如何在Haskell中减少多个eta [英] How to make multiple eta reductions in Haskell

问题描述

我有一个任务可以从 [[a]] 矩阵中获得一列。

I have a task to get a column from a [[a]] matrix.

解决方案将是

A simple solution would be

colFields :: Int -> [[a]] -> [a]
colFields n c = map (!! n) c

一个抽象层次将是

and when reduced by one level of abstraction it would be

colFields n = map (!! n)

我觉得我可以轻易摆脱 n ，但我无法做到这一点。

I sense that I could get rid of n easily, but I can't do it.

推荐答案

$ b $

What you're looking for is

colFields = map . flip (!!)

然而，这不是很清楚，我会离开<那里有code> n 参数。用 n 作为显式参数，我立即明白了函数的作用。没有它，为了理解定义，我必须考虑一下，即使是像这样的简单情况。

However, this is not very clear to read, I'd leave the n parameter in there. With the n as an explicit parameter, I understand immediately what the function does. Without it, I have to think for a minute in order to understand the definition, even for a simple case like this.

我非常简单地通过使用<一个href =http://hackage.haskell.org/package/pointfree> pointfree 工具，虽然有手动派生的方法。

I obtained this answer very simply by using the pointfree tool, although there are methods for deriving this by hand.

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