如何在Haskell中压缩多个列表？ [英] How to zip multiple lists in Haskell?

问题描述

在python zip 函数中可以接受任意数量的列表并将它们拉到一起。

In python zip function accepts arbitrary number of lists and zips them together.

>>> l1 = [1,2,3]
>>> l2 = [5,6,7]
>>> l3 = [7,4,8]
>>> zip(l1,l2,l3)
[(1, 5, 7), (2, 6, 4), (3, 7, 8)]
>>> 

如何将 zip haskell？

How can I zip together multiple lists in haskell?

推荐答案

可以使用应用符号。由于新类型的打包/解包使用起来有点不愉快，但如果你正在做一些无法用 zipWithn 完成的小n，你很可能已经处于一个足够高的抽象层次中，无论如何，这些标记性的痛苦都不存在。

A generalization of zip can be achieved using Applicative Notation. It's a bit unpleasant to use because of the newtype wrapping/unwrapping, but if you are doing something that can't be done with a zipWithn for reasonably small n, you are probably already at a high enough level of abstraction where the notational pains are absent anyway.

类型是 ZipList a ，并将其应用实例拉到一起列表。例如：

The type is ZipList a, and its applicative instance zips together lists. For example:

(+) <$> ZipList [1,2] <*> ZipList [3,4] == ZipList [4,6]

这概括为任意数组的函数并使用部分申请类型：

This generalizes to functions of arbitrary arity and type using partial application:

(+) <$> ZipList [1,2]  :: ZipList (Int -> Int)

部分适用于此？

如果您不喜欢到处添加ZipList和getZipList，您可以轻松地重新创建符号：

If you don't like adding ZipList and getZipList everywhere, you could recreate the notation easily enough:

(<$>) :: (a -> b) -> [a] -> [b]
(<$>) = map

(<*>) :: [a -> b] -> [a] -> [b]
(<*>) = zipWith ($)

符号 zipWith fabcd ... 为：

Then the notation for zipWith f a b c d ... is:

f <$> a <*> b <*> c <*> d <*> ...

应用符号是一种功能非常强大且通用的技术，其范围比广义拉链。有关应用符号的更多信息，请参阅 Typeclassopedia 。

Applicative notation is a very powerful and general technique that has much wider scope than just generalized zipping. See the Typeclassopedia for more on Applicative notation.

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