2012-09-27 44 views

回答

8

你接近

vector[!vector %in% vector2] 

,或者即使你說 「不使用setdiff」

setdiff(vector, vector2) 
1
vector1 <- letters[1:4] 
set.seed(001) 
vector2 <- sample(letters[1:15], 10, replace=TRUE) 

vector1 
[1] "a" "b" "c" "d" 
vector2 
[1] "d" "f" "i" "n" "d" "n" "o" "j" "j" "a" 

vector2 [!(vector2 %in% vector1)] # elements in vector2 that are not in vector1 
[1] "f" "i" "n" "n" "o" "j" "j" 
1

您可以定義一個新的運營商,

`%ni%` = Negate(`%in%`) 
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