Question

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

我对haskell相当新，所以这可能只是一些愚蠢的新秀错误

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，而在于我输入了什么

Answer 1

您可能正在寻找类似

的内容

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，则会导致程序崩溃。因此，如果xs为空，或x内的任何列表xs为空，则可能会崩溃。

（顺便说一句，如果xs为空，y将永远不会被评估，因此不会出现崩溃。但是，这并不容易发现，使代码更难阅读。）

假设我们要丢弃空x，则替代方案是。

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

根据列表列表中列表的最后一个元素和列表列表的最后列表过滤列表列表

1 个答案: