递归循环数 [英] Number of loops in recursion

问题描述

我想计算列表中正整数/元素的数量.这将返回具有正值的元素，我如何计算元素?想要构造类似count(array(...))的东西.

I would like to count the number of positive integers/elements in the list. This returns the elements with positive values, how can I count the elements? would like to construct something like count(array(...)).

我想查看带有 i ++ 和 foldl 的版本.那将非常有帮助.

I would like to see a version with i++ and foldl. That would be very helpful.

countPositivesRec :: [Int] -> [Int]
countPositivesRec [] = []
countPositivesRec (x:xs) | x >= 0 = x : tl
                         | otherwise = tl
                           where tl = countPositivesRec xs

推荐答案

您拥有

filtering p cons x r = if | p x -> cons x r | otherwise -> r

countPositives = length 
                   . filter (> 0)
               = foldr (\x r -> r + 1) 0                              -- r++
                   . foldr (filtering (> 0) (:)            ) []
               =     foldr (filtering (> 0) (\x r -> r + 1))  0

(由于通过组合其减速器变压器折叠 fuse ，a-la "fold用减速器操作替换了缺点，因此，如果要替换缺点，为什么要首先创建缺点无论如何" )和

(since folds fuse by composing their reducer transformers, a-la "fold replaces the cons with a reducer operation, so why create the cons in the first place if it gonna be replaced anyway"), and

filtering (> 0) (\x r -> r + 1) x r 
                     = if | (> 0) x -> (\x r -> r + 1) x r | otherwise -> r
                     = if | x > 0 -> r + 1 | otherwise -> r

，因此是您想要的具有 fold 和 increment 的版本，

and thus, a version with fold and increment that you wanted,

countPositives = foldr (\x r -> if | x > 0 -> r + 1 | otherwise -> r) 0  -- r++

您可以从这里拿走它.

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