Question

使用Enum.find_index/2，我们可以找到元素的索引。但是，如果多次出现相同的元素，我们该怎么做？

我想有这种行为：

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]

感谢任何想法。

Answer 1

我在库中找不到确切的函数，所以试图实现它。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]

Answer 2

或者，您可以使用0..length(list)范围压缩列表，并使用新项目过滤列表：

line = IO.read(:stdio, :all) 
|> String.split
|> Enum.zip(0..100)
|> Enum.filter(fn({_, x}) -> rem(x, 2) != 0 end)
|> Enum.map(fn({x, _}) -> "#{x}\n" end)

从stdin中过滤给定列表中的奇数元素。

请注意，范围（0..100）中的100必须是列表的长度。我以为我有100个项目的清单。

从Elixir中的列表中查找索引

2 个答案: