How to built an array with some keys of a dictionary ?

[Fred]

Hi,

I have many problems to build arrays with the keys of a dictionary except for the simple situation :

a = [uppercase(key) for key in keys(dict)]    # works fine

If I try to select the keys with more complex criteria like :

dict = Dict("a" => 1, "b" => 2, "c" => 3, "d" => 4, "e" => 5)

a = Array{String}

for k in keys(dict)
  if dict[k] < 2
    continue
  end
    push!(a, k) # building array from keys produces errors
end


ERROR: MethodError: `push!` has no method matching push!(::Type{Array{AbstractString,N}}, ::ASCIIString)

Thank you for your help !

[Steven G. Johnson]

On Monday, November 7, 2016 at 9:02:44 AM UTC-5, Fred wrote:

Hi,

I have many problems to build arrays with the keys of a dictionary except for the simple situation :

a = [uppercase(key) for key in keys(dict)]    # works fine

If I try to select the keys with more complex criteria like :

dict = Dict("a" => 1, "b" => 2, "c" => 3, "d" => 4, "e" => 5)

a = Array{String}

This is a mistake: you just assigned "a" to an array type, not an array instance.  It should be a = Array{String}(). 

In Julia 0.5, you can also just do:

[k for k in keys(dict) if dict[k] < 2]

[Fred]

Hi Steven,


I also tried  a = Array{String}() unfortunately it produces errors as well.

julia> a = Array{String}()
WARNING: Base.String is deprecated, use AbstractString instead.
  likely near no file:0
0-dimensional Array{AbstractString,0}:
#undef


julia> for k in keys(dict)
         if dict[k] < 2
           continue
         end
           push!(a, k)
       end
ERROR: MethodError: `push!` has no method matching push!(::Array{AbstractString,0}, ::ASCIIString)
Closest candidates are:
  push!(::Any, ::Any, ::Any)
  push!(::Any, ::Any, ::Any, ::Any...)
  push!(::Array{Any,1}, ::ANY)
  ...
 [inlined code] from none:5
 in anonymous at no file:0

[Mauro]

On Mon, 2016-11-07 at 15:27, Fred <fred.so...@gmail.com> wrote:
> Hi Steven,
>
>
> I also tried a = Array{String}() unfortunately it produces errors as well.

Array{String}(0)

works. The other creates a zero-dimensional vector (just like a scalar)
to which you cannot push.

[Fred]

Thank you very much Mauro, that was the mistake ! :)

[Steven G. Johnson]

On Monday, November 7, 2016 at 9:45:37 AM UTC-5, Mauro wrote:

On Mon, 2016-11-07 at 15:27, Fred <fred.so...@gmail.com> wrote:
> Hi Steven,
>
>
> I also tried a = Array{String}() unfortunately it produces errors as well.

Array{String}(0)

Whoops, sorry about the typo. 

[Dan]

a = Vector{String}

is also a good option.

[Dan]

I meant,

a = Vector{String}()

Dan

[Fred]

Thank you Dan, this solution works to ! I don't know which one is better :)

[Fabian Gans]

I usually use

a = String[]





Fabian

[David P. Sanders]

An alternative syntax is

[k for (k,v) in dict if v < 2]

which iterates over (key, value) pairs in the dictionary.