Let's say I have two vectors:
x <- c(1,16,20,7,2)
y <- c(1, 7, 5,2,4,16,20,10)
I want to remove elements in y that are not in x. That is, I want to remove elements 5, 4, 10 from y.
y
[1] 1 7 2 16 20
In the end, I want vectors x and y to have to same elements. Order does not matter.
My thoughts: The match function lists the indices of the where the two vectors contains a matching element but I need a function is that essentially the opposite. I need a function that displays the indices where the elements in the two vectors don't match.
# this lists the indices in y that match the elements in x
match(x,y)
[1] 1 6 7 2 4 # these are the indices that I want; I want to remove
# the other indices from y
Does anyone know how to do this? thank you
You are after intersect
intersect(x,y)
## [1] 1 16 20 7 2
If you want the indices for the elements of y in x, using which and %in% (%in% uses match internally, so you were on the right track here)
which(y %in% x)
## [1] 1 2 4 6 7
As @joran points out in the comments intersect will drop the duplicates, so perhaps a safe option, if you want to return true matches would be something like
intersection <- function(x,y){.which <- intersect(x,y)
.in <- x[which(x %in% y)]
.in}
x <- c(1,1,2,3,4)
y <- c(1,2,3,3)
intersection(x,y)
## [1] 1 1 2 3
# compare with
intersect(x,y)
## [1] 1 2 3
intersection(y,x)
## [1] 1 2 3 3
# compare with
intersect(y, x)
## [1] 1 2 3
You then need to be careful about ordering with this modified function (which is avoided with intersect as it drops duplicated elements )
If you want the index of those element of y not in x, simply prefix with ! as `%in% returns a logical vector
which(!y%in%x)
##[1] 3 5 8
Or if you want the elements use setdiff
setdiff(y,x)
## [1] 5 4 10