删除语法糖:Haskell 中的列表理解 [英] Removing syntactic sugar: List comprehension in Haskell

问题描述

我可以在这个表达式中取消对列表理解的糖化:

Can I unsugar list comprehension in this expression:

[(i,j) | i <- [1..4], j <- [i+1..4]]

这是输出:

[(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)]

我怎样才能用 map、filter 等写出那段代码?

How can I, with map, filter and so on, write that piece of code?

编辑

这里是另一个:

[(i,j,k) | i <- [1..6], j <- [i+1..6],k <- [j+1..6]]

这是输出:

[(1,2,3),(1,2,4),(1,2,5),(1,2,6),(1,3,4),(1,3,5),(1,3,6),(1,4,5),(1,4,6),(1,5,6),(2,3,4),(2,3,5),(2,3,6),(2,4,5),(2,4,6),(2,5,6),(3,4,5),(3,4,6),(3,5,6),(4,5,6)]

推荐答案

列表推导式(实际上是 Monad 推导式)可以脱糖为 do 表示法.

List comprehensions (in fact, Monad comprehensions) can be desugared into do notation.

do i <- [1..4]
   j <- [i+1..4]
   return (i,j)

可以像往常一样脱糖:

[1..4]   >>= i ->
[i+1..4] >>= j ->
return (i,j)

众所周知，a >>= x ->return b 与 fmap (x -> b) a 相同.所以中间脱糖步骤:

It is well known that a >>= x -> return b is the same as fmap (x -> b) a. So an intermediate desugaring step:

[1..4] >>= i -> 
fmap (j -> (i,j)) [i+1..4]

对于列表，(>>=) = flip concatMap 和 fmap = map

(flip concatMap) [1..4] (i -> map (j -> (i,j) [i+1..4])

flip 只是切换输入的顺序.

flip simply switches the order of the inputs.

concatMap (i -> map (j -> (i,j)) [i+1..4]) [1..4]

这就是你最终得到 Tsuyoshi 答案的方式.

And this is how you wind up with Tsuyoshi's answer.

第二个可以类似地脱糖为:

The second can similarly be desugared into:

concatMap (i ->
  concatMap (j ->
    map       (k ->
      (i,j,k))
    [j+1..6])
  [i+1..6])
[1..6]

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