One of the step in an iodoform reaction is the abstraction of a proton from the CH3 group.
However, in a 1,3 diketone, the proton present in between the 2 carbonyl groups (eg. the proton on the third carbon in pent-2,4-dione aka acetylacetone) would be more acidic, and hence more readily abstracted. How does a diketone then give the iodoform reaction?