Is there a good reason that there is no single-argument version of filter that returns a filter-function? I.e. why is the following (or something similar) not in Base filter(f::Function) = Base.Fix1(filter, f) ? This would be in line with the existence of <(x) = (y -> y < x) or occursin(haystack…

It’s included in Julia 1.9 :slight_smile:

Why no single argument `filter(f::function)`?

sijo January 9, 2023, 4:24pm 2

It’s included in Julia 1.9

8 Likes

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