vignettes/CombinatorialSampling.Rmd
CombinatorialSampling.RmdThis document covers topics in generating random samples of combinations, permutations, partitions, compositions and partition of groups. It is encouraged to read General Combinatorics first.
At the heart of sampling is the ability to efficiently generate the
nth lexicographical
result. The algorithms in RcppAlgos are flexible and
optimized, allowing for tackling this task with ease.
To illustrate this in base R, let us consider getting 5 random
combinations of the vector 1:20 of length 10. How should we
proceed?
A naive approach would be to generate all of the combinations using
combn and then call sample:
options(width = 90)
naive <- function(v, m, n, s) {
allCombs <- combn(v, m)
set.seed(s)
allCombs[, sample(ncol(allCombs), n)]
}
fiveRndCombs <- naive(20, 10, 5, 42)
t(fiveRndCombs)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> [1,] 1 3 5 10 11 14 15 16 18 20
#> [2,] 1 3 4 9 10 11 12 13 18 19
#> [3,] 2 3 4 6 9 10 12 13 15 19
#> [4,] 1 4 5 10 13 14 15 17 18 19
#> [5,] 1 3 4 5 7 8 13 15 18 19This is okay for this small example (there are only
choose(20, 10) = 184756 results), however what if we wanted
to find one hundred thousand random combinations from the vector
1:100 of length 20? Clearly, the approach above will not be
feasible as there are far too many results to generate
(choose(100, 20) = 5.359834e+20). Furthermore, there are
internal limitations on sample. If we try to pass
choose(100, 20), we will get an error:
We could also try calling sample(100, 20) a bunch of
times and hope we don’t get duplicate combinations. This is neither
promising nor elegant.
RcppAlgos provides five functions:
comboSample, permuteSample,
partitionsSample, compositionsSample, and
comboGroupsSample for seamlessly attacking these types of
problems. All functions provide the following:
nThreads or the Parallel parameters.sampleVec
or rely on the internal sampling functions. We call sample
when the total number of results is small and for larger cases, the
sampling is done in a very similar fashion to urand.bigz
from the gmp package.partitionsGeneral)seed parameter allows for generating reproducible
samples.seed parameter must
be set in order to have reproducible results (E.g.
set.seed()) has no effect in these cases).comboSample and permuteSample
Let’s first look at the first problem above (i.e. getting 5 random
combinations of the vector 1:20 of length 10):
library(RcppAlgos)
set.seed(42)
comboSample(20, 10, n = 5)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> [1,] 1 3 5 10 11 14 15 16 18 20
#> [2,] 1 3 4 9 10 11 12 13 18 19
#> [3,] 2 3 4 6 9 10 12 13 15 19
#> [4,] 1 4 5 10 13 14 15 17 18 19
#> [5,] 1 3 4 5 7 8 13 15 18 19
## Use the seed argument directly to produce the same output
comboSample(20, 10, n = 5, seed = 42)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> [1,] 1 3 5 10 11 14 15 16 18 20
#> [2,] 1 3 4 9 10 11 12 13 18 19
#> [3,] 2 3 4 6 9 10 12 13 15 19
#> [4,] 1 4 5 10 13 14 15 17 18 19
#> [5,] 1 3 4 5 7 8 13 15 18 19
## fiveRndCombs produced above
identical(t(fiveRndCombs),
comboSample(20, 10, n = 5, seed = 42))
#> [1] TRUEJust like with comboGeneral and
permuteGeneral, we can explore results with repetition.
comboSample(10, 8, TRUE, n = 3, seed = 84)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> [1,] 2 5 5 7 9 9 9 9
#> [2,] 4 5 8 8 8 10 10 10
#> [3,] 2 6 6 6 6 6 9 9
permuteSample(10, 8, TRUE, n = 3)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> [1,] 4 10 4 4 10 2 2 10
#> [2,] 1 4 5 10 5 5 2 2
#> [3,] 4 1 7 9 1 5 6 5
comboSample(10, 12, freqs = 1:10, n = 3)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#> [1,] 2 2 3 5 5 6 6 6 7 8 9 10
#> [2,] 1 2 3 3 5 5 6 7 9 9 9 9
#> [3,] 1 2 5 5 5 6 6 9 10 10 10 10
permuteSample(10, 12, freqs = 1:10, n = 3, seed = 123)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#> [1,] 2 8 7 4 8 9 10 10 7 1 8 2
#> [2,] 5 5 9 8 1 8 3 2 6 4 3 10
#> [3,] 10 3 8 8 4 8 8 6 10 6 3 8sampleVec
We can also utilize sampleVec to generate specific
results.
## E.g. the below generates the 1st, 5th, 25th, 125th, and
## 625th lexicographical combinations
comboSample(10, 8, TRUE, sampleVec = c(1, 5, 25, 125, 625))
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> [1,] 1 1 1 1 1 1 1 1
#> [2,] 1 1 1 1 1 1 1 5
#> [3,] 1 1 1 1 1 1 3 8
#> [4,] 1 1 1 1 1 3 6 9
#> [5,] 1 1 1 1 5 6 10 10
## Is the same as:
comboGeneral(10, 8, TRUE)[5^(0:4), ]
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> [1,] 1 1 1 1 1 1 1 1
#> [2,] 1 1 1 1 1 1 1 5
#> [3,] 1 1 1 1 1 1 3 8
#> [4,] 1 1 1 1 1 3 6 9
#> [5,] 1 1 1 1 5 6 10 10namedSample
Have you ever wondered which lexicographical
combinations/permutations are returned when sampling? No worries, simply
set namedSample = TRUE:
testInd <- permuteSample(30, 10, n = 3, seed = 100, namedSample = TRUE)
testInd
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> 86626302070118 24 26 7 29 3 21 20 9 16 28
#> 15871916538841 5 12 21 9 6 3 14 23 4 20
#> 87932455980012 25 6 20 23 18 10 27 30 19 29
## Same output as above
permuteSample(30, 10, sampleVec = row.names(testInd))
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> [1,] 24 26 7 29 3 21 20 9 16 28
#> [2,] 5 12 21 9 6 3 14 23 4 20
#> [3,] 25 6 20 23 18 10 27 30 19 29Just like the General counterparts, the sampling
functions utilize GMP to allow for exploration of
combinations/permutations of large vectors where the total number of
results is enormous. They also offer parallel options using
Parallel or nThreads.
## Uses min(stdThreadMax() - 1, 5) threads (in this case)
permuteSample(500, 10, TRUE, n = 5, seed = 123, Parallel = TRUE)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> [1,] 55 435 274 324 200 152 6 313 121 377
#> [2,] 196 166 331 154 443 329 155 233 354 442
#> [3,] 235 325 94 27 370 117 302 86 229 126
#> [4,] 284 104 464 104 207 127 117 9 390 414
#> [5,] 456 76 381 456 219 23 376 187 11 123
permuteSample(factor(state.abb), 15, n = 3, seed = 50, nThreads = 3)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15]
#> [1,] ME FL DE OK ND CA PA AL ID MO NM HI KY MT NJ
#> [2,] AZ CA AL CT ME SD ID SC OK NH HI TN ND IA MT
#> [3,] MD MO NC MT NH AL VA MA VT WV NJ NE MN MS MI
#> 50 Levels: AK AL AR AZ CA CO CT DE FL GA HI IA ID IL IN KS KY LA MA MD ME MI MN MO ... WY
permuteCount(factor(state.abb), 15)
#> Big Integer ('bigz') :
#> [1] 2943352142120754524160000The algorithms are incredibly efficient and offer tremendous gains over the naive approach above:
## the function "naive" is defined above
system.time(naive(25, 10, 5, 15))
#> user system elapsed
#> 1.497 0.011 1.509
system.time(comboSample(25, 10, n = 5, seed = 15))
#> user system elapsed
#> 0.000 0.000 0.001Even when dealing with extremely large numbers, these algorithms are very fast. And using the parallel options have even greater effects than we saw with the general counterparts (typically around ~2-3 times faster with the general functions, whereas with the last example below with sampling we see a nearly 5x improvement).
## Lightning fast even with examples involving many results
system.time(comboSample(2500, 100, n = 5, seed = 15))
#> user system elapsed
#> 0.001 0.000 0.000
## The total number of combinations has ~180 digits
gmp::log10.bigz(comboCount(2500, 100))
#> [1] 180.9525
## Still fast with larger samples
system.time(comboSample(2500, 100, n = 1e4, seed = 157))
#> user system elapsed
#> 1.252 0.001 1.253
## Using Parallel/nThreads in these cases has an even greater effect
system.time(comboSample(2500, 100, n = 1e4, seed = 157, nThreads = 8))
#> user system elapsed
#> 1.935 0.004 0.265Again, just as with the general functions, you can pass a custom
function to {combo|permute}Sample using the
FUN argument.
permuteSample(5000, 1000, n = 3, seed = 101, FUN = sd)
#> [[1]]
#> [1] 1431.949
#>
#> [[2]]
#> [1] 1446.859
#>
#> [[3]]
#> [1] 1449.272
## Example using complex numbers
myCplx <- as.complex(1:100 + rep(c(-1, 1), 50) * 1i)
permuteSample(myCplx, 10, freqs = rep(1:5, 20),
n = 3, seed = 101, FUN = function(x) {
sqrt(sum(x))
})
#> [[1]]
#> [1] 24.83948+0i
#>
#> [[2]]
#> [1] 20.9285+0.04778i
#>
#> [[3]]
#> [1] 22.20379+0.09007ipartitionsSample
The partitionsSample function allows one to draw a
random sample of partitions of a number. Many of the features present in
comboSample and permuteSample are available in
partitionsSample.
## Use the seed parameter to obtain reproducible results
partitionsSample(100, 8, TRUE, n = 3, seed = 42)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> [1,] 1 1 3 3 4 20 23 45
#> [2,] 1 1 2 7 14 14 29 32
#> [3,] 2 10 11 11 16 16 16 18
## Used namedSample to obtain the lexicographical indices
partitionsSample(100, 8, TRUE, n = 3, seed = 42, namedSample = TRUE)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> 61413 1 1 3 3 4 20 23 45
#> 54425 1 1 2 7 14 14 29 32
#> 623844 2 10 11 11 16 16 16 18
## Use sampleVec to obtain specific results
partitionsSample(100, 8, TRUE, sampleVec = c(61413, 54425, 623844))
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> [1,] 1 1 3 3 4 20 23 45
#> [2,] 1 1 2 7 14 14 29 32
#> [3,] 2 10 11 11 16 16 16 18
partitionsCount(2500, 10)
#> Big Integer ('bigz') :
#> [1] 2621914835336941325
## Algorithms are very efficient
system.time(serial <- partitionsSample(2500, 10, n = 1e3,
seed = 8128))
#> user system elapsed
#> 2.986 0.006 2.992
## Use nThreads for greater efficiency
system.time(multi <- partitionsSample(2500, 10, n = 1e3,
seed = 8128, nThreads = 8))
#> user system elapsed
#> 4.304 0.027 0.582
identical(multi, serial)
#> [1] TRUE
## Even works with non-standard setup
partitionsSample(17 + (1:10) * 3, 10, TRUE,
target = 320, n = 3, seed = 111)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> [1,] 23 23 26 26 29 29 38 38 41 47
#> [2,] 26 26 26 29 29 29 32 41 41 41
#> [3,] 20 23 23 26 26 35 38 41 44 44There are sampling algorithms available for most partition cases, but some cases are not covered. For example, with standard multisets, we are currently unable to efficiently generate the nth lexicographical result. Another example is when the source vector is not uniform (e.g. when the distance between each element is irregular).
Observe the following:
## No sampling algorithm available when the source vector is not uniform
partitionsSample(c(1, 4, 6, 7, 10, seq(11, 100, 7)), 10, n = 1, target = 340)
#> Error: Partition sampling not available for this case.
## As stated above, the standard multiset case doesn't work either
partitionsSample(0:50, 6, freqs = rep(1:3, 17), n = 2)
#> Error: Partition sampling not available for this case.
## If we use freqs to indicate that zeros can repeat,
## then we can obtain random samples
partitionsSample(0:50, 6, freqs = c(50, rep(1, 50)), n = 3, seed = 222)
#> [,1] [,2] [,3] [,4] [,5] [,6]
#> [1,] 0 0 1 4 9 36
#> [2,] 0 0 0 0 17 33
#> [3,] 2 4 5 6 8 25
## Even works when the vector is restricted in regards to the target
partitionsSample(0:50, 6, freqs = c(50, rep(1, 50)),
n = 3, seed = 222, target = 100)
#> [,1] [,2] [,3] [,4] [,5] [,6]
#> [1,] 0 1 6 15 29 49
#> [2,] 0 0 0 8 43 49
#> [3,] 4 7 17 19 22 31There is ongoing research in this area and our goal is to eventually be able to cover the standard multiset case.
compositionsSample
The compositionsSample function allows one to draw a
random sample of compositions of a number. Many of the features present
in comboSample and permuteSample are available
in compositionsSample.
## Use the seed parameter to obtain reproducible results
compositionsSample(100, 8, TRUE, n = 3, seed = 42)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> [1,] 5 4 34 31 7 3 1 15
#> [2,] 3 21 8 6 12 43 6 1
#> [3,] 6 6 1 36 17 18 10 6
## Used namedSample to obtain the lexicographical indices
compositionsSample(100, 8, TRUE, n = 3, seed = 42, namedSample = TRUE)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> 4024715585 5 4 34 31 7 3 1 15
#> 2756281572 3 21 8 6 12 43 6 1
#> 4873365553 6 6 1 36 17 18 10 6
## Use sampleVec to obtain specific results
compositionsSample(100, 8, TRUE,
sampleVec = c(4024715585, 2756281572, 4873365553))
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> [1,] 5 4 34 31 7 3 1 15
#> [2,] 3 21 8 6 12 43 6 1
#> [3,] 6 6 1 36 17 18 10 6
compositionsCount(2500, 10, TRUE)
#> Big Integer ('bigz') :
#> [1] 10324518903611974891453624
## Algorithms are very efficient...
## The below retrieves 10,000 compositions in under a second
system.time(serial <- compositionsSample(2500, 10, TRUE,
n = 1e4, seed = 8128))
#> user system elapsed
#> 0.262 0.000 0.263
## Use nThreads for greater efficiency
system.time(multi <- compositionsSample(2500, 10, TRUE, n = 1e4,
seed = 8128, nThreads = 8))
#> user system elapsed
#> 0.367 0.001 0.050
identical(multi, serial)
#> [1] TRUE
## Sample weak compositions
compositionsSample(0:100, 8, repetition = TRUE, weak = TRUE,
seed = 245659, n = 3, namedSample = TRUE)
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> 21547195040 23 0 0 8 5 22 25 17
#> 7649268748 5 0 13 36 42 4 0 0
#> 7598223523 4 54 21 4 6 0 8 3Currently, there are only sampling algorithms for most cases of compositions with repetition. There is ongoing work to expand these algorithms in the future.
comboGroupsSample
Just as we can generate random samples of combinations and permutations, we are also able to generate random samples of partitions of groups as well. There are many problems that present in this manner. Below, we examine one involving playing cards.
Let’s say we have 4 players and each player is to have 3 cards a piece. Given that the deck is shuffled, the dealer then distributes 12 cards.
What possible hands can each player have?
See Creating A Deck Of Cards In R Without Using While And Double For Loop (Credit to @MichaelChirico)
cards <- c(2:10, "J", "Q", "K", "A")
suits <- c("♠", "♥", "♦", "♣")
deck <- paste0(rep(cards, length(suits)), # card values
rep(suits, each = length(cards))) # suits
set.seed(1738)
shuffled <- factor(deck[sample(52)], levels = deck)
## Here are 2 possibilities
comboGroupsSample(shuffled[1:12], numGroups = 4, n = 2, seed = 13)
#> Grp1 Grp1 Grp1 Grp2 Grp2 Grp2 Grp3 Grp3 Grp3 Grp4 Grp4 Grp4
#> [1,] 8♦ 3♥ 5♦ 9♦ J♠ 7♥ 8♠ K♦ 10♦ A♦ J♥ 3♦
#> [2,] 8♦ K♦ 10♦ 9♦ J♥ 3♥ J♠ 8♠ 3♦ A♦ 5♦ 7♥
#> 52 Levels: 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠ A♠ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ J♥ ... A♣
comboGroupsSample(shuffled[1:12], numGroups = 4, retType = "3Darray",
n = 2, seed = 13, namedSample = TRUE)
#> , , Grp1
#>
#> [,1] [,2] [,3]
#> 13784 8♦ 3♥ 5♦
#> 9152 8♦ K♦ 10♦
#>
#> , , Grp2
#>
#> [,1] [,2] [,3]
#> 13784 9♦ J♠ 7♥
#> 9152 9♦ J♥ 3♥
#>
#> , , Grp3
#>
#> [,1] [,2] [,3]
#> 13784 8♠ K♦ 10♦
#> 9152 J♠ 8♠ 3♦
#>
#> , , Grp4
#>
#> [,1] [,2] [,3]
#> 13784 A♦ J♥ 3♦
#> 9152 A♦ 5♦ 7♥
#>
#> 52 Levels: 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠ A♠ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ J♥ ... A♣What if we add a twist on the problem above. What if instead we want to deal players 1 & 2 two cards, player 3 three cards, and player 4 five cards. How might we do this?
For this, we make use of the grpSizes parameter:
## Again, here are 2 possibilities
comboGroupsSample(shuffled[1:12], grpSizes = c(2, 2, 3, 5),
n = 2, seed = 13, namedSample = TRUE)
#> Grp1 Grp1 Grp2 Grp2 Grp3 Grp3 Grp3 Grp4 Grp4 Grp4 Grp4 Grp4
#> 81667 J♥ 10♦ 3♦ 3♥ 8♦ A♦ 5♦ 9♦ J♠ 8♠ K♦ 7♥
#> 71525 K♦ A♦ J♥ 5♦ 8♦ 8♠ 3♥ 9♦ J♠ 3♦ 10♦ 7♥
#> 52 Levels: 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠ A♠ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ J♥ ... A♣Ranking is the complement of sampling. That is, given a combination (or permutation/partition), determine which lexicographical result it is. As an example, consider all of the combinations of 5 choose 3:
comboGeneral(5, 3)
#> [,1] [,2] [,3]
#> [1,] 1 2 3
#> [2,] 1 2 4
#> [3,] 1 2 5
#> [4,] 1 3 4
#> [5,] 1 3 5
#> [6,] 1 4 5
#> [7,] 2 3 4
#> [8,] 2 3 5
#> [9,] 2 4 5
#> [10,] 3 4 5We can see that the rank of the combination: c(2, 3, 4)
is 7. That is, c(2, 3, 4) is the 7th combination
of 5 choose 3.
Just as we saw before, we could easily produce a brute force approach that would work well with small cases, but would become unmanageable very quickly. For example:
naive_rank <- function(v, m, comb) {
comb <- as.integer(comb)
which(apply(combn(v, m), 2, function(x) identical(x, comb)))
}
naive_rank(5, 3, 2:4)
#> [1] 7
## Larger example
comb = comboSample(25, 12, sampleVec = 2e6)[1, ]
system.time(print(naive_rank(25, 12, comb)))
#> [1] 2000000
#> user system elapsed
#> 6.048 0.061 6.110Similar to the sampling problem, RcppAlgos provides four
functions: comboRank, permuteRank,
partitionsRank, and compositionsRank
(currently there is not a ranking function for
comboGroups). These functions are very similar to their
sampling counterparts.
For both problems presented above, here is how you would attack them
with comboRank:
comboRank(2:4, v = 5)
#> [1] 7
## Since order doesn't matter with combinations, 4:2 should return 7 as well
comboRank(4:2, v = 5)
#> [1] 7
## comb was provided above
system.time(print(comboRank(comb, v = 25)))
#> [1] 2000000
#> user system elapsed
#> 0.001 0.000 0.000All that is needed is the original vector that was used to produce
the results and whether or not repetition is used via the
repetition or freqs arguments. The width is
determined automatically by the input.
A neat feature of the ranking functions is the ability to rank multiple inputs at once. We can either pass a single vector, multiple vectors, and we can even pass matrices.
comboRank
combs = comboSample(50, 8, n = 10, seed = 123, namedSample = TRUE)
combs
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> 161401295 3 4 6 13 40 44 48 50
#> 120710262 2 6 21 29 32 36 42 45
#> 220886245 3 22 23 25 35 36 48 50
#> 442099291 10 11 13 20 21 24 36 45
#> 334689371 6 9 13 20 28 41 43 48
#> 189241370 3 7 18 26 29 31 43 50
#> 168175018 3 4 20 21 27 42 44 49
#> 105740881 2 4 24 27 40 46 48 49
#> 36321127 1 5 10 15 18 32 40 46
#> 292830028 5 8 16 19 34 35 46 47
comboRank(combs, v = 50)
#> [1] 161401295 120710262 220886245 442099291 334689371 189241370 168175018 105740881
#> [9] 36321127 292830028permuteRank
perms_len_5 = permuteSample(100, 5, freqs = rep(1:5, 20),
n = 3, seed = 987, namedSample = TRUE)
perms_len_5
#> [,1] [,2] [,3] [,4] [,5]
#> 3474930553 36 47 93 7 32
#> 5793832271 60 12 27 39 99
#> 797663634 9 16 23 3 35
perms_len_8 = permuteSample(100, 8, freqs = rep(1:5, 20),
n = 3, seed = 123, namedSample = TRUE)
perms_len_8
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> 6897937625624040 73 96 48 62 19 39 60 31
#> 6937176899580966 74 37 42 44 34 80 18 94
#> 5771073522599470 62 7 60 69 14 60 7 30
## Note you can name the inputs
permuteRank(p5 = perms_len_5, p8 = perms_len_8,
v = 100, freqs = rep(1:5, 20))
#> $p5
#> [1] 3474930553 5793832271 797663634
#>
#> $p8
#> Big Integer ('bigz') object of length 3:
#> [1] 6897937625624040 6937176899580966 5771073522599470partitionsRank
parts = partitionsSample(50, 8, target = 100, repetition = TRUE,
n = 3, seed = 42, namedSample = TRUE)
parts
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> 61413 1 1 3 11 12 21 23 28
#> 54425 1 1 3 6 7 13 29 40
#> 623844 3 4 4 7 7 15 28 32
partitionsRank(parts, v = 50, target = 100, repetition = TRUE)
#> [1] 61413 54425 623844compositionsRank
comps = compositionsSample(50, 8, repetition = TRUE,
n = 3, seed = 42, namedSample = TRUE)
comps
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#> 36761573 4 6 2 1 13 3 15 6
#> 21025945 2 11 13 8 2 3 7 4
#> 71927012 11 4 17 1 2 5 5 5
compositionsRank(comps, v = 50, repetition = TRUE)
#> [1] 36761573 21025945 71927012