Create a dataframe of combinations of data with proper labels in R - r

Question

I am trying to create a data frame that pulls combinations of values from a data frame, and applies a function on the data, and finally assigning the label of the combinations as identifiers in the data frame.
To create a simple mockup of the data:
A = (1:20)
B = seq(1, 40, 2)
To create a mockup of each identifier:
L = c("q", "r", "q", "p", "q", "p", "r", "r", "r", "r", "r", "r", "r", "r", "o", "p", "q", "r", "r", "q")
The mockup data frame is then created with:
dat.example = data.frame(A, B, L)
Here, the data in each row is subtracted by the data from the previous row, with all possible combinations.
diffA = combn(dat.example$A, 2, FUN = diff)
diffB = combn(dat.example$B, 2, FUN = diff)
After all possible combination of differences within columns A and B are calculated, a function is applied on the generated output.
calc.diff = sqrt((diffA^2) + (diffB^2))
Now comes the tricky part, where I pull out the identifiers for the combinations.
(L.list = combn(dat.example$L, 2, simplify = F))
L.list.1 = sapply(L.list, "[[", 1)
L.list.2 = sapply(L.list, "[[", 2)
L.dat = data.frame(
X = L.list.1,
Y = L.list.2
)
L.dat$comp = with(
L.dat,
interaction(X, Y, sep = "_")
)
And combine them into a master dataset
master = data.frame(
calc.diff,
L.dat$comp
)
Is there a simpler way to perform this that I am missing?
Update: I just realised the labels I created in this fashion is duplicated, as p_q is equivalent to q_r as only the magnitude of the value matters after the calc.dist function. Therefore, this question still needs a better answer. My present solution is to simply substitute each occurrence with it's twin in the dataset, but this becomes substantially difficult the more labels there are.
Update 2: I have solved the issue of a pure R solution to properly assigning unique combination of labels to the dataset. Below, I shall paste the entire code for generating the above dataset, and assigning the right labels. Improvements are of course welcomed.
A = (1:20)
B = seq(1, 40, 2)
L = c("q", "r", "q", "p", "q", "p", "r", "r", "r", "r", "r", "r", "r", "r", "o", "p", "q", "r", "r", "q")
dat.example = data.frame(A, B, L)
diffA = combn(dat.example$A, 2, FUN = diff)
diffB = combn(dat.example$B, 2, FUN = diff)
calc.diff = sqrt((diffA^2) + (diffB^2))
L.dat.comp <- combn(dat.example$L, 2, paste0, collapse = "_", simplify=F)
master <- data.frame(
calc.diff,
id = unlist(L.dat.comp))
Function to help with combinations
expand.grid.unique <- function(x, y, include.equals=FALSE)
{
x <- unique(x)
y <- unique(y)
g <- function(i)
{
z <- setdiff(y, x[seq_len(i-include.equals)])
if(length(z)) cbind(x[i], z, deparse.level=0)
}
do.call(rbind, lapply(seq_along(x), g))
}
Create dataframe with unique combinations of id labels.
dat.combo =
as.data.frame(
expand.grid.unique(L, L, include.equals = T)
)
dat.combo$c1 = with(dat.combo, interaction(V1, V2, sep = "_"))
dat.combo$c2 = with(dat.combo, interaction(V2, V1, sep = "_"))
master$id2 = master$id
Where ids match dat.combo$c2, replace with dat.com$c1
master$id2[] = setNames(dat.combo$c1,
dat.combo$c2)[as.character(
unlist(
master$id2
)
)
]
Where NAs are present in id2, replace with values in id
master$id2[is.na(master$id2)] = master$id[is.na(master$id2)]

Answer 1

Based on Jay's comement for the "tricky part":
L.dat.comp <- combn(dat.example$L, 2, paste0, collapse = "_", simplify=F)
master <- data.frame(cbind(calc.diff,unlist(L.dat.comp)))
So the whole thing looks like:
A = (1:20)
B = seq(1, 40, 2)
L = c("q", "r", "q", "p", "q", "p", "r", "r", "r", "r", "r", "r", "r", "r", "o", "p", "q", "r", "r", "q")
dat.example = data.frame(A, B, L)
diffA = combn(dat.example$A, 2, FUN = diff)
diffB = combn(dat.example$B, 2, FUN = diff)
calc.diff = sqrt((diffA^2) + (diffB^2))
L.dat.comp <- combn(dat.example$L, 2, paste0, collapse = "_", simplify=F)
master <- data.frame(cbind(calc.diff,unlist(L.dat.comp)))