Create N random integers with no gaps - r

Question

For a clustering algorithm that I'm implementing, I would like to initialize the clusters assignments at random. However, I need that there are no gaps. That is, this is not ok:
set.seed(2)
K <- 10 # initial number of clusters
N <- 20 # number of data points
z_init <- sample(K,N, replace=TRUE) # initial assignments
z_init
# [1] 2 8 6 2 10 10 2 9 5 6 6 3 8 2 5 9 10 3 5 1
sort(unique(z_init))
# [1] 1 2 3 5 6 8 9 10
where labels 4 and 7 have not been used.
Instead, I would like this vector to be:
# [1] 2 6 5 2 8 8 2 7 4 5 5 3 6 2 4 7 8 3 4 1
where the label 5 has become 4 and so forth to fill the lower empty labels.
More examples:
The vector 1 2 3 5 6 8 should be ̀1 2 3 4 5 6 7
The vector 15,5,7,7,10 should be ̀1 2 3 3 4
Can it be done avoiding for loops? I don't need it to be fast, I prefer it to be elegant and short, since I'm doing it only once in the code (for label initialization).
My solution using a for loop
z_init <- c(3,2,1,3,3,7,9)
idx <- order(z_init)
for (i in 2:length(z_init)){
if(z_init[idx[i]] > z_init[idx[i-1]]){
z_init[idx[i]] <- z_init[idx[i-1]]+1
}
else{
z_init[idx[i]] <- z_init[idx[i-1]]
}
}
z_init
# 3 2 1 3 3 4 5

Answer 1

Edit: #GregSnow came up with the current shortest answer. I'm 100% convinced that this is the shortest possible way.
For fun, I decided to golf the code, i.e. write it as short as possible:
z <- c(3, 8, 4, 4, 8, 2, 3, 9, 5, 1, 4)
# solution by hand: 1 2 3 3 4 4 4 5 6 6 7
sort(c(factor(z))) # 18 bits, as proposed by #GregSnow in the comments
# [1] 1 2 3 3 4 4 4 5 6 6 7
Some other (functioning) attempts:
y=table(z);rep(seq(y),y) # 24 bits
sort(unclass(factor(z))) # 24 bits, based on #GregSnow 's answer
diffinv(diff(sort(z))>0)+1 # 26 bits
sort(as.numeric(factor(z))) # 27 bits, #GregSnow 's original answer
rep(seq(unique(z)),table(z)) # 28 bits
cumsum(c(1,diff(sort(z))>0)) # 28 bits
y=rle(sort(z))$l;rep(seq(y),y) # 30 bits
Edit2: Just to show that bits isn't everything:
z <- sample(1:10,10000,replace=T)
Unit: microseconds
expr min lq mean median uq max neval
sort(c(factor(z))) 2550.128 2572.2340 2681.4950 2646.6460 2729.7425 3140.288 100
{ y = table(z) rep(seq(y), y) } 2436.438 2485.3885 2580.9861 2556.4440 2618.4215 3070.812 100
sort(unclass(factor(z))) 2535.127 2578.9450 2654.7463 2623.9470 2708.6230 3167.922 100
diffinv(diff(sort(z)) > 0) + 1 551.871 572.2000 628.6268 626.0845 666.3495 940.311 100
sort(as.numeric(factor(z))) 2603.814 2672.3050 2762.2030 2717.5050 2790.7320 3558.336 100
rep(seq(unique(z)), table(z)) 2541.049 2586.0505 2733.5200 2674.0815 2760.7305 5765.815 100
cumsum(c(1, diff(sort(z)) > 0)) 530.159 545.5545 602.1348 592.3325 632.0060 844.385 100
{ y = rle(sort(z))$l rep(seq(y), y) } 661.218 684.3115 727.4502 724.1820 758.3280 857.412 100
z <- sample(1:100000,replace=T)
Unit: milliseconds
expr min lq mean median uq max neval
sort(c(factor(z))) 84.501189 87.227377 92.13182 89.733291 94.16700 150.08327 100
{ y = table(z) rep(seq(y), y) } 78.951701 82.102845 85.54975 83.935108 87.70365 106.05766 100
sort(unclass(factor(z))) 84.958711 87.273366 90.84612 89.317415 91.85155 121.99082 100
diffinv(diff(sort(z)) > 0) + 1 9.784041 9.963853 10.37807 10.090965 10.34381 17.26034 100
sort(as.numeric(factor(z))) 85.917969 88.660145 93.42664 91.542263 95.53720 118.44512 100
rep(seq(unique(z)), table(z)) 86.568528 88.300325 93.01369 90.577281 94.74137 118.03852 100
cumsum(c(1, diff(sort(z)) > 0)) 9.680615 9.834175 10.11518 9.963261 10.16735 14.40427 100
{ y = rle(sort(z))$l rep(seq(y), y) } 12.842614 13.033085 14.73063 13.294019 13.66371 133.16243 100

Answer 2

It seems to me that you are trying to randomly assign elements of a set (the numbers 1 to 20) to clusters, subject to the requirement that each cluster be assigned at least one element.
One approach that I could think of would be to select a random reward r_ij for assigning element i to cluster j. Then I would define binary decision variables x_ij that indicate whether element i is assigned to cluster j. Finally, I would use mixed integer optimization to select the assignment from elements to clusters that maximizes the collected reward subject to the following conditions:
Every element is assigned to exactly one cluster
Every cluster has at least one element assigned to it
This is equivalent to randomly selecting an assignment, keeping it if all clusters have at least one element, and otherwise discarding it and trying again until you get a valid random assignment.
In terms of implementation, this is pretty easy to accomplish in R using the lpSolve package:
library(lpSolve)
N <- 20
K <- 10
set.seed(144)
r <- matrix(rnorm(N*K), N, K)
mod <- lp(direction = "max",
objective.in = as.vector(r),
const.mat = rbind(t(sapply(1:K, function(j) rep((1:K == j) * 1, each=N))),
t(sapply(1:N, function(i) rep((1:N == i) * 1, K)))),
const.dir = c(rep(">=", K), rep("=", N)),
const.rhs = rep(1, N+K),
all.bin = TRUE)
(assignments <- apply(matrix(mod$solution, nrow=N), 1, function(x) which(x > 0.999)))
# [1] 6 5 3 3 5 6 6 9 2 1 3 4 7 6 10 2 10 6 6 8
sort(unique(assignments))
# [1] 1 2 3 4 5 6 7 8 9 10

Answer 3

You could do like this:
un <- sort(unique(z_init))
(z <- unname(setNames(1:length(un), un)[as.character(z_init)]))
# [1] 2 6 5 2 8 8 2 7 4 5 5 3 6 2 4 7 8 3 4 1
sort(unique(z))
# [1] 1 2 3 4 5 6 7 8
Here I replace elements of un in z_init with corresponding elements of 1:length(un).

Answer 4

A simple (but possibly inefficient) approach is to convert to a factor then back to numeric. Creating the factor will code the information as integers from 1 to the number of unique values, then add labels with the original values. Converting to numeric then drops the labels and leaves the numbers:
> x <- c(1,2,3,5,6,8)
> (x2 <- as.numeric(factor(x)))
[1] 1 2 3 4 5 6
>
> xx <- c(15,5,7,7,10)
> (xx2 <- as.numeric(factor(xx)))
[1] 4 1 2 2 3
> (xx3 <- as.numeric(factor(xx, levels=unique(xx))))
[1] 1 2 3 3 4
The levels = portion in the last example sets the numbers to match the order in which they appear in the original vector.