Hello everybody, Someone can explain me this result please? I have 2 list x = [“0001”, “0002”] y = [1,1] And I loop them like this: for x1 ← x do for y1<- y do IO.inspect x1 IO.inspect y1 end end But the result is like this: “0001” 1 “0001” 1 “0002” 1 “0002” 1 But I want result like this : “0001” 1 “0002” 1 Thanks! :smiley:

I want this result after loop: “0001” 1 “0002” 1 but I receive: “0001” 1 “0001” 1 “0002” 1 “0002” 1 the result is multiply by 2 , but I want also one result

[image] sabiwara: Thanks, it’s resolve my problem

If Enum.zip/2 solves your problem then consider using [head | tail] notation for lists like: defmodule Example do # when done or empty input do nothing returning empty list def sample([], []), do: [] # optional code in case you have more items in one of the list # def sample([], _non_empty_list), do: [] # def sample(_non_empty_list, []), do: [] # loop using [head | tail] notation together with your code def sample([head1 | tail1], [head2 | tail2]) do IO.puts(head1) IO.puts(head2) sample(tail1, tail2) end end x = ["0001", "0002"] y = [1, 1] Example.sample(x, y) This gave you exactly same result, but … Enum.zip is one loop and for is another one - this means that if you write a bit more code then using just one loop your code would be faster.

Indeed, it depends on the actual use case and what we want to return. We might also want to reach for Enum.zip_reduce/4 (for an arbitrary accumulator) or Enum.zip_with/3 (if we want to return a list). Enum.zip_reduce(x, y, :ok, fn x1, y1, _acc -> IO.puts(x1) IO.puts(y1) end)

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Multiple response from loop

Questions & Help Questions

sabiwara June 27, 2022, 10:01am 2

Do you mean something like Enum.zip/2?

for {x1, y1} <- Enum.zip(x, y) do