Le dimanche 29 novembre 2015 à 04:21 -0800, Dan a écrit :
> Since this situation comes up once in a while, it might be nice to
> have a copyfill which does a copy on each cell. In code:
>
> function copyfill!{T}(a::Array{T}, x)
> xT = convert(T, x)
> for i in eachindex(a)
> @inbounds a[i] = copy(xT)
> end
> return a
> end
>
> copyfill(v, dims::Dims) = copyfill!(Array(typeof(v), dims), v)
> copyfill(v, dims::Integer...) = copyfill!(Array(typeof(v), dims...),
> v)
>
> And then we have:
>
> julia> m = copyfill(Int[1],3,3)
> 3x3 Array{Array{Int64,1},2}:
> [1] [1] [1]
> [1] [1] [1]
> [1] [1] [1]
>
> julia> push!(m[1,1],2)
> 2-element Array{Int64,1}:
> 1
> 2
>
> julia> m
> 3x3 Array{Array{Int64,1},2}:
> [1,2] [1] [1]
> [1] [1] [1]
> [1] [1] [1]
>
> As some would expect. Is there a +1 on this?
This has been discussed in depth here:
https://github.com/JuliaLang/julia/pull/8759
One issue is that sometimes copy() is enough, but sometimes you want
deepcopy() instead (when the object itself contains references). Thus,
a more general solution would be to take as first argument a function
returning the value:
https://groups.google.com/d/msg/julia-users/L5fXkHPduBo/UNoq6NTQOnEJ
But as Stefan noted the comprehension form is quite good too. Maybe the
docs should point to it after the warning about shared references.
Regards
> On Sunday, November 29, 2015 at 1:36:03 PM UTC+2, Kristoffer Carlsson
> wrote:
> > I guess the simplest would be:
> >
> > [Int[] for i = 1:3, j=1:3, k=1:3]
> >
> >
> > And to repeat what Milan already said, you don't want fill! because
> > then all your arrays point to the same memory location.
> >
> > On Sunday, November 29, 2015 at 11:51:28 AM UTC+1, Aleksandr
> > Mikheev wrote:
> > > Hi all. Once again I have some questions about Julia.
> > >
> > > I know that there is a possibility to create a list of arrays.
> > > For exmple:
> > >
> > > s = fill(Array(Int64,1),4)
> > >
> > > And then I can do something like this:
> > >
> > > s[1] = [1; 2]
> > > s[1] = [s[1]; 5]
> > >
> > > By parity of reasoning I did this:
> > >
> > > s = fill(Array(Int64,1),4,4,4)
> > >
> > > And it worked fine. But in both cases I had initial elements in s
> > > (like when I construct arrays with Array{Int64}(m,n)):
> > >
> > > julia> s = fill(Array(Int64,1),3,3,3)
> > > 3x3x3 Array{Array{Int64,1},3}:
> > > [:, :, 1] =
> > > [2221816832] [2221816832] [2221816832]
> > > [2221816832] [2221816832] [2221816832]
> > > [2221816832] [2221816832] [2221816832]
> > >
> > >
> > > [:, :, 2] =
> > > [2221816832] [2221816832] [2221816832]
> > > [2221816832] [2221816832] [2221816832]
> > > [2221816832] [2221816832] [2221816832]
> > >
> > >
> > > [:, :, 3] =
> > > [2221816832] [2221816832] [2221816832]
> > > [2221816832] [2221816832] [2221816832]
> > > [2221816832] [2221816832] [2221816832]
> > >
> > >
> > > Is there something I could do to prevent this? I know I could
> > > easily fix it by:
> > >
> > >
> > > for i = 1:3
> > > for j = 1:3
> > > for k = 1:3
> > > s[i,j,k] = []
> > > end
> > > end
> > > end
> > >
> > > But I guess this is a weird solution.
> > >
> > > Thank you in advance!
> > >
> > >
