Elixir递归处理刮刮卡时出现KeyError：nil中无:owned_numbers键

问题场景 #

定义了如下Card结构体：

defmodule Card do
  defstruct id: nil, result_numbers: [], owned_numbers: []
end

通过CardEvaluator.evaluate_card/1计算卡片中owned_numbers与result_numbers的匹配数量，接着根据结果构建范围(card.id + 1)..(card.id + value)，递归调用process_card_reward/2处理该范围内的卡片，最终统计总卡片数。

实现代码如下：

def process_scratchcards(card_strings) do
  cards =
    card_strings
    |> Enum.map(&CardParser.parse_card/1)
    |> IO.inspect()

  total_reward = cards
    |> Enum.map(fn card -> process_card_reward(card, cards) end)
    |> Enum.sum()

  IO.puts(total_reward)
  total_reward
end

def process_card_reward(card, all_cards) do
  value =
    card
    |> CardEvaluator.evaluate_card()

  victory_range = (card.id + 1)..(card.id + value)
  IO.inspect(victory_range)

  num_children =
    Enum.reduce(victory_range, 0, fn n, acc ->
      acc + process_card_reward(Enum.at(all_cards,n), all_cards)
    end)

  IO.puts(num_children)

  num_children + 1
end

运行时出现以下异常：

test Processes test data correctly (ScratchcardsTest)
test/scratchcards_test.exs:5
** (KeyError) key :owned_numbers not found in: nil

If you are using the dot syntax, such as map.field, make sure the left-hand side of the dot is a map
code: |> Scratchcards.process_scratchcards()
stacktrace:
(scratchcards 0.1.0) lib/card_evaluator.ex:3: CardEvaluator.evaluate_card/1
(scratchcards 0.1.0) lib/scratchcards.ex:19: Scratchcards.process_card_reward/2
(scratchcards 0.1.0) lib/scratchcards.ex:26: anonymous fn/3 in Scratchcards.process_card_reward/2
(elixir 1.15.7) lib/enum.ex:4379: Enum.reduce/3
(scratchcards 0.1.0) lib/scratchcards.ex:25: Scratchcards.process_card_reward/2
...

错误原因 #

异常核心是Enum.at(all_cards, n)返回了nil，传入CardEvaluator.evaluate_card/1后，函数尝试访问nil的:owned_numbers字段导致KeyError。

触发原因有两点：

• 索引不匹配：卡片id通常从1开始，但Elixir列表是0-based索引，直接用n（卡片id）作为索引取元素，会导致索引越界或取到错误元素。
• 范围超出卡片总数：当card.id + value超过现有卡片的最大id时，Enum.at会返回nil。

修复方案 #

方案1：使用Map按id查找（推荐） #

将卡片列表转换为以id为键的Map，直接通过id安全查找卡片，同时过滤不存在的id：

修改后的代码：

def process_scratchcards(card_strings) do
  cards =
    card_strings
    |> Enum.map(&CardParser.parse_card/1)
    |> IO.inspect()

  # 将卡片列表转为id为键的Map
  cards_map = Map.new(cards, &{&1.id, &1})

  total_reward = cards
    |> Enum.map(fn card -> process_card_reward(card, cards_map) end)
    |> Enum.sum()

  IO.puts(total_reward)
  total_reward
end

def process_card_reward(card, cards_map) do
  value = CardEvaluator.evaluate_card(card)

  victory_range = (card.id + 1)..(card.id + value)
  IO.inspect(victory_range)

  num_children =
    victory_range
    # 过滤出存在的卡片id，避免传入nil
    |> Enum.filter(&Map.has_key?(cards_map, &1))
    |> Enum.reduce(0, fn id, acc ->
      acc + process_card_reward(cards_map[id], cards_map)
    end)

  IO.puts(num_children)

  num_children + 1
end

方案2：修正列表索引（仅适用于id连续从1开始的场景） #

若确定卡片id连续且从1开始，可将索引转换为n-1，同时添加越界检查：

def process_card_reward(card, all_cards) do
  value = CardEvaluator.evaluate_card(card)

  victory_range = (card.id + 1)..(card.id + value)
  IO.inspect(victory_range)

  num_children =
    Enum.reduce(victory_range, 0, fn n, acc ->
      # 转换为0-based索引，且检查索引是否在列表范围内
      index = n - 1
      if index >= 0 and index < length(all_cards) do
        acc + process_card_reward(Enum.at(all_cards, index), all_cards)
      else
        acc
      end
    end)

  IO.puts(num_children)

  num_children + 1
end

方案1优势在于查找效率更高（O(1)），且无需依赖id的连续性，更健壮。

内容的提问来源于stack exchange，提问作者ScottishTapWater