I have a vector (testCol_1) consisting of 3 letters. Each letter is binary and can be replaced by either 0 or 1.
THE RULE is, if one letter (e.g. S) is replaced by a number (e.g. 1), all the S in the vector will also be replaced by that same number.
I would like to find all the possible combinations; for the following reproducible example, there are 8 possible combinations.
test <- expand.grid(rep(list(c("S","E","F")),4))
testCol_1 <- test$Var1
My solution:
This is my own solution but as you can see this is rather novice and I am quite sure there are better way to solve this.
# Changing to integer
testCol_1 <- as.integer(testCol_1)
# Finding all the combinations first
temp.combo <- expand.grid(rep(list(0:1),3))
test.final.result <- list()
for (i in 1:nrow(temp.combo)){
testSol <- replace(testCol_1, testCol_1 == 1,temp.combo[i,1])
testSol <- replace(testSol, testSol == 2,temp.combo[i,2])
testSol <- replace(testSol, testSol == 3,temp.combo[i,3])
test.final.result[[i]] <- testSol
}
Here is a solution with no loops, use R's vectorized instructions to get the 0/1 directly from temp.combo, by subsetting it, run the following to see it:
temp.combo[, testCol_1]
Note also that nvals is not strictly needed, it only makes the code more readable.
test <- expand.grid(rep(list(c("S","E","F")), 4))
testCol_1 <- as.integer(test$Var1)
nvals <- length(unique(testCol_1))
temp.combo <- expand.grid(rep(list(0:1), nvals))
temp <- as.data.frame(t(temp.combo[, testCol_1]))
test.final.result <- as.list(temp)
test.final.result
#> $V1
#> [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#> [39] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
#> [77] 0 0 0 0 0
#>
#> $V2
#> [1] 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0
#> [39] 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1
#> [77] 0 0 1 0 0
#>
#> $V3
#> [1] 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1
#> [39] 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0
#> [77] 1 0 0 1 0
#>
#> $V4
#> [1] 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1
#> [39] 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1
#> [77] 1 0 1 1 0
#>
#> $V5
#> [1] 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0
#> [39] 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0
#> [77] 0 1 0 0 1
#>
#> $V6
#> [1] 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0
#> [39] 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1
#> [77] 0 1 1 0 1
#>
#> $V7
#> [1] 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1
#> [39] 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0
#> [77] 1 1 0 1 1
#>
#> $V8
#> [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
#> [39] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
#> [77] 1 1 1 1 1
Created on 2022-09-30 with reprex v2.0.2