How to produce combinations iteratively in R? [duplicate]

问题

So I am currently using the following code to generate my combinations:

combn(x,y)

But the thing is that function stores all of the possible combinations. I dont want to store them, I just want to produce them through like a loop or something. It would be way more efficient for my program. Is there a way to generate combinations through a for loop rather than storing them all?

I know I asked a similar question here: How do I find all possible subsets of a set iteratively in R?

But in that solution the combinations are still being stored...

Here is some more detail:

Lets say I want to find 4 choose 2. combn(4,2) would essentially store the following: ((1,4),(1,3),(1,2),(2,4),(2,3)(3,4))

What I want is this:

   loop{
       produces one combination at a time 
   }

回答1:

Here is a suggestion which allows to generate the combination for the current iteration of the loop based on the combination used in the previous iteration of the loop.

## Function definition
gen.next.cbn <- function(cbn, n){
    ## Generates the combination that follows the one provided as input
    cbn.bin      <- rep(0, n)
    cbn.bin[cbn] <- 1
    if (tail(cbn.bin, 1) == 0){
        ind <- tail(which(cbn.bin == 1), 1)
        cbn.bin[c(ind, ind+1)] <- c(0, 1)
    }else{
        ind <- 1 + tail(which(diff(cbn.bin) == -1), 1)
        nb  <- sum(cbn.bin[-c(1:ind)] == 1)
        cbn.bin[c(ind-1, (n-nb+1):n)] <- 0
        cbn.bin[ind:(ind+nb)]         <- 1
    }
    cbn <- which(cbn.bin == 1)
}

## Example parameters
n   <- 6
k   <- 3

## Iteration example
for (i in 1:choose(n, k)){
    if (i == 1){
        cbn <- 1:k
    }else{
        cbn <- gen.next.cbn(cbn, n)
    }
    print(cbn)
}

# [1] 1 2 3
# [1] 1 2 4
# [1] 1 2 5
# [1] 1 2 6
# [1] 1 3 4
# [1] 1 3 5
# [1] 1 3 6
# [1] 1 4 5
# [1] 1 4 6
# [1] 1 5 6
# [1] 2 3 4
# [1] 2 3 5
# [1] 2 3 6
# [1] 2 4 5
# [1] 2 4 6
# [1] 2 5 6
# [1] 3 4 5
# [1] 3 4 6
# [1] 3 5 6
# [1] 4 5 6

回答2:

If the aim is to use each combination as an input for some computations, you might want to use the FUN argument of combn, a la apply. It seems that this won't store the combinations, but will still return at once the result of the function applied to each combination.

Here is an example with a dummy function:

fct <- function(x, y){sum(x*y) + 2*x[1]}
y   <- 1:5
system.time(combn(1:20, 5, fct, y = y))
# user  system elapsed 
# 0.160   0.000   0.161 
system.time(apply(combn(1:20, 5), 2, fct, y = y))
# user  system elapsed 
# 0.224   0.000   0.222

回答3:

To return each of the possible combinations, one at a time, in a loop, do the following:

#Sample data:
x <- c(1,2,3,4)
y <- 2
all_combinations <- combn(x,y)

#Return each value:
for (i in 1:ncol(all_combinations)) {
  print(all_combinations[,i])
}

But I'm not sure why you want to do this in a for loop, given that it's pretty slow. Is there a desired final output beyond this application?

来源：https://stackoverflow.com/questions/17683370/how-to-produce-combinations-iteratively-in-r

标签

loops

iteration

combinations