你如何在Haskell中编写函数'pair'？

Question

对函数需要做这样的事情：

pairs [1, 2, 3, 4] -> [(1, 2), (2, 3), (3, 4)]

Answer 1

pairs [] = []
pairs xs = zip xs (tail xs)

Answer 2

You could go as far as

import Control.Applicative (<*>)
pairs = zip <*> tail

but

pairs xs = zip xs (tail xs)

is probably clearer.

Answer 3

Just for completeness, a more "low-level" version using explicit recursion:

pairs (x:xs@(y:_)) = (x, y) : pairs xs
pairs _          = []

The construct x:xs@(y:_) means "a list with a head x , and a tail xs that has at least one element y ". This is because y doubles as both the second element of the current pair and the first element of the next. Otherwise we'd have to make a special case for lists of length 1.

pairs [_] = []
pairs []  = []
pairs (x:xs) = (x, head xs) : pairs xs

Answer 4

Call to the Aztec god of consecutive numbers:

import Control.Monad (ap)
import Control.Monad.Instances() -- for Monad ((->) a)

foo = zip`ap`tail $ [1,2,3,4]