conditional sampling without replacement - r

Question

I am attempting to write a simulation that involves randomly re-assigning items to categories with some restrictions.
Lets say I have a collection of pebbles 1 to N distributed across buckets A through J:
set.seed(100)
df1 <- data.frame(pebble = 1:100,
bucket = sample(LETTERS[1:10], 100, T),
stringsAsFactors = F)
head(df1)
#> pebble bucket
#> 1 1 D
#> 2 2 C
#> 3 3 F
#> 4 4 A
#> 5 5 E
#> 6 6 E
I want to randomly re-assign pebbles to buckets. Without restrictions I could do it like so:
random.permutation.df1 <- data.frame(pebble = df1$pebble, bucket = sample(df1$bucket))
colSums(table(random.permutation.df1))
#> A B C D E F G H I J
#> 4 7 13 14 12 11 11 10 9 9
colSums(table(df1))
#> A B C D E F G H I J
#> 4 7 13 14 12 11 11 10 9 9
Importantly this re-assigns pebbles while ensuring that each bucket retains the same number (because we are sampling without replacement).
However, I have a set of restrictions such that certain pebbles cannot be assigned to certain buckets. I encode the restrictions in df2:
df2 <- data.frame(pebble = sample(1:100, 10),
bucket = sample(LETTERS[1:10], 10, T),
stringsAsFactors = F)
df2
#> pebble bucket
#> 1 33 I
#> 2 39 I
#> 3 5 A
#> 4 36 C
#> 5 55 J
#> 6 66 A
#> 7 92 J
#> 8 95 H
#> 9 2 C
#> 10 49 I
The logic here is that pebbles 33 and 39 cannot be placed in bucket I, or pebble 5 in bucket A, etc. I would like to permute which pebbles are in which bucket subject to these restrictions.
So far, I've thought of tackling it in a loop as below, but this does not result in buckets retaining the same number of pebbles:
perms <- character(0)
cnt <- 1
for (p in df1$pebble) {
perms[cnt] <- sample(df1$bucket[!df1$bucket %in% df2$bucket[df2$pebble==p]], 1)
cnt <- cnt + 1
}
table(perms)
#> perms
#> A B C D E F G H I J
#> 6 7 12 22 15 1 14 7 7 9
I then tried sampling positions, and then removing that position from the available buckets and the available remaining positions. This is also not working, and I suspect it is because I am sampling my way into branches of the tree that do not yield solutions.
set.seed(42)
perms <- character(0)
cnt <- 1
ids <- 1:nrow(df1)
bckts <- df1$bucket
for (p in df1$pebble) {
id <- sample(ids[!bckts %in% df2$bucket[df2$pebble==p]], 1)
perms[cnt] <- bckts[id]
bckts <- bckts[-id]
ids <- ids[ids!=id]
cnt <- cnt + 1
}
table(perms)
#> perms
#> A B C D E F G J
#> 1 1 4 1 2 1 2 2
Any thoughts or advice much appreciated (and apologies for the length).
EDIT:
I foolishly forgot to clarify that I was previously solving this by just resampling until I got a draw that didn't violate any of the conditions in df2, but I now have many conditions such that this would make my code take too long to run. I am still up for trying to force it if I could figure out a way to make forcing it faster.

Answer 1

I have a solution (I managed to write it in base R, but the data.table solution is easier to understand and write:
random.permutation.df2 <- data.frame(pebble = df1$pebble, bucket = rep(NA,length(df1$pebble)))
for(bucket in unique(df1$bucket)){
N <- length( random.permutation.df2$bucket[is.na(random.permutation.df2$bucket) &
!random.permutation.df2$pebble %in% df2$pebble[df2$bucket == bucket] ] )
random.permutation.df2$bucket[is.na(random.permutation.df2$bucket) &
!random.permutation.df2$pebble %in% df2$pebble[df2$bucket == bucket] ] <-
sample(c(rep(bucket,sum(df1$bucket == bucket)),rep(NA,N-sum(df1$bucket == bucket))))
}
The idea is to sample the authorised peeble for each bucket: those that are not in df2, and those that are not already filled. You sample then a vector of the good length, choosing between NAs (for the following buckets values) and the value in the loop, and voilà.
Now easier to read with data.table
library(data.table)
random.permutation.df2 <- setDT(random.permutation.df2)
df2 <- setDT(df2)
for( bucketi in unique(df1$bucket)){
random.permutation.df2[is.na(bucket) & !pebble %in% df2[bucket == bucketi, pebble],
bucket := sample(c(rep(bucketi,sum(df1$bucket == bucket)),rep(NA,.N-sum(df1$bucket == bucket))))]
}
it has the two conditions
> colSums(table(df1))
A B C D E F G H I J
4 7 13 14 12 11 11 10 9 9
> colSums(table(random.permutation.df2))
A B C D E F G H I J
4 7 13 14 12 11 11 10 9 9
To verify that there isn't any contradiction with df2
> df2
pebble bucket
1: 37 D
2: 95 H
3: 90 C
4: 80 C
5: 31 D
6: 84 G
7: 76 I
8: 57 H
9: 7 E
10: 39 A
> random.permutation.df2[pebble %in% df2$pebble,.(pebble,bucket)]
pebble bucket
1: 7 D
2: 31 H
3: 37 J
4: 39 F
5: 57 B
6: 76 E
7: 80 F
8: 84 B
9: 90 H
10: 95 D

Answer 2

Here a brute force approach where one simply tries long enough until a valid solution is found:
set.seed(123)
df1 <- data.frame(pebble = 1:100,
bucket = sample(LETTERS[1:10], 100, T),
stringsAsFactors = F)
df2 <- data.frame(pebble = sample(1:100, 10),
bucket = sample(LETTERS[1:10], 10, T),
stringsAsFactors = F)
random.permutation.df1 <- data.frame(pebble = df1$pebble, bucket = sample(df1$bucket))
Random permutation does not match the condition, so try new ones:
merge(random.permutation.df1, df2)
#> pebble bucket
#> 1 60 J
while(TRUE) {
random.permutation.df1 <- data.frame(pebble = df1$pebble, bucket = sample(df1$bucket))
if(nrow(merge(random.permutation.df1, df2)) == 0)
break;
}
New permutation matches the condition:
merge(random.permutation.df1, df2)
#> [1] pebble bucket
#> <0 Zeilen> (oder row.names mit Länge 0)
colSums(table(random.permutation.df1))
#> A B C D E F G H I J
#> 7 12 11 9 14 7 11 11 11 7
colSums(table(df1))
#> A B C D E F G H I J
#> 7 12 11 9 14 7 11 11 11 7