Haskell中如何对嵌套元组列表的特定值按行求和？

我有一个嵌套元组列表，每个元组格式为(a, b, c, d)，其中c是a对应的值，d是b对应的值，具体列表如下：

[
[("A","B",0,3),("A","C",0,3),("B","C",0,3)],
[("A","B",0,3),("A","C",0,3),("B","C",1,1)],
[("A","B",0,3),("A","C",0,3),("B","C",3,0)],
[("A","B",0,3),("A","C",1,1),("B","C",0,3)],
[("A","B",0,3),("A","C",1,1),("B","C",1,1)],
[("A","B",0,3),("A","C",1,1),("B","C",3,0)],
[("A","B",0,3),("A","C",3,0),("B","C",0,3)],
[("A","B",0,3),("A","C",3,0),("B","C",1,1)],
[("A","B",0,3),("A","C",3,0),("B","C",3,0)],
[("A","B",1,1),("A","C",0,3),("B","C",0,3)],
[("A","B",1,1),("A","C",0,3),("B","C",1,1)],
[("A","B",1,1),("A","C",0,3),("B","C",3,0)],
[("A","B",1,1),("A","C",1,1),("B","C",0,3)],
[("A","B",1,1),("A","C",1,1),("B","C",1,1)],
[("A","B",1,1),("A","C",1,1),("B","C",3,0)],
[("A","B",1,1),("A","C",3,0),("B","C",0,3)],
[("A","B",1,1),("A","C",3,0),("B","C",1,1)],
[("A","B",1,1),("A","C",3,0),("B","C",3,0)],
[("A","B",3,0),("A","C",0,3),("B","C",0,3)],
[("A","B",3,0),("A","C",0,3),("B","C",1,1)],
[("A","B",3,0),("A","C",0,3),("B","C",3,0)],
[("A","B",3,0),("A","C",1,1),("B","C",0,3)],
[("A","B",3,0),("A","C",1,1),("B","C",1,1)],
[("A","B",3,0),("A","C",1,1),("B","C",3,0)],
[("A","B",3,0),("A","C",3,0),("B","C",0,3)],
[("A","B",3,0),("A","C",3,0),("B","C",1,1)],
[("A","B",3,0),("A","C",3,0),("B","C",3,0)]
]

需要对每一行中的A、B、C对应的值分别求和：

• 示例1：处理列表[("A","B",3,0),("A","C",1,1),("B","C",3,0)]，结果为A=>4，B=>3，C=>1
• 示例2：处理列表[("A","B",3,0),("A","C",3,0),("B","C",3,0)]，结果为A=>6，B=>3，C=>0

实现方案 #

方法1：使用Map实现通用求和

这种方法适用于元素不限于A、B、C的场景，通过Data.Map动态累加每个元素的总和：

import Data.Map (Map)
import qualified Data.Map as Map

-- 处理单个行的求和逻辑
sumRow :: [(String, String, Int, Int)] -> Map String Int
sumRow = foldl updateSum Map.empty
  where
    updateSum acc (elemX, elemY, valX, valY) =
      Map.insertWith (+) elemX valX $ Map.insertWith (+) elemY valY acc

-- 处理整个嵌套列表
sumAllRows :: [[(String, String, Int, Int)]] -> [Map String Int]
sumAllRows = map sumRow

测试示例：

testRow1 = [("A","B",3,0),("A","C",1,1),("B","C",3,0)]
-- sumRow testRow1 → fromList [("A",4),("B",3),("C",1)]

testRow2 = [("A","B",3,0),("A","C",3,0),("B","C",3,0)]
-- sumRow testRow2 → fromList [("A",6),("B",3),("C",0)]

方法2：针对A、B、C的专用求和

如果确定只有A、B、C三个元素，用三元组直接累加会更高效，不需要依赖Map：

-- 单个行求和，返回(A总和, B总和, C总和)
sumRow' :: [(String, String, Int, Int)] -> (Int, Int, Int)
sumRow' = foldl update (0, 0, 0)
  where
    update (aSum, bSum, cSum) (x, y, valX, valY) =
      case (x, y) of
        ("A", "B") -> (aSum + valX, bSum + valY, cSum)
        ("A", "C") -> (aSum + valX, bSum, cSum + valY)
        ("B", "C") -> (aSum, bSum + valX, cSum + valY)
        -- 若有其他元素组合，可在此扩展分支

测试示例：

-- sumRow' testRow1 → (4, 3, 1)
-- sumRow' testRow2 → (6, 3, 0)

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