Find vector overlap from the start - r

Question

I am looking for an efficient way to get the first k elements that are the same between two vectors in R.
For example:
orderedIntersect(c(1,2,3,4), c(1,2,5,4))
# [1] 1 2
orderedIntersect(c(1,2,3), c(1,2,3,4))
# [1] 1 2 3
This is the same as the intersect behavior, but any values after the first mismatch should be dropped.
I also want this to work for strings.
So far, the solution that I have is this:
orderedIntersect <- function(a,b) {
a <- as.vector(a)
NAs <- is.na(match(a, as.vector(b)))
last <- ifelse(any(NAs), min(which(NAs)) - 1, length(a))
a[1:last]
}
I am troubled by the fact that I have to iterate over n input elements 6 times: match, is.na, any, which, min, and the subset [].
Clearly, it would be faster to write an external C function (with a for loop and a break), but I am wondering if there is any clever R trick I can use here.

Answer 1

You can compare the values of your vectors and drop elements when the first FALSE is reached:
orderedIntersect <- function(a,b) {
# check the lengths are equal and if not, "cut" the vectors so they are (to avoid warnings)
l_a <- length(a) ; l_b <- length(b)
if(l_a != l_b) {m_l <- min(l_a, l_b) ; a <- a[1:m_l] ; b <- b[1:m_l]}
# compare the elements : they are equal if both are not NA and have the same value or if both are NA
comp <- (!is.na(a) & !is.na(b) & a==b) | (is.na(a) & is.na(b))
# return the right vector : nothing if the first elements do not match, everything if all elements match or just the part that match
if(!comp[1]) return(c()) else if (all(comp)) return(a) else return(a[1:(which(!comp)[1]-1)])
}
orderedIntersect(c(1,2,3,4), c(1,2,5,4))
#[1] 1 2
orderedIntersect(c(1,2,3), c(1,2,3,4))
#[1] 1 2 3
orderedIntersect(c(1,2,3), c(2,3,4))
#NULL

Answer 2

The simple C solution (for integers) isn't really any longer than the R version, but it would be a little more work to extend to all the other classes.
library(inline)
orderedIntersect <- cfunction(
signature(x='integer', y='integer'),
body='
int i, l = length(x) > length(y) ? length(y) : length(x),
*xx = INTEGER(x), *yy = INTEGER(y);
SEXP res;
for (i = 0; i < l; i++) if (xx[i] != yy[i]) break;
PROTECT(res = allocVector(INTSXP, i));
for (l = 0; l < i; l++) INTEGER(res)[l] = xx[l];
UNPROTECT(1);
return res;'
)
## Tests
a <- c(1L,2L,3L,4L)
b <- c(1L,2L,5L,4L)
c <- c(1L,2L,8L,9L,9L,9L,9L,3L)
d <- c(9L,0L,0L,8L)
orderedIntersect(a,b)
# [1] 1 2
orderedIntersect(a,c)
# [1] 1 2
orderedIntersect(a,d)
# integer(0)
orderedIntersect(a, integer())
# integer(0)

Answer 3

This might work:
#test data
a <- c(1,2,3,4)
b <- c(1,2,5,4)
c <- c(1,2,8,9,9,9,9,3)
d <- c(9,0,0,8)
empty <- c()
string1 <- c("abc", "def", "ad","k")
string2 <- c("abc", "def", "c", "lds")
#function
orderedIntersect <- function(a, b) {
l <- min(length(a), length(b))
if (l == 0) return(numeric(0))
a1 <- a[1:l]
comp <- a1 != b[1:l]
if (all(!comp)) return(a1)
a1[ 0:(min(which(comp)) - 1) ]
}
#testing
orderedIntersect(a,b)
# [1] 1 2
orderedIntersect(a,c)
# [1] 1 2
orderedIntersect(a,d)
# numeric(0)
orderedIntersect(a, empty)
# numeric(0)
orderedIntersect(string1,string2)
# [1] "abc" "def"