I'd use logical indexing for this: a[isnan(a)] = 42 clean_a = a[!isnan(a)]
On Thursday, April 7, 2016 at 9:20:05 AM UTC-4, Kristoffer Carlsson wrote: > > to replace with 42, maybe: > > map!((i) -> isnan(i)? 42 : i, a) > > to delete: > > deleteat!(a, find(isnan, a)) > > > > > On Thursday, April 7, 2016 at 2:33:56 PM UTC+2, jw3126 wrote: >> >> I have an array, which has some NaNs and I want to do things like replace >> them by 42 or delete them. What is the canonical way to do this? >> >
