Elixir:没有语法糖的情况下，多生成器列表理解是什么样的? [英] Elixir: What does a multiple-generator list comprehension look like without the syntax sugar?

问题描述

我试图了解Elixir中的列表理解.

I'm trying to understand list comprehensions in Elixir.

我正在查看的示例是从

The example I'm looking at is producing the permutations of a string from this answer.

def shuffle([], _), do: [[]]
def shuffle(_,  0), do: [[]]
def shuffle(list, i) do
  for x <- list, y <- shuffle(list, i-1), do: [x|y]
end

在没有理解的情况下重写时，这种双生成器的理解如何?我做了一个尝试自己实现该算法，但是我的实现是追加到列表中，而不是像理解中那样在前面.我想编写没有理解但行为相同的算法.

How does this double-generator comprehension look when re-written without the comprehension? I made an attempt to implement the algorithm myself, but my implementation is appending to the list, rather than prepending as in the comprehension. I want to write the algorithm without the comprehension but with identical behaviour.

推荐答案

不带过滤器的理解可以转换为Enum.flat_map和Enum.map的序列.具体来说，除了最后一个以外的所有都将变为flat_map，最后一个将变为map.这是您的代码的翻译:

A comprehension without filters can be converted into a sequence of Enum.flat_map and Enum.map. Specifically, all but the last one will become flat_map and the last one will become map. Here's a translation of your code:

list
|> Enum.flat_map(fn x ->
  shuffle(list, i - 1)
  |> Enum.map(fn y ->
    [x | y]
  end)
end)

我用A.shuffle([1, 2, 3, 4, 5], 2)进行了测试，输出看起来与该问题中的原始代码相同.

I tested with A.shuffle([1, 2, 3, 4, 5], 2) and the output looks identical to the original code in that question.

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