如何压缩不同长度的列表？

Question

我怎么能zip两个像

["Line1","Line2","Line3"]
["Line4","Line5"]

不丢弃第一个列表中的休息元素？

如果可以的话,我想用空列表压缩额外的元素.

Answer 1

zipWithPadding :: a -> b -> [a] -> [b] -> [(a,b)]
zipWithPadding a b (x:xs) (y:ys) = (x,y) : zipWithPadding a b xs ys
zipWithPadding a _ []     ys     = zip (repeat a) ys
zipWithPadding _ b xs     []     = zip xs (repeat b)

只要有元素,我们就可以简单地压缩它们.一旦我们用完了元素,我们只需用填充元素的无限列表压缩剩余的列表.

在您的情况下,您将使用它作为

zipWithPadding "" "" ["Line1","Line2","Line3"] ["Line4","Line5"]
-- result: [("Line1","Line4"),("Line2","Line5"),("Line3","")]

Answer 2

另一种解决方案是制作一个zip函数,该函数适用于monoid并使用mempty填充缺失值:

import Data.Monoid

mzip :: (Monoid a, Monoid b) => [a] -> [b] -> [(a, b)]
mzip (a:as) (b:bs) = (a, b) : mzip as bs
mzip []     (b:bs) = (mempty, b) : mzip [] bs
mzip (a:as) []     = (a, mempty) : mzip as []
mzip _      _      = []

> mzip ["Line1","Line2","Line3"] ["Line4","Line5"]
[("Line1","Line4"),("Line2","Line5"),("Line3","")]