Returning vector of values shared across 3 vectors?

try this:

vec1<- c(4,5,6,7,8,9,10,11,12,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81)
vec2<- c (1,2,3,4,5,6,7,8,9,10,11,12,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66)
vec3<- c (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,52)
  intersect(vec1,intersect(vec2, vec3))

  [1]  4  5  6  7  8  9 10 11 12 52

On Sat, Oct 1, 2011 at 2:00 AM, Chris Conner<connerpharmd at yahoo.com>  wrote:

Help-Rs,

I've got three vectors representing participants:

vec1<- c(4,5,6,7,8,9,10,11,12,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81)
vec2<- c (1,2,3,4,5,6,7,8,9,10,11,12,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66)
vec3<- c (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,52)

I'd like to return a vector that contains only the values that are shared across ALL THREE vectors. So the statement would return a vector that looked like this:
4,5,6,7,8,9,10,11,12,52

For some reason I initially thought that a cbind and a unique() would handle it, but then common sense sunk in.  I think the sleep deprivation is starting to take it's toll.  I've got to believe that there is a simple solution to this dilema.

Thanks in adance for any help!
C
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