How can I quickly see if any elements of more than

问题:

(I expect this has already been asked/answered. If so, sorry, I'm failing to locate the answer.)

Let's say I have 6 vectors. How can I quickly check whether any element for each vector is equal to any element of all the other vectors?

I know I could do the following, and it feels really cumbersome/pre-historic/error-prone:

any(vec1 %in% vec2, vec1 %in% vec3, vec1 %in% vec4, vec1 %in% vec5, vec1 %in% vec6,
    vec2 %in% vec3, vec2 %in% vec4, vec2 %in% vec5, vec2 %in% vec6,
    vec3 %in% vec4, vec3 %in% vec5, vec3 %in% vec6,
    vec4 %in% vec5, vec4 %in% vec6,
    vec5 %in% vec6)

Thanks.

By the way, I checked How to find common elements from multiple vectors? and that appears to be asking for how to identify elements that are present in each vector, rather than if any elements from any of the vectors are equal.

回答1:

If you put your vectors in a list, they'll be substantially easier to work with:

# make sample data
set.seed(47)
x <- replicate(6, rpois(3, 10), simplify = FALSE) 

str(x)
# List of 6
#  $ : int [1:3] 16 12 10
#  $ : int [1:3] 9 10 6
#  $ : int [1:3] 10 14 4
#  $ : int [1:3] 7 6 4
#  $ : int [1:3] 12 8 7
#  $ : int [1:3] 7 11 8

Now iterate with lapply:

lapply(x, function(y){sapply(x, function(z){y %in% z})})

## [[1]]
##      [,1]  [,2]  [,3]  [,4]  [,5]  [,6]
## [1,] TRUE FALSE FALSE FALSE FALSE FALSE
## [2,] TRUE FALSE FALSE FALSE  TRUE FALSE
## [3,] TRUE  TRUE  TRUE FALSE FALSE FALSE
## 
## [[2]]
##       [,1] [,2]  [,3]  [,4]  [,5]  [,6]
## [1,] FALSE TRUE FALSE FALSE FALSE FALSE
## [2,]  TRUE TRUE  TRUE FALSE FALSE FALSE
## [3,] FALSE TRUE FALSE  TRUE FALSE FALSE
## ...    ...  ...   ...   ...   ...   ...

which is a matrix for each vector, where the rows are the elements of that respective vector and the columns are each of the vectors in the list, and the values indicate whether that element is in that vector. Obviously each will match with itself, so the first column of the first element is all TRUE, as is the second column of the second element, etc. Other TRUEs indicate cross-vector matches. If lengths are inconsistent, it will return a nested list of the same information instead of a matrix. If you'd rather have a nested list anyway, change sapply to lapply.

Alternately, if you just want a vector of matches for each vector,

str(lapply(x, function(y){which(sapply(x, function(z){any(y %in% z)}))}))

## List of 6
##  $ : int [1:4] 1 2 3 5
##  $ : int [1:4] 1 2 3 4
##  $ : int [1:4] 1 2 3 4
##  $ : int [1:5] 2 3 4 5 6
##  $ : int [1:4] 1 4 5 6
##  $ : int [1:3] 4 5 6

where each element still contains itself as a match. Take out which for Booleans instead of indices.

回答2:

We can use combn to find the combination of vector strings, get the data, compare them with %in%, wrap with any and unlist if needed

v1 <- unlist(combn(paste0("vec", 1:6), 2, FUN = function(x) 
             any(get(x[1]) %in% get(x[2])), simplify = FALSE))
names(v1) <- combn(paste0("vec", 1:6), 2, FUN = paste, collapse="-")

As the OP mentioned about efficiency, a faster version of combn can be used if needed.

---

Also, combn can be applied directly to list. So, the vectors can be placed in a list and then do the combn

v2 <- combn(mget(paste0("vec", 1:6)), 2, FUN = function(x) any(x[[1]] %in% x[[2]]))
names(v2) <- names(v1)

Also, as the OP is wrapping any over all the comparisons. we can also do it with one any

any(combn(mget(paste0("vec", 1:6)), 2, FUN = function(x) x[[1]] %in% x[[2]]))

but I am not sure whether that is a correct way.

data

vec1 <- 1:6
vec2 <- 2:3
vec3 <- 5:7
vec4 <- 6:8
vec5 <- 9:10
vec6 <- 11:12