根据列表列表中列表的最后一个元素和列表列表的最后一个列表过滤列表列表

Question

我正在尝试为我的输入列表创建一个过滤器，以删除最后一个元素不是列表的最后一个列表的元素的列表

我对 Haskell 还很陌生，所以这可能只是一些愚蠢的菜鸟错误

FilterBoi xs = filter (\x -> elem (x) y) xs
                        Where x = last (x:xs)
                                     y = last xs

返回错误

Occurs check: cannot construct the infinite type: 
a ~ t0 a
Expected type: [t0 a]
   Actual type: [a]
In the first argument of `last' , namely `xs'
In the expression: last xs
In an equation for `y' : y = last xs
Relevant bindings include
    y :: t0 a (bound at filter.hs:3:22)
    xs :: [a] (bound at filter.hs:1:11)
    FilterBoi :: [a] -> [a] (bound at filter.hs:1:1)
Failed, modules loaded: none.

请注意，我是在手机上输入的，所以问题不在于我输入的方式，而在于我输入的内容

Answer 1

您可能正在寻找类似的东西

filterBoi :: [[a]] -> [[a]]
filterBoi xs = filter (\x -> elem (last x) y) xs
   where y = last xs

请注意，last 是危险的，如果应用于空列表会导致程序崩溃。因此，如果xs 为空，或者xs 中的任何列表x 为空，则可能会崩溃。

（顺便说一句，如果xs 为空，则永远不会评估y，因此不会发生崩溃。但是，这并非易事，使代码更难阅读。）

假设我们要丢弃空的x，另一种选择是。

filterBoi :: [[a]] -> [[a]]
filterBoi [] = []   -- not needed, but clarifies the intent
filterBoi xs = filter (\x -> not (empty x) && elem (last x) y) xs
   where y = last xs