Haskell groupBy函数：它是如何工作的？

Question

I experienced the following behavior:

ghci :m +Data.List

ghci> groupBy (\x y -> succ x == y) [1..6]
[[1,2], [3,4], [5,6]]
ghci> groupBy (\x y -> succ x /= y) [1..6]
[[1], [2], [3], [4], [5], [6]]

ghci :m +Data.Char   -- just a test to verify that nothing is broken with my ghc

ghci> groupBy (const isAlphaNum) "split this"
["split"," this"]

which surprised me, I thought, based on the example down below, that groupBy splits a list whenever the predicate evaluates to True for two successive elements supplied to a predicate. But in my second example above it splits the list on every element, but the predicate should evaluate to False . I framed my assumption of how it works in Haskell as well, just so everybody understands how I believed it to work:

groupBy :: (a -> a -> Bool) -> [a] -> [[a]]
groupBy p l@(x:y:ys)
  | p x y     = (x:y:(head l)) ++ (tail l)       -- assuming l has a tail, unwilling to
  | otherwise = [x] ++ (y:(head l)) ++ (tail l)  -- to debug this right now, I guess you
groupBy _ [x] = [[x]]                            -- already got the hang of it ;)

Which brought me to the conclusion, that it works somewhat different. So my question is, how does that function actually work?

Answer 1

But in my second example above it splits the list on every element, but the predicate should evaluate to False .

In the second example it also evaluates every two consecutive elements. The function on which it works is const isAlphaNum . So that means that the type is:

const isAlphaNum :: b -> Char -> Bool

It thus calls the function with the beginning of the group and the element, but it takes only the second element into account .

So if we call it with: groupBy (const isAlphaNum) "split this" , it will evaluate:

succs   2nd    const isAlphaNum
-------------------------------
"sp"    'p'    True
"sl"    'l'    True
"si"    'i'    True
"st"    't'    True
"s "    ' '    False
" t"    't'    True
" h"    'h'    True
" i"    'i'    True
" s"    's'    True

Every time const isAlphaNum is True , it will append the character to the current sequence. So in case we evaluate "t " , const isAlphaNum , it will evaluate to False , groupBy will start a new group.

So here we thus construct two groups since there is only one False .

We can also obtain this result if we analyze the function source code :

 groupBy :: (a -> a -> Bool) -> [a] -> [[a]] groupBy _ [] = [] groupBy eq (x:xs) = (x:ys) : groupBy eq zs where (ys,zs) = span (eq x) xs 

So here groupBy will return the empty list if the given list is empty. In case it is not an empty list (x:xs) then we will construct a new sequence. The sequence starts with x , and contains furthermore all the elements of ys . ys is the first element of the 2-tuple constructed by span .

span :: (a -> Bool) -> [a] -> ([a],[a]) constructs a 2-tuple where the first element is the longest possible prefix of the list that satisfies the predicate, the predicate here is eq x so we keep adding elements to the group, as long as eq xy (with y an element) holds.

With the remaining part of the list ys , we construct a new group until the input is completely exhausted.