match

match는 2nd argument와 일치하는 1st argument의 위치값vector를 리턴
%in%은 T/F값을 리턴

## The intersection of two sets can be defined via match():
## Simple version:
5:10 %in% c(7:15)       # [1] FALSE FALSE  TRUE  TRUE  TRUE  TRUE

match(5:10, 7:15)       # [1] NA NA  1  2  3  4
match(7:15, 5:10)       # [1]  3  4  5  6 NA NA NA NA NA

## intersect <- function(x, y) y[match(x, y, nomatch = 0)]
intersect(5:10, 7:15)   # [1]  7  8  9 10

sstr <- c("c","ab","B","bba","c",NA,"@","bla","a","Ba","%")
alphabet <- c(letters, LETTERS)
sstr[sstr %in% alphabet ]                     # [1] "c" "B" "c" "a"
sstr[which(sstr %in% alphabet)]               # [1] "c" "B" "c" "a"
alphabet[ match(sstr, alphabet, nomatch=0) ]  # [1] "c" "B" "c" "a"

sstr[match(alphabet, sstr, nomatch=0)]        # [1] "a" "c" "B"

"%w/o%" <- function(x, y) x[!x %in% y] #--  x without y
(1:5) %w/o% c(2,3,12)   # [1] 1 4 5
## Note that setdiff() is very similar and typically makes more sense:
        c(1:6,7:2) %w/o% c(3,7,12)  # -> keeps duplicates
setdiff(c(1:6,7:2),      c(3,7,12)) # -> unique values