如何創建按第二個元素排序並分組的對列表的列表？

Question

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

這就是我現在所擁有的，sorteerOpY (2,3)結果：

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

但是，這是我想它會導致什麼：

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

什麼我需要改變，使我的列表被正確創建？

Answer 1

你能想到的列表內涵爲[[(a, b) | b <- [0..y-1]]| a<-[0..x-1]]這樣的：

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 

所以，如果a在(a, b)應該改變更快，你需要更換這些線路：

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 

這直接導致[[(a,b) | a <- [0..x-1]]| b <- [0..y-1]] 。