如何结合过滤条件 [英] How do you combine filter conditions

问题描述

函数的过滤器类采用条件(a-> Bool)并在过滤时应用该条件.

The filter class of functions takes a condition (a -> Bool) and applies it when filtering.

在多种情况下使用过滤器的最佳方法是什么?

What is the best way to use a filter on when you have multiple conditions?

使用了应用功能 liftA2而不是liftM2，因为我出于某种原因不了解liftM2是如何在纯代码中工作的.

Used the applicative function liftA2 instead of liftM2 because I for some reason didn't understand how liftM2 worked within pure code.

推荐答案

可以在Reader monad中使用liftM2组合器，以更实用"的方式进行此操作:

The liftM2 combinator can be used in the Reader monad to do this in a 'more functional' way:

import Control.Monad
import Control.Monad.Reader

-- ....

filter (liftM2 (&&) odd (> 100)) [1..200]

请注意，进口很重要； Control.Monad.Reader提供Monad(e->)实例，以使所有这些工作正常进行.

Note that the imports are important; Control.Monad.Reader provides the Monad (e ->) instance that makes this all work.

之所以如此，是因为读者monad在某些环境e下才有效(e->).因此，布尔谓词是一个0元单子函数，在与其自变量相对应的环境中返回布尔值.然后，我们可以使用liftM2在两个这样的谓词上分布环境.

The reason this works is the reader monad is just (e ->) for some environment e. Thus, a boolean predicate is a 0-ary monadic function returning bool in an environment corresponding to its argument. We can then use liftM2 to distribute the environment over two such predicates.

或者，用更简单的术语来说，当类型确定时，liftM2会像这样工作:

Or, in simpler terms, liftM2 will act a bit like this when the types work out:

liftM2 f g h a = f (g a) (h a)

如果您希望能够轻松地将它们链接起来，并且/或者不希望与liftM2发生冲突，那么还可以定义一个新的组合器:

You can also define a new combinator if you want to be able to chain these easily, and/or don't want to mess with liftM2:

(.&&.) :: (a -> Bool) -> (a -> Bool) -> (a -> Bool)
(.&&.) f g a = (f a) && (g a)
-- or, in points-free style:
(.&&.) = liftM2 (&&)    

filter (odd .&&. (> 5) .&&. (< 20)) [1..100]

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