7

有了Enum.find_index/2，我们可以找到一个元素的索引。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]

感谢您的任何想法。

4

2 回答 2

3

我在库中找不到确切的函数，所以尝试实现它。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]
于 2013-09-01T13:28:32.493 回答
1

或者，您可以使用以下范围压缩列表0..length(list)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)

从标准输入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。我假设我有一个包含 100 个项目的列表。

于 2017-11-05T21:51:59.497 回答