According to A Gentle Introduction to Haskell …
In Haskell the partial application of an infix operator is called a section.
Consider the Haskell expression filter (\n -> n > 0) [-3,-4,5,6,-7,8], which evaluates to [5,6,8] .
Using a section, this may be re-written in Haskell as filter (>0) [-3,-4,5,6,-7,8] .
In Julia, one may write filter( n -> n > 0, [-3,-4,5,6,-7,8] ) .
Can this last be re-written in Julia using an equivalent of the Haskell section (>0) ?
The following yields a syntax error …
filter( (>0), [-3,-4,5,6,-7,8] )
Also, in Haskell one can re-write …
filter (\list -> length list > 2) [ [2,3], [5,7,11], [13], [17,19,23,29] ]
… as …
filter ((>2).length) [ [2,3], [5,7,11], [13], [17,19,23,29] ]
In Julia, can one similarly re-write, using a section and function composition?
For your first exemple, you can write:
julia> filter(>(0), [-3,-4,5,6,-7,8])
3-element Vector{Int64}:
5
6
8
This works because according to the help:
julia> ?
help?> >
>(x)
Create a function that compares its argument to x using >, i.e. a function equivalent to y -> y > x. The returned function is of type Base.Fix2{typeof(>)}, which can be used to implement specialized methods.
│ Julia 1.2
│
│ This functionality requires at least Julia 1.2.
So if you want something similar for your second exemple, you may need to define yourself a similar function like this:
julia> length_sup(x) = y -> length(y) > x
length_sup (generic function with 1 method)
And then you can do:
julia> filter(length_sup(2), [ [2,3], [5,7,11], [13], [17,19,23,29] ])
2-element Vector{Vector{Int64}}:
[5, 7, 11]
[17, 19, 23, 29]
However, whether it is a good idea or not to create custom functions just for some sugar-syntaxing will be up to you. You may eventually want to code macro to simplify this task.