owaisqayum
How can we create a nested keyword list
How can we create a nested keyword list from the following list. The list length can change with time.
list
[:a, :b, :c, :d, :e, :f]
how can we convert it to
[a: [b: [c: [d: [e: [f: [ true ] ]]]]]]
Is it possible with elixir as I tried to use recursion but it’s not working as expected.
Thanks
Marked As Solved
kokolegorille
iex> l |> Enum.reverse() |> Enum.reduce([true], fn x, acc -> Keyword.new([{x, acc}]) end)
[a: [b: [c: [d: [e: [f: [true]]]]]]]
# or the short version...
iex> l |> Enum.reverse() |> Enum.reduce([true], &Keyword.new([{&1, &2}]))
Also Liked
gregvaughn
Perhaps a bit 
ish, but I’ve never actually used foldr before, so I put this together 
iex(40)> :lists.foldr(&[{&1, &2}], [true], [:a, :b, :c, :d, :e, :f])
[a: [b: [c: [d: [e: [f: [true]]]]]]]
eksperimental
You have List.foldr/3 if you want to make it more Elixirish.
kokolegorille
When You have an Enumerable, and want to get one value… the usual suspect is Enum.reduce.
iex> l = [:a, :b, :c, :d, :e, :f]
iex> Enum.reduce l, [], fn x, acc -> Keyword.put([], x, acc) end
# or
iex> Enum.reduce l, [], fn x, acc -> Keyword.new([{x, acc}]) end
[f: [e: [d: [c: [b: [a: []]]]]]]
This code does not solve your problem, but shows how to nest… I would reverse the list first, and add the final true to get it work.
owaisqayum
Enum.reduce is extremely powerful and it has reduced the code to just one line.
kokolegorille
Yes, keyword new is not required 
BTW parens are not required too…
iex> [:a, :b, :c, :d, :e, :f] |> Enum.reverse() |> Enum.reduce([true], &[{&1, &2}])
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