Re: [julia-users] Re: Use reference of array comprehension internal variables?
OMG that's awesome, we need more docs about this feature, thank you very
much Stefan!
El miércoles, 10 de agosto de 2016, 12:02:34 (UTC-5), Stefan Karpinski
escribió:
>
> julia> [(x, y, z) for x = 1:n for y = x:n for z = y:n if x^2 + y^2 == z^2]
> 6-element Array{Tuple{Int64,Int64,Int64},1}:
> (3,4,5)
> (5,12,13)
> (6,8,10)
> (8,15,17)
> (9,12,15)
> (12,16,20)
>
> On Wed, Aug 10, 2016 at 12:58 PM, Ismael Venegas Castelló <
> ismael...@gmail.com > wrote:
>
>> In order to be a little more specific I wanted to add, that it seems
>> weird that I can use the variables for the if clause, but not for creating
>> the other ranges, it's just that I don't know how to express myself
>> correctly, I hope you can understand me.
>>
>>
>> El miércoles, 10 de agosto de 2016, 11:56:00 (UTC-5), Ismael Venegas
>> Castelló escribió:
>>>
>>> Is there a way to make reference of the internal variables of an array
>>> comprehension? I'm trying to improve this Rosetta Code task:
>>>
>>>
>>>- https://rosettacode.org/wiki/List_comprehensions#Julia
>>>
>>>
>>> const n = 20
>>> sort(filter(x -> x[1] < x[2] && x[1]^2 + x[2]^2 == x[3]^2, [(a, b, c)
>>> for a=1:n, b=1:n, c=1:n]))
>>>
>>>
>>> In Python it's:
>>>
>>> In [2]: n = 20
>>>
>>> In [3]: [(x,y,z) for x in xrange(1,n+1) for y in xrange(x,n+1) for z in
>>> xrange(y,n+1) if x**2 + y**2 == z**2]
>>> Out[3]: [(3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (9, 12, 15), (
>>> 12, 16, 20)]
>>>
>>>
>>> I'll update the task with:
>>>
>>> julia> [(x, y, z) for x = 1:n, y = 1:n, z = 1:n if x < y && x^2 + y^2 ==
>>> z^2] |> sort
>>> 6-element Array{Tuple{Int64,Int64,Int64},1}:
>>> (3,4,5)
>>> (5,12,13)
>>> (6,8,10)
>>> (8,15,17)
>>> (9,12,15)
>>> (12,16,20)
>>>
>>> But I tried this and it doesn't work, I wonder why?
>>>
>>> julia> [(x, y, z) for x = 1:n, y = x:n, z = y:n if x^2 + y^2 == z^2]
>>> ERROR: UndefVarError: x not defined
>>>
>>> Is there a way to do this? Thanks in advance!
>>>
>>
>
Re: [julia-users] Re: Use reference of array comprehension internal variables?
julia> [(x, y, z) for x = 1:n for y = x:n for z = y:n if x^2 + y^2 == z^2]
6-element Array{Tuple{Int64,Int64,Int64},1}:
(3,4,5)
(5,12,13)
(6,8,10)
(8,15,17)
(9,12,15)
(12,16,20)
On Wed, Aug 10, 2016 at 12:58 PM, Ismael Venegas Castelló <
ismael.vc1...@gmail.com> wrote:
> In order to be a little more specific I wanted to add, that it seems weird
> that I can use the variables for the if clause, but not for creating the
> other ranges, it's just that I don't know how to express myself correctly,
> I hope you can understand me.
>
>
> El miércoles, 10 de agosto de 2016, 11:56:00 (UTC-5), Ismael Venegas
> Castelló escribió:
>>
>> Is there a way to make reference of the internal variables of an array
>> comprehension? I'm trying to improve this Rosetta Code task:
>>
>>
>>- https://rosettacode.org/wiki/List_comprehensions#Julia
>>
>>
>> const n = 20
>> sort(filter(x -> x[1] < x[2] && x[1]^2 + x[2]^2 == x[3]^2, [(a, b, c) for
>> a=1:n, b=1:n, c=1:n]))
>>
>>
>> In Python it's:
>>
>> In [2]: n = 20
>>
>> In [3]: [(x,y,z) for x in xrange(1,n+1) for y in xrange(x,n+1) for z in
>> xrange(y,n+1) if x**2 + y**2 == z**2]
>> Out[3]: [(3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (9, 12, 15), (
>> 12, 16, 20)]
>>
>>
>> I'll update the task with:
>>
>> julia> [(x, y, z) for x = 1:n, y = 1:n, z = 1:n if x < y && x^2 + y^2 ==
>> z^2] |> sort
>> 6-element Array{Tuple{Int64,Int64,Int64},1}:
>> (3,4,5)
>> (5,12,13)
>> (6,8,10)
>> (8,15,17)
>> (9,12,15)
>> (12,16,20)
>>
>> But I tried this and it doesn't work, I wonder why?
>>
>> julia> [(x, y, z) for x = 1:n, y = x:n, z = y:n if x^2 + y^2 == z^2]
>> ERROR: UndefVarError: x not defined
>>
>> Is there a way to do this? Thanks in advance!
>>
>
[julia-users] Re: Use reference of array comprehension internal variables?
In order to be a little more specific I wanted to add, that it seems weird
that I can use the variables for the if clause, but not for creating the
other ranges, it's just that I don't know how to express myself correctly,
I hope you can understand me.
El miércoles, 10 de agosto de 2016, 11:56:00 (UTC-5), Ismael Venegas
Castelló escribió:
>
> Is there a way to make reference of the internal variables of an array
> comprehension? I'm trying to improve this Rosetta Code task:
>
>
>- https://rosettacode.org/wiki/List_comprehensions#Julia
>
>
> const n = 20
> sort(filter(x -> x[1] < x[2] && x[1]^2 + x[2]^2 == x[3]^2, [(a, b, c) for
> a=1:n, b=1:n, c=1:n]))
>
>
> In Python it's:
>
> In [2]: n = 20
>
> In [3]: [(x,y,z) for x in xrange(1,n+1) for y in xrange(x,n+1) for z in
> xrange(y,n+1) if x**2 + y**2 == z**2]
> Out[3]: [(3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (9, 12, 15), (12
> , 16, 20)]
>
>
> I'll update the task with:
>
> julia> [(x, y, z) for x = 1:n, y = 1:n, z = 1:n if x < y && x^2 + y^2 == z
> ^2] |> sort
> 6-element Array{Tuple{Int64,Int64,Int64},1}:
> (3,4,5)
> (5,12,13)
> (6,8,10)
> (8,15,17)
> (9,12,15)
> (12,16,20)
>
> But I tried this and it doesn't work, I wonder why?
>
> julia> [(x, y, z) for x = 1:n, y = x:n, z = y:n if x^2 + y^2 == z^2]
> ERROR: UndefVarError: x not defined
>
> Is there a way to do this? Thanks in advance!
>
