如何创建按第二个元素排序和分组的对列表？

Question

sorteerOpY :: (Int, Int) -> [[(Int, Int)]]
sorteerOpY (x,y) = [[(a,b)]|b<-[0..y-1],a<-[0..x-1]]

这就是我现在所拥有的，sorteerOpY (2,3)导致：

[[(0,0)],[(1,0)],[(0,1)],[(1,1)],[(0,2)],[(1,2)]]

但这就是我想要的结果：

[[(0,0),(1,0)],[(0,1),(1,1)],[(0,2),(1,2)]]

我需要更改哪些内容才能正确创建列表？

Answer 1

您可以将列表推导视为[[(a, b) | b <- [0..y-1]]| a<-[0..x-1]]，如下所示：

for every b <- [0..y-1]       
   xs <- for every a <- [0..x-1]
      add (a, b) to the resulting list
   add xs to the resulting list

因此，如果a中的(a, b)变化更快，则需要交换这些行：

for every a <- [0..x-1]
   xs <- for every b <- [0..y-1]
      add (a, b) to the resulting list
   add xs to the resulting list

直接导致[[(a,b) | a <- [0..x-1]]| b <- [0..y-1]]。