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

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}]))

[image] owaisqayum: Is it possible with elixir Yes it is possible. It is even the main purpose of functions in FP … to take something as input, and transform it to any output. [image] owaisqayum: I tried to use recursion but it’s not working as expected. Please show what You have tri…

actually, i have tried this defp identifier_handling([], key_value), do: key_value defp identifier_handling([{head, []}], key_value), do: key_value defp identifier_handling(list, key_value) do [key] = Keyword.keys(list) |> IO.inspect() [head | tail] = Keyword.values(lis…

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: []]]]]]] Thi…

[image] kokolegorille: When You have an Enumerable, and want to get one value… the usual suspect is Enum.reduce. That makes sense now, so whenever one value is related to the other one, I should use Enum.reduce. How would to put the true in the final list, do I have to loop through the final…

Enum.reduce is extremely powerful and it has reduced the code to just one line.

[image] owaisqayum: [a: [b: [c: [d: [e: [f: [ true ] ]]]]]] Here’s my take: iex> [:a, :b, :c, :d, :e, :f] |> Enum.reverse() |> Enum.reduce([true], &([{&1, &2}])) [a: [b: [c: [d: [e: [f: [true]]]]]]]

Yes, keyword new is not required :slight_smile: BTW parens are not required too… iex> [:a, :b, :c, :d, :e, :f] |> Enum.reverse() |> Enum.reduce([true], &[{&1, &2}])

Perhaps a bit :golf: ish, but I’ve never actually used foldr before, so I put this together :grin: iex(40)> :lists.foldr(&[{&1, &2}], [true], [:a, :b, :c, :d, :e, :f]) [a: [b: [c: [d: [e: [f: [true]]]]]]]

If you want to use recursion, it should be as simple as: defmodule Solution do def run([]), do: [true] def run([head | tail]), do: [{head, run(tail)}] end Solution.run([:a, :b, :c, :d, :e, :f]) # [a: [b: [c: [d: [e: [f: [true]]]]]]]

How can we create a nested keyword list

Questions & Help Questions

eksperimental May 25, 2021, 7:01pm 11

You have List.foldr/3 if you want to make it more Elixirish.