Dear Barry, thank you very much for this information. This looks pretty interesting! Best Simon On Jan 31, 2013, at 4:09 PM, Barry Rowlingson <b.rowlingson at lancaster.ac.uk> wrote: >> it lets you do: >> >> (a~b~c) = foo() > > Mistook. should be: > > (a~b~c) %=% foo() > > because it defines the %=% operator. > > Barry