[ncl-talk] How to find the same elements in two matrixes

[Dennis Shea]

I do not think you can do it without a loop.

function get_dup_elements(a[*], b[*])
begin
   dima = dimsizes(a)
   dimb = dimsizes(b)
   ndup = -1

   if (dima.le.dimb) then
       dimd = new( dima, typeof(a))
       do n=0,dima-1
           ii := ind(b.eq.a(n))
           if (.not.ismissing(ii(0))) then
               ndup =  ndup+1
               dimd(ndup) = a(n)
           end if
       end do
   else
       dimd = new( dimb, typeof(a))
       do n=0,dimb-1
          ii := ind(a.eq.b(n))
          if (.not.ismissing(ii(0))) then
               ndup = ndup+1
               dimd(ndup) = b(n)
          end if
       end do

   end if

   return(dimd(0:ndup))
end

A = (/ 1, 2, 3, 4, 5, 6, 7, 8, 9/)
B = (/ 2, 14, 7, 1/)

dup = get_dup_elements(A,B)
print(dup)

;; sort(dup)  ; ascending order




On Wed, Dec 17, 2014 at 1:59 AM, Lin Wang <wanglin at post.iap.ac.cn> wrote:

> Dear all,
>
> Suppose:
> A = (/ 1, 2, 3, 4, 5, 6, 7, 8, 9/)
> B = (/ 2, 14, 7, 1/)
>
> If don't use loop, how to find the same elements between A and B such as
> (/1, 2, 7/)? Thanks.
>
> Best wishes,
> Lin Wang
>
>
>
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