Question

因此，我想将榆木中的List ItemModel分成List (List ItemModel)。 List.partition仅将列表分成两个列表。

我写了一些代码，将列表分成了我想要的部分（下面的代码）。

但这并不是我想要的解决方案，而且既然这似乎是许多人都会遇到的问题，我想知道这样做的更好例子吗？

partition : List (ItemModel -> Bool) -> List ItemModel -> List (List ItemModel)
partition filters models =
    let
        filterMaybe =
            List.head filters
    in
        case filterMaybe of
            Just filter ->
                let
                    part =
                        Tuple.first (List.partition filter models)
                in
                    part :: (partition (List.drop 1 filters) models)

            Nothing ->
                []

Answer 1

返回的列表直接从filters参数映射，因此实际上仅使用List.map和List.filter即可做到这一点（这是您实际上一直在做的事情，重新丢弃从List.partition返回的余数列表：

multifilter : List (a -> Bool) -> List a -> List (List a)
multifilter filters values =
    filters |> List.map(\filter -> List.filter filter values)

Answer 2

重复的分区需要使用每个步骤的剩余数据作为下一步的输入。这与通过多个过滤器对相同序列进行简单的重复过滤不同。

在Haskell（此问题最初也被标记为）

filter([1, 2, 3, 4, 5, 6], &is_odd)

也就是说，

partitions :: [a -> Bool] -> [a] -> [[a]]

partitions preds xs = go preds xs
  where
  go [] xs = []
  go (p:ps) xs = let { (a,b) = partition p xs } in (a : go ps b)

或

partitions preds xs = foldr g (const []) preds xs
  where
  g p r xs = let { (a,b) = partition p xs } in (a : r b)

测试：

-- mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])
partitions preds xs = snd $ mapAccumL (\xs p -> partition (not . p) xs) xs preds

与重复过滤不同，

> partitions [ (<5), (<10), const True ] [1..15]
[[1,2,3,4],[5,6,7,8,9],[10,11,12,13,14,15]]

分区列表分为两部分

2 个答案: