Parallelized sampling utilizing exponential variates

As a part of our current work to help weighted sampling of Spark knowledge frames in sparklyr, we launched into a journey looking for algorithms that may carry out weighted sampling, particularly sampling with out substitute, in environment friendly and scalable methods inside a distributed cluster-computing framework, equivalent to Apache Spark.

Within the curiosity of brevity, “weighted sampling with out substitute” shall be shortened into SWoR for the rest of this weblog submit.

Within the following sections, we’ll clarify and illustrate what SWoR means probability-wise, briefly define some various options now we have thought-about however weren’t fully glad with, after which deep-dive into exponential variates, a easy mathematical assemble that made the best answer for this downside potential.

Should you can’t wait to leap into motion, there’s additionally a part through which we showcase instance usages of sdf_weighted_sample() in sparklyr. As well as, you may study the implementation element of sparklyr::sdf_weighted_sample() on this pull request.

How it began

Our journey began from a Github problem inquiring about the potential for supporting the equal of dplyr::sample_frac(..., weight = <weight_column>) for Spark knowledge frames in sparklyr. For instance,

dplyr::sample_frac(mtcars, 0.25, weight = gear, exchange = FALSE)

##                    mpg cyl  disp  hp drat    wt  qsec vs am gear carb
## Merc 280C         17.8   6 167.6 123 3.92 3.440 18.90  1  0    4    4
## Chrysler Imperial 14.7   8 440.0 230 3.23 5.345 17.42  0  0    3    4
## Fiat X1-9         27.3   4  79.0  66 4.08 1.935 18.90  1  1    4    1
## Hornet Sportabout 18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2
## Valiant           18.1   6 225.0 105 2.76 3.460 20.22  1  0    3    1
## Porsche 914-2     26.0   4 120.3  91 4.43 2.140 16.70  0  1    5    2
## Maserati Bora     15.0   8 301.0 335 3.54 3.570 14.60  0  1    5    8
## Ferrari Dino      19.7   6 145.0 175 3.62 2.770 15.50  0  1    5    6

will randomly choose one-fourth of all rows from a R knowledge body named “mtcars” with out substitute, utilizing mtcars$gear as weights. We had been unable to search out any operate implementing the weighted variations of dplyr::sample_frac amongst Spark SQL built-in capabilities in Spark 3.0 or in earlier variations, which implies a future model of sparklyr might want to run its personal weighted sampling algorithm to help such use circumstances.

What precisely is SWoR

The aim of this part is to mathematically describe the likelihood distribution generated by SWoR by way of (w_1, dotsc, w_N), in order that readers can clearly see that the exponential-variate based mostly algorithm offered in a subsequent part in actual fact samples from exactly the identical likelihood distribution. Readers already having a crystal-clear psychological image of what SWoR entails ought to most likely skip most of this part. The important thing take-away right here is given (N) rows (r_1, dotsc, r_N) and their weights (w_1, dotsc, w_N) and a desired pattern dimension (n), the likelihood of SWoR deciding on ((r_1, dotsc, r_n)) is (prodlimits_{j = 1}^{n} left( {w_j} center/ {sumlimits_{ok = j}^{N}{w_k}} proper)).

SWOR is conceptually equal to a (n)-step course of of choosing 1 out of ((n – j + 1)) remaining rows within the (j)-th step for (j in {1, dotsc, n}), with every remaining row’s chance of getting chosen being linearly proportional to its weight in any of the steps, i.e.,

samples := {}
inhabitants := {r[1], ..., r[N]}

for j = 1 to n
  choose r[x] from inhabitants with likelihood
    (w[x] / TotalWeight(inhabitants))
  samples := samples + {r[x]}
  inhabitants := inhabitants - {r[x]}

Discover the end result of a SWoR course of is in actual fact order-significant, which is why on this submit it is going to at all times be represented as an ordered tuple of components.

Intuitively, SWoR is analogous to throwing darts at a bunch of tiles. For instance, let’s say the dimensions of our pattern area is 5:

• Think about (r_1, r_2, dotsc, r_5) as 5 rectangular tiles laid out contiguously on a wall with widths (w_1, w_2, dotsc, w_5), with (r_1) protecting ([0, w_1)), (r_2) covering ([w_1, w_1 + w_2)), …, and (r_5) covering (left[sumlimits_{j = 1}^{4} w_j, sumlimits_{j = 1}^{5} w_jright))

• Equate drawing a random sample in each step to throwing a dart uniformly randomly within the interval covered by all tiles that are not hit yet

• After a tile is hit, it gets taken out and remaining tiles are re-arranged so that they continue to cover a contiguous interval without overlapping

If our sample size is 3, then we shall ask ourselves what is the probability of the dart hitting ((r_1, r_2, r_3)) in that order?

In step (j = 1), the dart will hit (r_1) with probability (left. w_1 middle/ left(sumlimits_{k = 1}^{N}w_kright) right.)

.

After deleting (r_1) from the sample space after it’s hit, step (j = 2) will look like this:

,

and the probability of the dart hitting (r_2) in step 2 is (left. w_2 middle/ left(sumlimits_{k = 2}^{N}w_kright) right.) .

Finally, moving on to step (j = 3), we have:

,

with the probability of the dart hitting (r_3) being (left. w_3 middle/ left(sumlimits_{k = 3}^{N}w_kright) right.).

So, combining all of the above, the overall probability of selecting ((r_1, r_2, r_3)) is (prodlimits_{j = 1}^{3} left( {w_j} middle/ {sumlimits_{k = j}^{N}{w_k}} right)).

Naive approaches for implementing SWoR

This section outlines some possible approaches that were briefly under consideration. Because none of these approaches scales well to a large number of rows or a non-trivial number of partitions in a Spark data frame, we decided to avoid all of them in sparklyr.

A tree-base approach

One possible way to accomplish SWoR is to have a mutable data structure keeping track of the sample space at each step.

Continuing with the dart-throwing analogy from the previous section, let us say initially, none of the tiles has been taken out yet, and a dart has landed at some point (x in left[0, sumlimits_{k = 1}^{N} w_kright)). Which tile did it hit? This can be answered efficiently if we have a binary tree, pictured as the following (or in general, some (b)-ary tree for integer (b ge 2))

To find the tile that was hit given the dart’s position (x), we simply need to traverse down the tree, going through the box containing (x) in each level, incurring a (O(log(N))) cost in time complexity for each sample. To take a tile out of the picture, we update the width of the tile to (0) and propagate this change upwards from leaf level to root of the tree, again incurring a (O(log(N))) cost in time complexity, making the overall time complexity of selecting (n) samples (O(n cdot log(N))), which is not so great for large data sets, and also, not parallelizable across multiple partitions of a Spark data frame.

Rejection sampling

Another possible approach is to use rejection sampling. In term of the previously mentioned dart-throwing analogy, that means not removing any tile that is hit, hence avoiding the performance cost of keeping the sample space up-to-date, but then having to re-throw the dart in each of the subsequent rounds until the dart lands on a tile that was not hit previously. This approach, just like the previous one, would not be performant, and would not be parallelizable across multiple partitions of a Spark data frame either.

A solution that has proven to be much better than any of the naive approaches turns out to be a numerical stable variant of the algorithm described in “Weighted Random Sampling” (Efraimidis and Spirakis 2016) by Pavlos S. Efraimidis and Paul G. Spirakis.

A version of this sampling algorithm implemented by sparklyr does the following to sample (n) out of (N) rows from a Spark data frame (X):

• For each row (r_j in X), draw a random number (u_j) independently and uniformly randomly from ((0, 1)) and compute the key of (r_j) as (k_j = ln(u_j) / w_j), where (w_j) is the weight of (r_j). Perform this calulation in parallel across all partitions of (X).
• Select (n) rows with largest keys and return them as the result. This step is also mostly parallelizable: for each partition of (X), one can select up to (n) rows having largest keys within that partition as candidates, and after selecting candidates from all partitions in parallel, simply extract the top (n) rows among all candidates, and return them as the (n) chosen samples.

There are at least 4 reasons why this solution is highly appealing and was chosen to be implemented in sparklyr:

• It is a one-pass algorithm (i.e., only need to iterate through all rows of a data frame exactly once).
• Its computational overhead is quite low (as selecting top (n) rows at any stage only requires a bounded priority queue of max size (n), which costs (O(log(n))) per update in time complexity).
• More importantly, most of its required computations can be performed in parallel. In fact, the only non-parallelizable step is the very last stage of combining top candidates from all partitions and choosing the top (n) rows among those candidates. So, it fits very well into the world of Spark / MapReduce, and has drastically better horizontal scalability compared to the naive approaches.
• Bonus: It is also suitable for weighted reservoir sampling (i.e., can sample (n) out of a possibly infinite stream of rows according to their weights such that at any moment the (n) samples will be a weighted representation of all rows that have been processed so far).

Why does this algorithm work

As an interesting aside, some readers have probably seen this technique presented in a slightly different form under another name. It is in fact equivalent to a generalized version of the Gumbel-max trick which is commonly referred to as the Gumbel-top-k trick. Readers familiar with properties of the Gumbel distribution will no doubt have an easy time convincing themselves the algorithm above works as expected.

In this section, we will also present a proof of correctness for this algorithm based on elementary properties of probability density function (shortened as PDF from now on), cumulative distribution function (shortened as CDF from now on), and basic calculus.

First of all, to make sense of all the (ln(u_j) / w_j) calculations in this algorithm, one has to understand inverse transform sampling. For each (j in {1, dotsc, N}), consider the probability distribution defined on ((-infty, 0)) with CDF (F_j(x) = e^{w_j cdot x}). In order to pluck out a value (y) from this distribution, we first sample a value (u_j) uniformly randomly out of ((0, 1)) that determines the percentile of (y) (i.e., how our (y) value ranks relative to all possible (y) values, a.k.a, the “overall population,” from this distribution), and then apply (F_j^{-1}) to (u_j) to find (y), so, (y = F_j^{-1}(u_j) = ln(u_j) / w_j).

Secondly, after defining all the required CDF functions (F_j(x) = e^{w_j cdot x}) for (j in {1, dotsc, N}), we can also easily derive their corresponding PDF functions (f_j): [f_j(x) = frac{d F_j(x)}{dx} = w_j e^{w_j cdot x}].

Lastly, with a transparent understanding of the household of likelihood distributions concerned, one can show the likelihood of this algorithm deciding on a given sequence of rows ((r_1, dotsc, r_n)) is the same as (prodlimits_{j = 1}^{n} left( {w_j} center/ {sumlimits_{ok = j}^{N}{w_k}} proper)), similar to the likelihood beforehand talked about within the “What precisely is SWoR” part, which means the potential outcomes of this algorithm will observe precisely the identical likelihood distribution as that of a (n)-step SWoR.

With a purpose to not deprive our expensive readers the pleasure of finishing this proof by themselves, now we have determined to not inline the remainder of the proof (which boils right down to a calculus train) inside this weblog submit, however it’s out there in this file.

Whereas all earlier sections targeted solely on weighted sampling with out substitute, this part will briefly focus on how the exponential-variate method can even profit the weighted-sampling-with-replacement use case (which will probably be shortened as SWR to any extent further).

Though SWR with pattern dimension (n) may be carried out by (n) unbiased processes every deciding on (1) pattern, parallelizing a SWR workload throughout all partitions of a Spark knowledge body (let’s name it (X)) will nonetheless be extra performant if the variety of partitions is way bigger than (n) and greater than (n) executors can be found in a Spark cluster.

An preliminary answer we had in thoughts was to run SWR with pattern dimension (n) in parallel on every partition of (X), after which re-sample the outcomes based mostly on relative whole weights of every partition. Regardless of sounding deceptively easy when summarized in phrases, implementing such an answer in follow could be a reasonably sophisticated process. First, one has to use the alias methodology or comparable with the intention to carry out weighted sampling effectively on every partition of (X), and on prime of that, implementing the re-sampling logic throughout all partitions accurately and verifying the correctness of such process will even require appreciable effort.

As compared, with the assistance of exponential variates, a SWR carried out as (n) unbiased SWoR processes every deciding on (1) pattern is way less complicated to implement, whereas nonetheless being similar to our preliminary answer by way of effectivity and scalability. An instance implementation of it (which takes fewer than 60 strains of Scala) is offered in samplingutils.scala.

How do we all know sparklyr::sdf_weighted_sample() is working as anticipated? Whereas the rigorous reply to this query is offered in full within the testing part, we thought it could even be helpful to first present some histograms that may assist readers visualize what that take a look at plan is. Subsequently on this part, we’ll do the next:

• Run dplyr::slice_sample() a number of instances on a small pattern area, with every run utilizing a special PRNG seed (pattern dimension will probably be decreased to (2) right here in order that there’ll fewer than 100 potential outcomes and visualization will probably be simpler)
• Do the identical for sdf_weighted_sample()
• Use histograms to visualise the distribution of sampling outcomes

All through this part, we’ll pattern (2) components out of ({0, dotsc, 7}) with out substitute in keeping with some weights, so, step one is to arrange the next in R:

library(sparklyr)

sc <- spark_connect(grasp = "native")

# `octs` will probably be our pattern area
octs <- knowledge.body(
  x = seq(0, 7),
  weight = c(1, 4, 2, 8, 5, 7, 1, 4)
)
# `octs_sdf` will probably be our pattern area copied right into a Spark knowledge body
octs_sdf <- copy_to(sc, octs)

sample_size <- 2

With a purpose to tally up and visualize the sampling outcomes effectively, we will map every potential final result to an octal quantity (e.g., (6, 7) will get mapped to (6 cdot 8^0 + 7 cdot 8^1)) utilizing a helper operate to_oct in R:

to_oct <- operate(pattern) sum(8 ^ seq(0, sample_sz - 1) * pattern$x)

We additionally must tally up sampling outcomes from dplyr::slice_sample() and sparklyr::sdf_weighted_sample() in 2 separate arrays:

max_possible_outcome <- to_oct(checklist(x = seq(8 - sample_sz, 7)))

sdf_weighted_sample_outcomes <- rep(0, max_possible_outcome)
dplyr_slice_sample_outcomes <- rep(0, max_possible_outcome)

Lastly, we will run each dplyr::slice_sample() and sparklyr::sdf_weighted_sample() for arbitrary variety of iterations and evaluate tallied outcomes from each:

num_sampling_iters <- 1000  # really we'll range this worth from 500 to 5000

for (x in seq(num_sampling_iters)) {
  sample1 <- octs_sdf %>%
    sdf_weighted_sample(
      ok = sample_size, weight_col = "weight", substitute = FALSE, seed = seed
    ) %>%
    acquire() %>%
    to_oct()
  sdf_weighted_sample_outcomes[[sample1]] <-
      sdf_weighted_sample_outcomes[[sample1]] + 1

  seed <- x * 97
  set.seed(seed) # set random seed for dplyr::sample_slice()
  sample2 <- octs %>%
    dplyr::slice_sample(
      n = sample_size, weight_by = weight, exchange = FALSE
    ) %>%
    to_oct()
  dplyr_slice_sample_outcomes[[sample2]] <-
      dplyr_slice_sample_outcomes[[sample2]] + 1
}

After all of the onerous work above, we will now get pleasure from plotting the sampling outcomes from dplyr::slice_sample() and people from sparklyr::sdf_weighted_sample() after 500, 1000, and 5000 iterations and observe how the distributions of each begin converging after numerous iterations.

Sampling outcomes after 500, 1000, and 5000 iterations, proven in 3 histograms:

(you’ll most likely must view it in a separate tab to see the whole lot clearly)

Whereas parallelized sampling based mostly on exponential variates appears improbable on paper, there are nonetheless loads of potential pitfalls in terms of translating such concept into code, and as typical, a great testing plan is critical to make sure implementation correctness.

As an example, numerical instability points from floating level numbers come up if (ln(u_j) / w_j) had been changed by (u_j ^ {1 / w_j}) within the aforementioned computations.

One other extra refined supply of error is the utilization of PRNG seeds. For instance, contemplate the next:

  def sampleWithoutReplacement(
    rdd: RDD[Row],
    weightColumn: String,
    sampleSize: Int,
    seed: Lengthy
  ): RDD[Row] = {
    val sc = rdd.context
    if (0 == sampleSize) {
      sc.emptyRDD
    } else {
      val random = new Random(seed)
      val mapRDDs = rdd.mapPartitions { iter =>
        for (row <- iter) {
          val weight = row.getAs[Double](weightColumn)
          val key = scala.math.log(random.nextDouble) / weight
          <after which make sampling determination for `row` based mostly on its `key`,
           as described within the earlier part>
        }
        ...
      }
      ...
    }
  }

Despite the fact that it’d look OK upon first look, rdd.mapPartitions(...) from the above will trigger the identical sequence of pseudorandom numbers to be utilized to a number of partitions of the enter Spark knowledge body, which is able to trigger undesired bias (i.e., sampling outcomes from one partition can have non-trivial correlation with these from one other partition when such correlation ought to be negligible in an accurate implementation).

The code snippet under is an instance implementation through which every partition of the enter Spark knowledge body is sampled utilizing a special sequence of pseudorandom numbers:

  def sampleWithoutReplacement(
    rdd: RDD[Row],
    weightColumn: String,
    sampleSize: Int,
    seed: Lengthy
  ): RDD[Row] = {
    val sc = rdd.context
    if (0 == sampleSize) {
      sc.emptyRDD
    } else {
      val mapRDDs = rdd.mapPartitionsWithIndex { (index, iter) =>
        val random = new Random(seed + index)

        for (row <- iter) {
          val weight = row.getAs[Double](weightColumn)
          val key = scala.math.log(random.nextDouble) / weight
          <after which make sampling determination for `row` based mostly on its `key`,
           as described within the earlier part>
        }

        ...
      }
    ...
  }
}

An instance take a look at case through which a two-sided Kolmogorov-Smirnov take a look at is used to check distribution of sampling outcomes from dplyr::slice_sample() with that from sparklyr::sdf_weighted_sample() is proven in this file. Such assessments have confirmed to be efficient in surfacing non-obvious implementation errors equivalent to those talked about above.

Please observe the sparklyr::sdf_weighted_sample() performance is just not included in any official launch of sparklyr but. We’re aiming to ship it as a part of sparklyr 1.4 in about 2 to three months from now.

In the mean time, you may strive it out with the next steps:

First, be certain that remotes is put in, after which run

to put in sparklyr from supply.

Subsequent, create a take a look at knowledge body with numeric weight column consisting of non-negative weight for every row, after which copy it to Spark (see code snippet under for instance):

library(sparklyr)

sc <- spark_connect(grasp = "native")

example_df <- knowledge.body(
  x = seq(100),
  weight = c(
    rep(1, 50),
    rep(2, 25),
    rep(4, 10),
    rep(8, 10),
    rep(16, 5)
  )
)
example_sdf <- copy_to(sc, example_df, repartition = 5, overwrite = TRUE)

Lastly, run sparklyr::sdf_weighted_sample() on example_sdf:

sample_size <- 5

samples_without_replacement <- example_sdf %>%
  sdf_weighted_sample(
    weight_col = "weight",
    ok = sample_size,
    substitute = FALSE
  )

samples_without_replacement %>% print(n = sample_size)

## # Supply: spark<?> [?? x 2]
##       x weight
##   <int>  <dbl>
## 1    48      1
## 2    22      1
## 3    78      4
## 4    56      2
## 5   100     16

samples_with_replacement <- example_sdf %>%
  sdf_weighted_sample(
    weight_col = "weight",
    ok = sample_size,
    substitute = TRUE
  )

samples_with_replacement %>% print(n = sample_size)

## # Supply: spark<?> [?? x 2]
##       x weight
##   <int>  <dbl>
## 1    86      8
## 2    97     16
## 3    91      8
## 4   100     16
## 5    65      2

Firstly, the writer needs to thank @ajing for reporting the weighted sampling use circumstances weren’t correctly supported but in sparklyr 1.3 and suggesting it ought to be a part of some future model of sparklyr on this Github problem.

Particular thanks additionally goes to Javier (@javierluraschi) for reviewing the implementation of all exponential-variate based mostly sampling algorithms in sparklyr, and to Mara (@batpigandme), Sigrid (@Sigrid), and Javier (@javierluraschi) for his or her priceless editorial solutions.

We hope you could have loved studying this weblog submit! Should you want to be taught extra about sparklyr, we suggest visiting sparklyr.ai, spark.rstudio.com, and a few of the earlier launch posts equivalent to sparklyr 1.3 and sparklyr 1.2. Additionally, your contributions to sparklyr are greater than welcome. Please ship your pull requests by way of right here and file any bug report or function request in right here.

Thanks for studying!