contrasts(x) contrasts(x, how.many=<<see below>>) <- value
contrasts(temp) <- contr.poly(3)contrasts(treatment) <- contr.sum(5)[, 1:3] # resulting contrast has 4 columns, probably not what was intended
contrasts(treatment, 3) <- contr.sum(5)[, 1:3] # resulting contrast has 3 columns