从Elixir中的列表中查找索引

Question

有了Enum.find_index/2,我们可以找到元素的索引.但是,如果同一个元素多次出现,我们该怎么办？

我想有这样的行为:

iex> find_indexes(["a", "b", "c", "b", "b"], fn(x) -> x == "a" end)
[0]

iex> find_indexes(["a", "b", "c", "b", "b"], fn(x) -> x == "c" end)
[2]

iex> find_indexes(["a", "b", "c", "b", "b"], fn(x) -> x == "b" end)
[1, 3, 4]

谢谢你的任何想法.

Answer 1

我在库中找不到确切的函数，因此尝试实现它。希望它可以帮助一些人。

defmodule Sample1 do
  # combining Enum functions
  def find_indexes(collection, function) do
    Enum.filter_map(Enum.with_index(collection), fn({x, _y}) -> function.(x) end, elem(&1, 1))
  end
end

defmodule Sample2 do
  # implementing as similar way as Enum.find_index
  def find_indexes(collection, function) do
    do_find_indexes(collection, function, 0, [])
  end

  def do_find_indexes([], _function, _counter, acc) do
    Enum.reverse(acc)
  end

  def do_find_indexes([h|t], function, counter, acc) do
    if function.(h) do
      do_find_indexes(t, function, counter + 1, [counter|acc])
    else
      do_find_indexes(t, function, counter + 1, acc)
    end
  end
end

IO.puts "Sample1"
IO.inspect Sample1.find_indexes(["a", "b", "c", "b", "b"], fn(x) -> x == "a" end)
IO.inspect Sample1.find_indexes(["a", "b", "c", "b", "b"], fn(x) -> x == "c" end)
IO.inspect Sample1.find_indexes(["a", "b", "c", "b", "b"], fn(x) -> x == "b" end)

IO.puts "Sample2"
IO.inspect Sample2.find_indexes(["a", "b", "c", "b", "b"], fn(x) -> x == "a" end)
IO.inspect Sample2.find_indexes(["a", "b", "c", "b", "b"], fn(x) -> x == "c" end)
IO.inspect Sample2.find_indexes(["a", "b", "c", "b", "b"], fn(x) -> x == "b" end)

执行如下，

% elixir find.ex
Sample1
[0]
[2]
[1, 3, 4]
Sample2
[0]
[2]
[1, 3, 4]