I have a vector like: a = c(1:10) and I need to remove multiple values, like: 2, 3, 5
How to delete those numbers (they are NOT the positions in the vector) in the vector?
at the moment i loop the vector and do something like:
a[!a=NUMBER_TO_REMOVE] But I think there is a function that does it automatically.
The %in% operator tells you which elements are among the numers to remove:
> a <- sample (1 : 10) > remove <- c (2, 3, 5) > a [1] 10 5 2 7 1 6 3 4 8 9 > a %in% remove [1] FALSE TRUE TRUE FALSE FALSE FALSE TRUE FALSE FALSE FALSE > a [! a %in% remove] [1] 10 7 1 6 4 8 9 Note that this will silently remove incomparables (stuff like NA or Inf) as well (while it will keep duplicate values in a as long as they are not listed in remove).
If a can contain incomparables, but remove will not, we can use match, telling it to return 0 for non-matches and incomparables (%in% is a conventient shortcut for match):
> a <- c (a, NA, Inf) > a [1] 10 5 2 7 1 6 3 4 8 9 NA Inf > match (a, remove, nomatch = 0L, incomparables = 0L) [1] 0 3 1 0 0 0 2 0 0 0 0 0 > a [match (a, remove, nomatch = 0L, incomparables = 0L) == 0L] [1] 10 7 1 6 4 8 9 NA Inf incomparables = 0 is not needed as incomparables will anyways not match, but I'd include it for the sake of readability.
This is, btw., what setdiff does internally (but without the unique to throw away duplicates in a which are not in remove).
If remove contains incomparables, you'll have to check for them individually, e.g.
if (any (is.na (remove))) a <- a [! is.na (a)] (This does not distinguish NA from NaN but the R manual anyways warns that one should not rely on having a difference between them)
For Inf/ -Inf you'll have to check both sign and is.finite
setdiff is better, as it does everything in one operation, and references the amended vector only once.
a that are not in remove as well. If that's not a problem, you can also use setdiff. setdiff, btw, uses match for which %in% is a shortcut.You can use setdiff.
Given
a <- sample(1:10) remove <- c(2, 3, 5) Then
> a [1] 10 8 9 1 3 4 6 7 2 5 > setdiff(a, remove) [1] 10 8 9 1 4 6 7 
a is the result of another function so you can do things in one line instead of 3 and a temp variable
%in% solution if the input vector contains duplicates (in which case setdiff will only return the unique set, i.e. without duplicates)
fsetdiff of data.table package has an all flag (default F) that allows to keep duplicates in the input vector.You can do it as follows:
> x<-c(2, 4, 6, 9, 10) # the list > y<-c(4, 9, 10) # values to be removed > idx = which(x %in% y ) # Positions of the values of y in x > idx [1] 2 4 5 > x = x[-idx] # Remove those values using their position and "-" operator > x [1] 2 6 Shortly
> x = x[ - which(x %in% y)] 
which here. It's basically the same as @cbeleites answer.which returns indexes of TRUE values. So minus sign can be used to say "the indexes other than these indexes". Also which is more readable since it is closer to the natural language.instead of
x <- x[! x %in% c(2,3,5)] using the packages purrr and magrittr, you can do:
your_vector %<>% discard(~ .x %in% c(2,3,5)) this allows for subsetting using the vector name only once. And you can use it in pipes :)
First we can define a new operator,
"%ni%" = Negate( "%in%" ) Then, its like x not in remove
x <- 1:10 remove <- c(2,3,5) x <- x[ x %ni% remove ] or why to go for remove, go directly
x <- x[ x %ni% c(2,3,5)] 
There is also subset which might be useful sometimes:
a <- sample(1:10) bad <- c(2, 3, 5) > subset(a, !(a %in% bad)) [1] 9 7 10 6 8 1 4 
UPDATE:
All of the above answers won't work for the repeated values, @BenBolker's answer using duplicated() predicate solves this:
full_vector[!full_vector %in% searched_vector | duplicated(full_vector)] Original Answer: here I write a little function for this:
exclude_val<-function(full_vector,searched_vector){ found=c() for(i in full_vector){ if(any(is.element(searched_vector,i))){ searched_vector[(which(searched_vector==i))[1]]=NA } else{ found=c(found,i) } } return(found) } so, let's say full_vector=c(1,2,3,4,1) and searched_vector=c(1,2,3).
exclude_val(full_vector,searched_vector) will return (4,1), however above answers will return just (4).
full_vector[!full_vector %in% searched_vector | duplicated(full_vector)] ?
full_vector = c(1,1,1,2,3); searched_vector = c(1,1,3); - that produces 1, 1, 2 instead of the correct answer 1, 2.removeif <- function(from, where) { for (i in where) if (i %in% from) {from = from[-match(i, from)]}; from}q <- c(1,1,2,2,3,3,3,4,4,5,5,7,7) rm <- q[11] remove(rm) q q[13] = NaN q q %in% 7 This sets the 13 in a vector to not a number(NAN) it shows false remove(q[c(11,12,13)]) if you try this you will see that remove function don't work on vector number. you remove entire vector but maybe not a single element.
Try this function
seq.int.exclude <- function(excluded, ...) { x <- seq.int(...) return(x[!(x %in% excluded)]) } Call examples:
seq.int.exclude(from = 10L, to = 20L, excluded = c(12L, 30L, 19L)) seq.int.exclude(from = 10L, to = 20L, excluded = 15L) 