I'm trying to create a list comprehension to create a list of tuples of length k
, and I figured that macro is the way to do it.
An example for k = 2 of the list comprehension i'm trying to get (for some given n
):
[[i, j] for i=1:(n-k+1) for j=(k+i-1):(n-k+2)]
The m-th index i_m would go from i_{m-1}+1 to n-k+m.
I tried creating the following macro, but this seems way off:
macro create_indices(n, k)
iterator_expressions = map(1:k) do s
Expr(
:(=),
Symbol("i", s),
quote $(esc(s)):$(esc(s - k + n)) end
)
end
Expr(
:comprehension,
[esc(Symbol("i", s)) for s=1:k],
iterator_expressions...,
)
end
for some reason i'm only getting a matrix of expressions, and I have no idea what i'm doing wrong. Please help!
Why do you need a macro for that?
julia> create_indices(n,k) = [(i,j) for i=1:(n-k+1) for j=(k+i-1):(n-k+2)]
create_indices (generic function with 1 method)
julia> create_indices(3,2)
3-element Vector{Tuple{Int64, Int64}}:
(1, 2)
(1, 3)
(2, 3)
because for k=3
I need
[[i, j, l] for i=1:(n-k+1) for j=(k+i-2):(n-k+2) for l=(k+j-2):(n-k+3)]
collect(zip([s:s-k+n for s in 1:k]...))
?
This will not generate the correct indices, but very likely you can adapt it to do it:
julia> function create_indices(n,k)
[ ntuple(k) do i
j-k+i
end
for j in 1:n
]
end
create_indices (generic function with 1 method)
julia> create_indices(3,4)
3-element Vector{NTuple{4, Int64}}:
(-2, -1, 0, 1)
(-1, 0, 1, 2)
(0, 1, 2, 3)
(in any case I'm sure a macro will only make things more complicated. If you can write the indexing correctly in a macro, it will be easier with a standar function)
Leandro Martínez said:
This will not generate the correct indices, but very likely you can adapt it to do it:
julia> function create_indices(n,k) [ ntuple(k) do i j-k+i end for j in 1:n ] end create_indices (generic function with 1 method) julia> create_indices(3,4) 3-element Vector{NTuple{4, Int64}}: (-2, -1, 0, 1) (-1, 0, 1, 2) (0, 1, 2, 3)
(in any case I'm sure a macro will only make things more complicated. If you can write the indexing correctly in a macro, it will be easier with a standar function)
I'm not sure this is possible as the indices depend on each other:
The m-th index i_m would go from i_{m-1}+1 to n-k+m.
Getting that kind of thing right is tricky, but it is trickier with a macro. In either case you need to figure out which is the (possible recursive) algorithm that will generate the correct sequence. And then, implementing that as a function is easier than as a macro.
I think, you can write iterator with a somewhat complicated state. Update of the state can have some overhead of course, but usually it should be overshadowed by main calculations.
@Or Golan It's not very clear to me what you're trying to do here, but here's me fixing the syntax errors in your macro. It at least correctly parses now, but I don't know if this is exactly the iteration layout you wanted, I left all the actual logic how you wrote it.
macro create_indices(n, k)
iterator_expressions = map(1:k) do s
:($(Symbol("i", s)) = $s:$(s - k + n))
end
Expr(
:comprehension,
Expr(:tuple, (Symbol("i", s) for s=1:k)...),
(iterator_expressions)...,
) |> esc
end
julia> @create_indices(4, 2)
3×3 Matrix{Tuple{Int64, Int64}}:
(1, 2) (1, 3) (1, 4)
(2, 2) (2, 3) (2, 4)
(3, 2) (3, 3) (3, 4)
julia> @create_indices(4, 3)
2×2×2 Array{Tuple{Int64, Int64, Int64}, 3}:
[:, :, 1] =
(1, 2, 3) (1, 3, 3)
(2, 2, 3) (2, 3, 3)
[:, :, 2] =
(1, 2, 4) (1, 3, 4)
(2, 2, 4) (2, 3, 4)
I suspect this isn't the logic you actually intended though because the example you have [[i, j] for i=1:(n-k+1) for j=(k+i-1):(n-k+2)]
is different
But also that example you gave contracts what you said about "The m-th index i_m would go from i_{m-1}+1 to n-k+m."
I'm trying to get all k
-combinations from some collection of length n
Mason Protter said:
I suspect this isn't the logic you actually intended though because the example you have
[[i, j] for i=1:(n-k+1) for j=(k+i-1):(n-k+2)]
is different
that's actually the same for k=2
- k+i-1 = i+1
.
I started with a recursive function that does this, but I'm trying to reduce the number of allocations, so I thought that generating the list comprehension using a macro would be more efficient. Here is the recursive function:
function combinations(collection, k)
n = length(collection)
combs = []
for i in 1:n
el = collection[i]
if k == 1
push!(combs, [el])
else
rest = collection[i+1:end]
subcombs = combinations(rest, k-1)
for subcomb in subcombs
push!(combs, [el; subcomb])
end
end
end
combs
end
Your combs
is a Vector{Any}
, you should give it a concrete eltype
Just for fun and not in any way efficient:
julia> create_indices(n, k) = reverse.(collect(Iterators.Filter(x -> issorted(x, lt = !<), Iterators.product(fill(1:n, k)...))))
create_indices (generic function with 1 method)
julia> create_indices(5, 3)
10-element Vector{Tuple{Int64, Int64, Int64}}:
(1, 2, 3)
(1, 2, 4)
(1, 2, 5)
(1, 3, 4)
(1, 3, 5)
(1, 4, 5)
(2, 3, 4)
(2, 3, 5)
(2, 4, 5)
(3, 4, 5)
Regarding the macro approach,
julia> dump(:([[i, j, l] for i=1:(n-k+1) for j=(k+i-2):(n-k+2) for l=(k+j-2):(n-k+3)]))
Expr
head: Symbol comprehension
args: Array{Any}((1,))
1: Expr
head: Symbol flatten
args: Array{Any}((1,))
1: Expr
head: Symbol generator
args: Array{Any}((2,))
1: Expr
head: Symbol flatten
args: Array{Any}((1,))
1: Expr
2: Expr
head: Symbol =
args: Array{Any}((2,))
1: Symbol i
2: Expr
indicates that you would need to nest flatten
and generator
expressions.
But I tend to agree with the opinion that an iterator should be the best approach for this problem.
Or you could just use Combinatorics.combinations
of course. :-)
Or Golan said:
I'm trying to get all
k
-combinations from some collection of lengthn
Mason Protter said:
I suspect this isn't the logic you actually intended though because the example you have
[[i, j] for i=1:(n-k+1) for j=(k+i-1):(n-k+2)]
is differentthat's actually the same for
k=2
-k+i-1 = i+1
.
Are you sure?
julia> let n = 5, k = 2
[[i, j] for i=1:(n-k+1) for j=(k+i-1):(n-k+2)]
end
10-element Vector{Vector{Int64}}:
[1, 2]
[1, 3]
[1, 4]
[1, 5]
[2, 3]
[2, 4]
[2, 5]
[3, 4]
[3, 5]
[4, 5]
julia> macro create_indices(n, k)
iterator_expressions = map(1:k) do s
:($(Symbol("i", s)) = $s:$(s - k + n))
end
Expr(
:comprehension,
Expr(:tuple, (Symbol("i", s) for s=1:k)...),
(iterator_expressions)...,
) |> esc
end
@create_indices (macro with 1 method)
julia> @create_indices(5, 2)
4×4 Matrix{Tuple{Int64, Int64}}:
(1, 2) (1, 3) (1, 4) (1, 5)
(2, 2) (2, 3) (2, 4) (2, 5)
(3, 2) (3, 3) (3, 4) (3, 5)
(4, 2) (4, 3) (4, 4) (4, 5)
Looks like you want the upper-triangle of what the macro is making
Yeah, the macro I tried writing was wrong, but in two dimensions. Thanks to you it is wrong only in one way, but when I tried to correct the logic I'm still way off:
macro create_indices(n, k)
local i0 = 0
iterator_expressions = map(1:k) do s
:($(Symbol("i", s)) = ($(Symbol("i", s-1)) + 1):$(s - k + n))
end
Expr(
:comprehension,
Expr(:tuple, (Symbol("i", s) for s=1:k)...),
(iterator_expressions)...,
) |> esc
end
It's telling me that i0
is not defined, and if I define it outside, it says that i1
is not defined. I think there is something to do with hygiene here?
Here's how I'd do this:
julia> @generated function create_indices(n, ::Val{k}) where {k}
ex = foldr(1:k, init=:(push!(out, $(Expr(:tuple, (Symbol(:i_, s) for s ∈ 1:k)...))))) do s, old
lower = s == 1 ? 1 : :(k + $(Symbol(:i_, s-1)) - 1)
quote
for $(Symbol(:i_, s)) ∈ $lower:(n - k + $s)
$old
end
end
end
quote
out = NTuple{$k, Int}[]
$ex
out
end
end
julia> create_indices(5, Val(2))
10-element Vector{Tuple{Int64, Int64}}:
(1, 2)
(1, 3)
(1, 4)
(1, 5)
(2, 3)
(2, 4)
(2, 5)
(3, 4)
(3, 5)
(4, 5)
julia> create_indices(8, Val(3))
20-element Vector{Tuple{Int64, Int64, Int64}}:
(1, 3, 5)
(1, 3, 6)
(1, 3, 7)
(1, 3, 8)
(1, 4, 6)
(1, 4, 7)
(1, 4, 8)
(1, 5, 7)
(1, 5, 8)
(1, 6, 8)
(2, 4, 6)
(2, 4, 7)
(2, 4, 8)
(2, 5, 7)
(2, 5, 8)
(2, 6, 8)
(3, 5, 7)
(3, 5, 8)
(3, 6, 8)
(4, 6, 8)
Also something along these lines may help:
Base.Cartesian.@nloops 3 i m->(k+i_{m+1}-2:n-k+m) begin
push!(res, Base.Cartesian.@ntuple 3 i)
end
Thanks @Mason Protter ! I was able to build exactly what I needed from this code.
In case anyone else is interested:
@generated function create_indices(n, ::Val{k}) where {k}
ex = foldr(1:k, init=:(push!(out, $(Expr(:vect, (Symbol(:i_, s) for s ∈ 1:k)...))))) do s, old
lower = s == 1 ? 1 : :($(Symbol(:i_, s-1)) + 1)
quote
for $(Symbol(:i_, s)) ∈ $lower:(n - k + $s)
$old
end
end
end
quote
out = Vector{Int}[]
$ex
out
end
end
By the way, why did you need to use Val
?
Why are you storing vectors instead of Tuples?
By the way, why did you need to use Val?
That's because generated functions operate on types rather than values, so to generate a different number of loops depending on k
, I needed to lift k
into the type domain using Val
So how did the performance end up compared to Combinatorics.combinations
?
julia> using Combinatorics
julia> @generated function create_indices_tup(n, ::Val{k}) where {k}
ex = foldr(1:k, init=:(push!(out, $(Expr(:tuple, (Symbol(:i_, s) for s ∈ 1:k)...))))) do s, old
lower = s == 1 ? 1 : :(k + $(Symbol(:i_, s-1)) - 1)
quote
for $(Symbol(:i_, s)) ∈ $lower:(n - k + $s)
$old
end
end
end
quote
out = NTuple{$k, Int}[]
$ex
out
end
end;
julia> @generated function create_indices_vec(n, ::Val{k}) where {k}
ex = foldr(1:k, init=:(push!(out, $(Expr(:vect, (Symbol(:i_, s) for s ∈ 1:k)...))))) do s, old
lower = s == 1 ? 1 : :($(Symbol(:i_, s-1)) + 1)
quote
for $(Symbol(:i_, s)) ∈ $lower:(n - k + $s)
$old
end
end
end
quote
out = Vector{Int}[]
$ex
out
end
end;
julia> let n = Ref(5), k = 3
@btime collect(Combinatorics.combinations(1:$n[], $k))
@btime create_indices_vec($n[], Val($k))
@btime create_indices_tup($n[], Val($k))
end;
356.250 ns (22 allocations: 1.30 KiB)
326.026 ns (13 allocations: 1.23 KiB)
122.733 ns (2 allocations: 272 bytes)
This is of course under the assumption that you can statically know the value of k
.
I used vec since I want to index a collection with it
Last updated: Nov 22 2024 at 04:41 UTC