haskell从列表列表中删除给定值的所有匹配项

我希望从列表列表中删除给定值的所有引用。例如 输入：

输出：

 ["bc", "bc", ""]

到目前为止：

remove :: Eq a => a -> [[a ]] -> [[a ]]
remove y xs = filter(\x -> x/= y) xs

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我有个错误，先谢谢你

您需要映射外部列表

remove y xs = map (filter(\x -> x/= y)) xs

你实际上不需要兰姆达，尼斯：

remove y xs = map (filter(/=y)) xs

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您还可以在两侧放置

xs

，甚至可以放置

map。过滤器。（/=）

remove y xs = map (filter(/=y)) xs