm L[2] improves upon this algorithm by computing how many items are discarded before the next item enters the reservoir. The key observation is that this number follows a geometric distribution and can therefore be computed in constant time.

(* S has items to sample, R will contain the result *)
ReservoirSample(S[1..n], R[1..k])
// fill the reservoir array
for i = 1 to k
R[i] := S[i]

(* random() generates a uniform (0,1) random number *)
W := exp(log(random())/k)

while i <= n
i := i + floor(log(random())/log(1-W)) + 1
if i <= n
(* replace a random item of the reservoir with item i *)
R[randomInteger(1,k)] := S[i]  // random index between 1 and k, inclusive
W := W * exp(log(random())/k)


This algorithm computes three random numbers for each item that becomes part of the reservoir, and does not spend any time on items that do not. Its expected running time is thus ${\displaystyle O(k(1+log(n/k)))}$,[2] which is optimal.[1] At the same time, it is simple to implement efficiently and does not depend on random deviates from exotic or hard-to-compute distributions.

## With random sort

If we associate with each item of the input a uniformly generated random number, the k items with the largest (or, equivalently, smallest) associated values form a simple random sample.[3] A simple reservoir-sampling thus maintains the k items with the currently largest associated values in a priority queue.

(*
S is a stream of items to sample
S.Current returns current item in stream
S.Next advances stream to next position
min-priority-queue supports:
Count -> number of items in priority queue
Minimum -> returns minimum key value of all items
Extract-Min() -> Remove the item with minimum key
Insert(key, Item) -> Adds item with specified key
*)
ReservoirSample(S[1..?])
H := new min-priority-queue
while S has data
r := random()   // uniformly random between 0 and 1, exclusive
if H.Count < k
H.Insert(r, S.Current)
else
// keep k items with largest associated keys
if r > H.Minimum
H.Extract-Min()
H.Insert(r, S.Current)
S.Next
return items in H


The expected running time of this algorithm is ${\displaystyle O(n+k\log k\log(n/k))}$ and it is relevant mainly because it can easily be extended to items with weights.

## Weighted random sampling

Some applications require items' sampling probabilities to be according to weights associated with each item. For example, it might be required to sample queries in a search engine with weight as number of times they were performed so that the sample can be analyzed for overall impact on user experience. Let the weight of item i be ${\displaystyle w_{i}}$, and the sum of all weights be W. There are two ways to interpret weights assigned to each item in the set:[4]

1. In each round, the probability of every unselected item to be selected in that round is proportional to its weight relative to the weights of all unselected items. If X is the current sample, then the probability of an item ${\displaystyle i\notin X}$ to be selected in the current round is ${\displaystyle \textstyle w_{i}/(W-\sum _{j\in X}{w_{j}})}$.
2. The probability of each item to be included in the random sample is proportional to its relative weight, i.e., ${\displaystyle w_{i}/W}$. Note that this interpretation might not be achievable in some cases, e.g., ${\displaystyle k=n}$.

### Algorithm A-Res

The following algorithm was given by Efraimidis and Spirakis that uses interpretation 1:[5]

(*
S is a stream of items to sample
S.Current returns current item in stream
S.Weight  returns weight of current item in stream
S.Next advances stream to next position
The power operator is represented by ^
min-priority-queue supports:
Count -> number of items in priority queue
Minimum() -> returns minimum key value of all items
Extract-Min() -> Remove the item with minimum key
Insert(key, Item) -> Adds item with specified key
*)
ReservoirSample(S[1..?])
H := new min-priority-queue
while S has data
r := random() ^ (1/S.Weight)   // random() produces a uniformly random number in (0,1)
if H.Count < k
H.Insert(r, S.Current)
else
// keep k items with largest associated keys
if r > H.Minimum
H.Extract-Min()
H.Insert(r, S.Current)
S.Next
return items in H


This algorithm is identical to the algorithm given in Reservoir Sampling with Random Sort except for the generation of the items' keys. The algorithm is equivalent to assigning each item a key ${\displaystyle r^{1/w_{i}}}$ where r is the random number and then selecting the k items with the largest keys. Equivalently, a more numerically stable formulation of this algorithm computes the keys as ${\displaystyle -\ln(r)/w_{i}}$ and select the k items with the smallest keys.[6][failed verification]

### Algorithm A-ExpJ

The following algorithm is a more efficient version of A-Res, also given by Efraimidis and Spirakis:[5]

(*
S is a stream of items to sample
S.Current returns current item in stream
S.Weight  returns weight of current item in stream
S.Next advances stream to next position
The power operator is represented by ^
min-priority-queue supports:
Count -> number of items in the priority queue
Minimum -> minimum key of any item in the priority queue
Extract-Min() -> Remove the item with minimum key
Insert(Key, Item) -> Adds item with specified key
*)
ReservoirSampleWithJumps(S[1..?])
H := new min-priority-queue
while S has data and H.Count < k
r := random() ^ (1/S.Weight)   // random() produces a uniformly random number in (0,1)
H.Insert(r, S.Current)
S.Next
X := log(random()) / log(H.Minimum) // this is the amount of weight that needs to be jumped over
while S has data
X := X - S.Weight
if X <= 0
t := H.Minimum ^ S.Weight
r := random(t, 1) ^ (1/S.Weight) // random(x, y) produces a uniformly random number in (x, y)

H.Extract-Min()
H.Insert(r, S.Current)

X := log(random()) / log(H.Minimum)
S.Next
return items in H


This algorithm follows the same mathematical properties that are used in A-Res, but instead of calculating the key for each item and checking whether that item should be inserted or not, it calculates an exponential jump to the next item which will be inserted. This avoids having to create random variates for each item, which may be expensive. The number of random variates required is reduced from ${\displaystyle O(n)}$ to ${\displaystyle O(k\log(n/k))}$ in expectation, where ${\displaystyle k}$ is the reservoir size, and ${\displaystyle n}$ is the number of items in the stream.[5]

### Algorithm A-Chao

Following algorithm was given by M. T. Chao uses interpretation 2:[7]

(*
S has items to sample, R will contain the result
S[i].Weight contains weight for each item
*)
WeightedReservoir-Chao(S[1..n], R[1..k])
WSum := 0
// fill the reservoir array
for i := 1 to k
R[i] := S[i]
WSum := WSum + S[i].Weight
for i := k+1 to n
WSum := WSum + S[i].Weight
p := S[i].Weight / WSum // probability for this item
j := random();          // uniformly random between 0 and 1
if j <= p               // select item according to probability
R[randomInteger(1,k)] := S[i]  //uniform selection in reservoir for replacement


For each item, its relative weight is calculated and used to randomly decide if the item will be added into the reservoir. If the item is selected, then one of the existing items of the reservoir is uniformly selected and replaced with the new item. The trick here is that, if the probabilities of all items in the reservoir are already proportional to their weights, then by selecting uniformly which item to replace, the probabilities of all items remain proportional to their weight after the replacement.

#### Algorithm A-Chao with Jumps

Similar to the other algorithms, it is possible to compute a random weight j and subtract items' probability mass values, skipping them while j > 0, reducing the number of random numbers that have to be generated.[4]

(*
S has items to sample, R will contain the result
S[i].Weight contains weight for each item
*)
WeightedReservoir-Chao(S[1..n], R[1..k])
WSum := 0
// fill the reservoir array
for i := 1 to k
R[i] := S[i]
WSum := WSum + S[i].Weight
j := random() // uniformly random between 0 and 1
pNone := 1 // probability that no item has been selected so far (in this jump)
for i := k+1 to n
WSum := WSum + S[i].Weight
p := S[i].Weight / WSum // probability for this item
j -= p * pNone
pNone := pNone * (1 - p)
if j <= 0
R[randomInteger(1,k)] := S[i]  //uniform selection in reservoir for replacement
j = random()
pNone := 1


## Relation to Fisher–Yates shuffle

Suppose one wanted to draw k random cards from a deck of cards. A natural approach would be to shuffle the deck and then take the top k cards. In the general case, the shuffle also needs to work even if the number of cards in the deck is not known in advance, a condition which is satisfied by the inside-out version of the Fisher–Yates shuffle:[8]

(* S has the input, R will contain the output permutation *)
Shuffle(S[1..n], R[1..n])
R[1] := S[1]
for i from 2 to n do
j := randomInteger(1, i)  // inclusive range
R[i] := R[j]
R[j] := S[i]


Note that although the rest of the cards are shuffled, only the first k are important in the present context. Therefore, the array R need only track the cards in the first k positions while performing the shuffle, reducing the amount of memory needed. Truncating R to length k, the algorithm is modified accordingly:

(* S has items to sample, R will contain the result *)
ReservoirSample(S[1..n], R[1..k])
R[1] := S[1]
for i from 2 to k do
j := randomInteger(1, i)  // inclusive range
R[i] := R[j]
R[j] := S[i]
for i from k + 1 to n do
j := randomInteger(1, i)  // inclusive range
if (j <= k)
R[j] := S[i]


Since the order of the first k cards is immaterial, the first loop can be removed and R can be initialized to be the first k items of the input. This yields Algorithm R.

## Statistical properties

Probabilities of selection of the reservoir methods are discussed in Chao (1982)[7] and Tillé (2006).[9] While the first-order selection probabilities are equal to ${\displaystyle k/n}$ (or, in case of Chao's procedure, to an arbitrary set of unequal probabilities), the second order selection probabilities depend on the order in which the records are sorted in the original reservoir. The problem is overcome by the cube sampling method of Deville and Tillé (2004).[10]

## Limitations

Reservoir sampling makes the assumption that the desired sample fits into main memory, often implying that k is a constant independent of n. In applications where we would like to select a large subset of the input list (say a third, i.e. ${\displaystyle k=n/3}$), other methods need to be adopted. Distributed implementations for this problem have been proposed.[11]

## References

1. ^ a b Vitter, Jeffrey S. (1 March 1985). "Random sampling with a reservoir" (PDF). ACM Transactions on Mathematical Software. 11 (1): 37–57. CiteSeerX 10.1.1.138.784. doi:10.1145/3147.3165. S2CID 17881708.
2. ^ a b Li, Kim-Hung (4 December 1994). "Reservoir-Sampling Algorithms of Time Complexity O(n(1+log(N/n)))". ACM Transactions on Mathematical Software. 20 (4): 481–493. doi:10.1145/198429.198435. S2CID 15721242.
3. ^ Fan, C.; Muller, M.E.; Rezucha, I. (1962). "Development of sampling plans by using sequential (item by item) selection techniques and digital computers". Journal of the American Statistical Association. 57 (298): 387–402. doi:10.1080/01621459.1962.10480667. JSTOR 2281647.
4. ^ a b Efraimidis, Pavlos S. (2015). "Weighted Random Sampling over Data Streams". Algorithms, Probability, Networks, and Games. Lecture Notes in Computer Science. 9295: 183–195. arXiv:1012.0256. doi:10.1007/978-3-319-24024-4_12. ISBN 978-3-319-24023-7. S2CID 2008731.
5. ^ a b c Efraimidis, Pavlos S.; Spirakis, Paul G. (2006-03-16). "Weighted random sampling with a reservoir". Information Processing Letters. 97 (5): 181–185. doi:10.1016/j.ipl.2005.11.003.
6. ^ Arratia, Richard (2002). Bela Bollobas (ed.). "On the amount of dependence in the prime factorization of a uniform random integer". Contemporary Combinatorics. 10: 29–91. CiteSeerX 10.1.1.745.3975. ISBN 978-3-642-07660-2.
7. ^ a b Chao, M. T. (1982). "A general purpose unequal probability sampling plan". Biometrika. 69 (3): 653–656. doi:10.1093/biomet/69.3.653.
8. ^ National Research Council (2013). Frontiers in Massive Data Analysis. The National Academies Press. p. 121. ISBN 978-0-309-28781-4.
9. ^ Tillé, Yves (2006). Sampling Algorithms. Springer. ISBN 978-0-387-30814-2.
10. ^ Deville, Jean-Claude; Tillé, Yves (2004). "Efficient balanced sampling: The cube method" (PDF). Biometrika. 91 (4): 893–912. doi:10.1093/biomet/91.4.893.
11. ^ Reservoir Sampling in MapReduce