如果谓词成立，则从列表中删除项目 [英] Removing items from a list if a predicate holds

问题描述

我试图做一个函数，把一个谓词和一个列表作为输入。并从谓词所保存的列表中删除所有元素。到目前为止，我的功能如下：

  removeif :: func-> [a]  - > [a] 
 removeif [] = [] 
 removeif func（h：t）= if func then delete h（h：t）else removeif func t 
  

我对 func-> [a]中的 func 部分感到困惑， - > [a] 因为我不知道该如何判断它是一个谓词。

例如，我想要的是我给终端这个命令

  removeif threefolds [1,2,3,4,5,6,7,8， 9] 
  

threefolds 是一个局部函数I有，它返回一个无限列表的三倍数，我想从输入列表中删除这些倍数。我正在考虑让三倍一个bool函数，但不确定是否有必要。

解决方案 div>

谓词应该是 a - >>类型的函数。布尔。因此， removeif 的签名应该是

  removeif ::（a - > Bool） - > [a]  - > [a] 
  

如果 threefolds code> a - > Bool ，你会没事的： - ）

I am trying to make a function which takes as input a predicate and a list. and removes all elements from the list for which the predicate holds. What I have so far is the following function:

removeif :: func->[a]->[a]
removeif [] = []
removeif func (h:t)= if func then delete h (h:t) else removeif func t

I am confused about the func part of the func->[a]->[a] because I don't know how should I tell that its a predicate.

For example what I want is that I give from the terminal this command

removeif threefolds [1,2,3,4,5,6,7,8,9]

threefolds is a local function I have and it returns an infinite list with multiples of three and I want to remove those multiples from the input list. I am thinking of making threefold a bool function but not sure if that is necessary.

解决方案

A predicate should be a function of type a -> Bool. Therefore, the signature of removeif should be

removeif :: (a -> Bool) -> [a] -> [a]

If threefolds has type a -> Bool, you'll be fine :-)

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