How to create matrix of all 2^n binary sequences of length n using recursion in R?

问题

I know I can use expand.grid for this, but I am trying to learn actual programming. My goal is to take what I have below and use a recursion to get all 2^n binary sequences of length n.

I can do this for n = 1, but I don't understand how I would use the same function in a recursive way to get the answer for higher dimensions.

Here is for n = 1:

binseq <- function(n){
  binmat <- matrix(nrow = 2^n, ncol = n)
  r <- 0 #row counter
  for (i in 0:1) {
        r <- r + 1
        binmat[r,] <- i
    }
  return(binmat)
  }

I know I have to use probably a cbind in the return statement. My intuition says the return statement should be something like cbind(binseq(n-1), binseq(n)). But, honestly, I'm completely lost at this point.

The desired output should basically recursively produce this for n = 3:


binmat <- matrix(nrow = 8, ncol = 3)
r <- 0 # current row of binmat
for (i in 0:1) {   
for (j in 0:1) {
for (k in 0:1) {
r <- r + 1
binmat[r,] <- c(i, j, k)}   
} 
}
binmat

It should just be a matrix as binmat is being filled recursively.

回答1:

I quickly wrote this function to generate all N^K permutations of length K for given N characters. Hope it will be useful.

gen_perm <- function(str=c(""), lst=5, levels = c("0", "1", "2")){
  if (nchar(str) == lst){
    cat(str, "\n")
    return(invisible(NULL))
  }
  for (i in levels){
    gen_perm(str = paste0(str,i), lst=lst, levels=levels)
  }
}

# sample call
gen_perm(lst = 3, levels = c("x", "T", "a"))

I will return to your problem when I get more time.

UPDATE I modified the code above to work for your problem. Note that the matrix being populated lives in the global environment. The function also uses the tmp variable to pass rows to the global environment. This was the easiest way for me to solve the problem. Perhaps, there are other ways.

levels <- c(0,1)
nc <- 3

m <- matrix(numeric(0), ncol = nc)

gen_perm <- function(row=numeric(), lst=nc, levels = levels){
  if (length(row) == lst){
    assign("tmp", row, .GlobalEnv)
    with(.GlobalEnv, {m <- rbind(m, tmp); rownames(m) <- NULL})
    return(invisible(NULL))
  }
  for (i in levels){
    gen_perm(row=c(row,i), lst=lst, levels=levels)
  }
}

gen_perm(lst=nc, levels=levels)

UPDATE 2 To get the expected output you provided, run

m <- matrix(numeric(0), ncol = 3)
gen_perm(lst = 3, levels = c(0,1))
m

levels specifies a range of values to generate (binary in our case) to generate permutations, m is an empty matrix to fill up, gen_perm generates rows and adds them to the matrix m, lst is a length of the permutation (matches the number of columns in the matrix).

来源：https://stackoverflow.com/questions/58399137/how-to-create-matrix-of-all-2n-binary-sequences-of-length-n-using-recursion-in

标签

recursion

binary