Common elements of vectors with multiple elements

问题:

How to efficiently find common elements of two vectors with duplicate elements?

Example:

v1 <- c(1, 1, 2, 3, 3, 4)  
v2 <- c(1, 1, 1, 3, 4, 5)  
commonElements <- c(1, 1, 3, 4)

intersect doesn't handle duplicate elements well.

回答1:

I like intersect and tables, so...

tv1 <- table(v1)
tv2 <- table(v2)
comvals <- intersect(names(tv1),names(tv2))
comtab <- apply(rbind(tv1[comvals],tv2[comvals]),2,min)

The information is still there, but in (what I view as) a nicer format:

> comtab
1 3 4 
2 1 1 

EDIT: If you really want that vector, though, it's: as.numeric(rep(names(comtab),comtab)).

回答2:

Here is another option:

common <- function(v1, v2) {
  lvls <- unique(c(v1, v2))
  v1a <- factor(v1, levels=lvls)
  v2a <- factor(v2, levels=lvls)
  v <- pmin(table(v1a), table(v2a))
  as.numeric(rep(names(v), v))
}

common(rep(1:3, 1:3), rep(1:2, 1:2))
[1] 1 2 2

common(rep(c(1,3,5), 1:3), rep(c(5,2), 2))
[1] 5 5

EDIT: wrap a function, demonstrate different cases and speed up per @Dason's comment

回答3:

I'm sure there are many ways to do this but I opted to sort it and use rle to get the values and counts. table could probably accomplish the same task as well.

common <- function(v1, v2){
  r1 <- rle(sort(v1))
  r2 <- rle(sort(v2))
  vals <- intersect(r1$values, r2$values)
  l1 <- r1$lengths[r1$values %in% vals]
  l2 <- r2$lengths[r2$values %in% vals]
  rep(vals, pmin(l1, l2))
}

common(v1, v2)

some examples

> common(v1, v2)
[1] 1 1 3 4
> common(c(1,1), c(3,2,1,3,1))
[1] 1 1
> common(c(1,2,3,2), c(1,2,3))
[1] 1 2 3