Re: [julia-users] help with @generated function call please?

[Florian Oswald]

Yes I think you are right about that. Using a comprehension is just as good
for my case. But part of me would like to just finally understand that part
of Julia! Ok will have to wait for the next opportunity. :-)

Thanks anyway.
Florian

On Saturday, 15 October 2016, Erik Schnetter  wrote:

> A generated function is only useful if you perform a non-trivial
> calculation based on the argument types. You don't do that here, so I
> wonder whether simply using the Cartesian indexing macros by themselves
> would be sufficient.
>
> Note also that you don't need to write `$N` in your code; using `N`
> directly has the same effect here.
>
> I'm not saying that generated functions should be avoided at all costs,
> but if it isn't necessary here you might as well skip the associated
> complications.
>
> -erik
>
> On Fri, Oct 14, 2016 at 11:51 AM, Florian Oswald  > wrote:
>
>> hi all,
>>
>> I want to evaluate a function at each index of an array. There is a N
>> dimensional function, and I want to map it onto an N-dimensional array:
>>
>> fpoly(x::Array{Real,5}) = x[1] + x[2]^2 + x[3] + x[4]^2 + x[5]
>>
>> want to do
>>
>> a = rand(2,2,2,2,2);
>> b = similar(a)
>>
>> for i1 in indices(a,1)
>> for i2 in indices(a,2)
>> ...
>> b[i1,i2,i3,i4,i5] = fpoly(a[i1,i2,i3,i4,i5])
>> end
>> end...
>>
>> I tried:
>> # actually want to do it inplace
>> @generated function set_poly!{T,N}(a::Array{T,N})
>> quote
>> @nloops $N i a begin
>> @nref $N a i = @ncall $N fpoly i->a[i]
>> end
>> end
>> end
>>
>> but that fails. I dont get further than:
>>
>> macroexpand(:(@nloops 3 j a begin
>> x = @ncall 3 fpoly i->a[j]
>> end))
>>
>> *quote  # cartesian.jl, line 62:*
>>
>> *for j_3 = indices(a,3) # cartesian.jl, line 63:*
>>
>> *nothing # cartesian.jl, line 64:*
>>
>> *begin  # cartesian.jl, line 62:*
>>
>> *for j_2 = indices(a,2) # cartesian.jl, line 63:*
>>
>> *nothing # cartesian.jl, line 64:*
>>
>> *begin  # cartesian.jl, line 62:*
>>
>> *for j_1 = indices(a,1) # cartesian.jl, line 63:*
>>
>> *nothing # cartesian.jl, line 64:*
>>
>> *begin  # REPL[145], line 2:*
>>
>> *x = fpoly(a[j],a[j],a[j])*
>>
>> *end # cartesian.jl, line 65:*
>>
>> *nothing*
>>
>> *end*
>>
>> *end # cartesian.jl, line 65:*
>>
>> *nothing*
>>
>> *end*
>>
>> *end # cartesian.jl, line 65:*
>>
>> *nothing*
>>
>> *end*
>>
>> *end*
>>
>>
>>
>> *which is a start but how can I get the LHS right the indices of a right?*
>>
>>
>>
>
>
> --
> Erik Schnetter  >
> http://www.perimeterinstitute.ca/personal/eschnetter/
>

Re: [julia-users] help with @generated function call please?

[Erik Schnetter]

A generated function is only useful if you perform a non-trivial
calculation based on the argument types. You don't do that here, so I
wonder whether simply using the Cartesian indexing macros by themselves
would be sufficient.

Note also that you don't need to write `$N` in your code; using `N`
directly has the same effect here.

I'm not saying that generated functions should be avoided at all costs, but
if it isn't necessary here you might as well skip the associated
complications.

-erik

On Fri, Oct 14, 2016 at 11:51 AM, Florian Oswald 
wrote:

> hi all,
>
> I want to evaluate a function at each index of an array. There is a N
> dimensional function, and I want to map it onto an N-dimensional array:
>
> fpoly(x::Array{Real,5}) = x[1] + x[2]^2 + x[3] + x[4]^2 + x[5]
>
> want to do
>
> a = rand(2,2,2,2,2);
> b = similar(a)
>
> for i1 in indices(a,1)
> for i2 in indices(a,2)
> ...
> b[i1,i2,i3,i4,i5] = fpoly(a[i1,i2,i3,i4,i5])
> end
> end...
>
> I tried:
> # actually want to do it inplace
> @generated function set_poly!{T,N}(a::Array{T,N})
> quote
> @nloops $N i a begin
> @nref $N a i = @ncall $N fpoly i->a[i]
> end
> end
> end
>
> but that fails. I dont get further than:
>
> macroexpand(:(@nloops 3 j a begin
> x = @ncall 3 fpoly i->a[j]
> end))
>
> *quote  # cartesian.jl, line 62:*
>
> *for j_3 = indices(a,3) # cartesian.jl, line 63:*
>
> *nothing # cartesian.jl, line 64:*
>
> *begin  # cartesian.jl, line 62:*
>
> *for j_2 = indices(a,2) # cartesian.jl, line 63:*
>
> *nothing # cartesian.jl, line 64:*
>
> *begin  # cartesian.jl, line 62:*
>
> *for j_1 = indices(a,1) # cartesian.jl, line 63:*
>
> *nothing # cartesian.jl, line 64:*
>
> *begin  # REPL[145], line 2:*
>
> *x = fpoly(a[j],a[j],a[j])*
>
> *end # cartesian.jl, line 65:*
>
> *nothing*
>
> *end*
>
> *end # cartesian.jl, line 65:*
>
> *nothing*
>
> *end*
>
> *end # cartesian.jl, line 65:*
>
> *nothing*
>
> *end*
>
> *end*
>
>
>
> *which is a start but how can I get the LHS right the indices of a right?*
>
>
>


-- 
Erik Schnetter 
http://www.perimeterinstitute.ca/personal/eschnetter/

[julia-users] help with @generated function call please?

[Florian Oswald]

hi all, 

I want to evaluate a function at each index of an array. There is a N 
dimensional function, and I want to map it onto an N-dimensional array:

fpoly(x::Array{Real,5}) = x[1] + x[2]^2 + x[3] + x[4]^2 + x[5] 

want to do 

a = rand(2,2,2,2,2);
b = similar(a)

for i1 in indices(a,1)
for i2 in indices(a,2)
...
b[i1,i2,i3,i4,i5] = fpoly(a[i1,i2,i3,i4,i5])
end
end...

I tried:
# actually want to do it inplace
@generated function set_poly!{T,N}(a::Array{T,N})
quote
@nloops $N i a begin
@nref $N a i = @ncall $N fpoly i->a[i]
end
end
end

but that fails. I dont get further than:

macroexpand(:(@nloops 3 j a begin
x = @ncall 3 fpoly i->a[j]
end))

*quote  # cartesian.jl, line 62:*

*for j_3 = indices(a,3) # cartesian.jl, line 63:*

*nothing # cartesian.jl, line 64:*

*begin  # cartesian.jl, line 62:*

*for j_2 = indices(a,2) # cartesian.jl, line 63:*

*nothing # cartesian.jl, line 64:*

*begin  # cartesian.jl, line 62:*

*for j_1 = indices(a,1) # cartesian.jl, line 63:*

*nothing # cartesian.jl, line 64:*

*begin  # REPL[145], line 2:*

*x = fpoly(a[j],a[j],a[j])*

*end # cartesian.jl, line 65:*

*nothing*

*end*

*end # cartesian.jl, line 65:*

*nothing*

*end*

*end # cartesian.jl, line 65:*

*nothing*

*end*

*end*



*which is a start but how can I get the LHS right the indices of a right?*