I have a binary string like this:
0010000
I'd like to have all these permutations:
1010000
0110000
0011000
0010100
0010010
0010001
Is there anybody know which function in R could give me these results?
3 Answers
R has functions for bitwise operations, so we can get the desired numbers with bitwOr:
bitwOr(16, 2^(6:0))
#> [1] 80 48 16 24 20 18 17
...or if we want to exclude the original,
setdiff(bitwOr(16, 2^(6:0)), 16)
#> [1] 80 48 24 20 18 17
However, it only works in decimal, not binary. That's ok, though; we can build some conversion functions:
bin_to_int <- function(bin){
vapply(strsplit(bin, ''),
function(x){sum(as.integer(x) * 2 ^ seq(length(x) - 1, 0))},
numeric(1))
}
int_to_bin <- function(int, bits = 32){
vapply(int,
function(x){paste(as.integer(rev(head(intToBits(x), bits))), collapse = '')},
character(1))
}
Now:
input <- bin_to_int('0010000')
output <- setdiff(bitwOr(input, 2^(6:0)),
input)
output
#> [1] 80 48 24 20 18 17
int_to_bin(output, bits = 7)
#> [1] "1010000" "0110000" "0011000" "0010100" "0010010" "0010001"
1 Comment
Somayeh Ghazalbash
That's what I was looking for. Thank you.
library(stringr)
bin <- '0010000'
ones <- str_locate_all(bin, '1')[[1]][,1]
zeros <- (1:str_length(bin))[-ones]
sapply(zeros, function(x){
str_sub(bin, x, x) <- '1'
bin
})
[1] "1010000" "0110000" "0011000" "0010100" "0010010" "0010001"
Comments
We assume that the problem is to successively replace each 0 in the input with a 1 for an input string of 0's and 1's.
Replace each character successively with a "1", regardless of its value and then remove any components of the result equal to the input. No packages are used.
input <- "0010000"
setdiff(sapply(1:nchar(input), function(i) `substr<-`(input, i, i, "1")), input)
## [1] "1010000" "0110000" "0011000" "0010100" "0010010" "0010001"
Update: Have completely revised answer.
Comments
bitwOr(16, 2^(0:6)), though that prints in decimalsapply(bitwOr(16, 2^(0:6)),function(x){ as.integer(intToBits(x))})and you can slice and dice it however you'd like after that.