Hasimbola

Hasimbola

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:

Showing Posts 1 to 5

sabiwara

sabiwara

Elixir Core Team

Do you mean something like Enum.zip/2?

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

Hasimbola OP

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

Hasimbola

Hasimbola OP

Thanks, it’s resolve my problem

Eiji

Eiji

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.

sabiwara

sabiwara

Elixir Core Team

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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