Haskell 使用数据构造函数过滤嵌套列表

Question

Let's say I have the following data type

data Number = Positive Integer | Negative Integer
     deriving (Eq, Show)

I have a function definition of (NOTE THAT I CANNOT CHANGE THIS DEFINITION, OR WORK AROUND IT IN ANY WAY. I have to work with the nested list and modify it in some way)

removePos :: [[Number]] -> [[Number]]

So we have a nested list of Number. An example would be

[[Positive 1, Negative 1], [Positive 2, Negative 2, Positive 1], [Positive 1]]

How can I write removePos so that it removes all lists in the nested list that contain Positive x, where x is A SPECIFIC Integer? The function is essentially looking at the first element in the first list, if it is Positive then remove all lists that contain Positive x.

Essentially, if we took a look at the example above, the output would be

[[]]

Note that the user performs the following function call

removePos [[Positive 1, Negative 1], [Positive 2, Negative 2, Positive 1], [Positive 1]]

Since each list in the nested list above contains Positive 1, the output is simply an empty nested list (All lists with Positive x are removed). However, if the first element in the first list was Positive 10, the output would be

[[Positive 2, Negative 2, Positive 1], [Positive 1]]

(Because the first list would have [Positive 10, Negative 1], which would get removed)

Any ideas?

EDIT:

For further examples, lets say I have a basket with bowls of fruits in it. The basket is the nested list, the bowls are lists within it. Now, I take a look in the basket. I check the first bowl. I look at the first fruit in the bowl, and determine that I don't want to eat that fruit from any of the bowls. So I throw out all bowls that contain that fruit, and give you back your basket.

Answer 1

One way is pattern matching on the nested list to peek at the first element of the first list to figure out what you need to filter:

-- Note you could give this the more general type Eq a => [[a]] -> [[a]]
-- (as well as a more appropriate name)
removePos :: [[Number]] -> [[Number]]
removePos [] = [] -- Empty list case.
removePos xss@[[]:_] = xss -- If the first inner list is empty,
                           -- return the whole thing unchanged
removePos ((x:_):xss) = filter (notElem x) xss