The following code used to work (some 1 year ago…)
collect(Iterators.flatten((prod(u) for u in with_replacement_combinations(x,i)) for i in 0:3))
Now, the code works if I change the i range to 1:3, but then I miss value 1 in the resulting vector:
[I used the code to generate multivariate basis functions for regression…]
julia> x = [1,2,3,4,5];
julia> collect(Iterators.flatten((prod(u) for u in with_replacement_combinations(x,i)) for i in 0:3))
56-element Vector{Int64}:
It works for me 
can you post more details about the error you’re seeing?
Sorry… my info was incomplete…
using Symbolics, Combinatorics
@variables x[1:2]
collect(Iterators.flatten((prod(u) for u in with_replacement_combinations(x,i)) for i in 0:3))
Ah… seems like it is due to a change in Symbolics…
If I instead do:
using Symbolics, Combinatorics
@variables x[1:2]
x = collect(x)
collect(Iterators.flatten((prod(u) for u in with_replacement_combinations(x,i)) for i in 0:3))
Apparently, now @variables x[1:2] creates a symbolic array variable, but I need an array of symbolic variables — which is achieved by x = collect(x). [A trick I learned from @baggepinnen in a totally different setting.]
ti potrebbe essere utile questo?
julia> using DynamicPolynomials
julia>
julia> x,y,z = @polyvar x y z
(x, y, z)
julia> monomials([x,y], 2)
3-element MonomialVector{true}:
x²
xy
y²
julia> monomials([x,y], 0:3)
10-element MonomialVector{true}:
x³
x²y
xy²
y³
x²
xy
y²
x
y
1
julia> monomials([x,y,z], 1:2)
9-element MonomialVector{true}:
x²
xy
xz
y²
yz
z²
x
y
z
My Italian is not good, but Google helps… I’ll check it.