Combining vectors of unequal length and non-unique values - r

Question

I would like to do the following:
combine into a data frame, two vectors that
have different length
contain sequences found also in the other vector
contain sequences not found in the other vector
sequences that are not found in other vector are never longer than 3 elements
always have same first element
The data frame should show the equal sequences in the two vectors aligned, with NA in the column if a vector lacks a sequence present in the other vector.
For example:
vector 1 vector 2 vector 1 vector 2
1 1 a a
2 2 g g
3 3 b b
4 1 or h a
1 2 a g
2 3 g b
5 4 c h
5 c
should be combined into data frame
1 1 a a
2 2 g g
3 3 b b
4 NA h NA
1 1 or a a
2 2 g g
NA 3 NA b
NA 4 NA h
5 5 c c
What I did, is to search for merge, combine, cbind, plyr examples but was not able to find solutions. I am afraid I will need to start write a function with nested for loops to solve this problem.

Answer 1

Note - this was proposed as an answer to the first version of the OP. The question has been modified since then but the problem is still not well-defined in my opinion.
Here is a solution that works with your integer example and would also work with numeric vectors. I am also assuming that:
both vectors contain the same number of sequences
a new sequence starts where value[i+1] <= value[i]
If your vectors are non-numeric or if one of my assumptions does not fit your problem, you'll have to clarify.
v1 <- c(1,2,3,4,1,2,5)
v2 <- c(1,2,3,1,2,3,4,5)
v1.sequences <- split(v1, cumsum(c(TRUE, diff(v1) <= 0)))
v2.sequences <- split(v2, cumsum(c(TRUE, diff(v2) <= 0)))
align.fun <- function(s1, s2) { #aligns two sequences
s12 <- sort(unique(c(s1, s2)))
cbind(ifelse(s12 %in% s1, s12, NA),
ifelse(s12 %in% s2, s12, NA))
}
do.call(rbind, mapply(align.fun, v1.sequences, v2.sequences))
# [,1] [,2]
# [1,] 1 1
# [2,] 2 2
# [3,] 3 3
# [4,] 4 NA
# [5,] 1 1
# [6,] 2 2
# [7,] NA 3
# [8,] NA 4
# [9,] 5 5

Answer 2

I maintain that your problem might be solved in terms of the shortest common supersequence. It assumes that your two vectors each represent one sequence. Please give the code below a try.
If it still does not solve your problem, you'll have to explain exactly what you mean by "my vector contains not one but many sequences": define what you mean by a sequence and tell us how sequences can be identified by scanning through your two vectors.
Part I: given two sequences, find the longest common subsequence
LongestCommonSubsequence <- function(X, Y) {
m <- length(X)
n <- length(Y)
C <- matrix(0, 1 + m, 1 + n)
for (i in seq_len(m)) {
for (j in seq_len(n)) {
if (X[i] == Y[j]) {
C[i + 1, j + 1] = C[i, j] + 1
} else {
C[i + 1, j + 1] = max(C[i + 1, j], C[i, j + 1])
}
}
}
backtrack <- function(C, X, Y, i, j) {
if (i == 1 | j == 1) {
return(data.frame(I = c(), J = c(), LCS = c()))
} else if (X[i - 1] == Y[j - 1]) {
return(rbind(backtrack(C, X, Y, i - 1, j - 1),
data.frame(LCS = X[i - 1], I = i - 1, J = j - 1)))
} else if (C[i, j - 1] > C[i - 1, j]) {
return(backtrack(C, X, Y, i, j - 1))
} else {
return(backtrack(C, X, Y, i - 1, j))
}
}
return(backtrack(C, X, Y, m + 1, n + 1))
}
Part II: given two sequences, find the shortest common supersequence
ShortestCommonSupersequence <- function(X, Y) {
LCS <- LongestCommonSubsequence(X, Y)[c("I", "J")]
X.df <- data.frame(X = X, I = seq_along(X), stringsAsFactors = FALSE)
Y.df <- data.frame(Y = Y, J = seq_along(Y), stringsAsFactors = FALSE)
ALL <- merge(LCS, X.df, by = "I", all = TRUE)
ALL <- merge(ALL, Y.df, by = "J", all = TRUE)
ALL <- ALL[order(pmax(ifelse(is.na(ALL$I), 0, ALL$I),
ifelse(is.na(ALL$J), 0, ALL$J))), ]
ALL$SCS <- ifelse(is.na(ALL$X), ALL$Y, ALL$X)
ALL
}
Your Example:
ShortestCommonSupersequence(X = c("a","g","b","h","a","g","c"),
Y = c("a","g","b","a","g","b","h","c"))
# J I X Y SCS
# 1 1 1 a a a
# 2 2 2 g g g
# 3 3 3 b b b
# 9 NA 4 h <NA> h
# 4 4 5 a a a
# 5 5 6 g g g
# 6 6 NA <NA> b b
# 7 7 NA <NA> h h
# 8 8 7 c c c
(where the two updated vectors are in columns X and Y.)