问题
I don't understand what "lifting" is. Should I first understand monads before understanding what a "lift" is? (I'm completely ignorant about monads, too :) Or can someone explain it to me with simple words?
回答1:
Lifting is more of a design pattern than a mathematical concept (although I expect someone around here will now refute me by showing how lifts are a category or something).
Typically you have some data type with a parameter. Something like
data Foo a = Foo { ...stuff here ...}
Suppose you find that a lot of uses of Foo take numeric types (Int, Double etc) and you keep having to write code that unwraps these numbers, adds or multiplies them, and then wraps them back up. You can short-circuit this by writing the unwrap-and-wrap code once. This function is traditionally called a "lift" because it looks like this:
liftFoo2 :: (a -> b -> c) -> Foo a -> Foo b -> Foo c
In other words you have a function which takes a two-argument function (such as the (+) operator) and turns it into the equivalent function for Foos.
So now you can write
addFoo = liftFoo2 (+)
Edit: more information
You can of course have liftFoo3, liftFoo4 and so on. However this is often not necessary.
Start with the observation
liftFoo1 :: (a -> b) -> Foo a -> Foo b
But that is exactly the same as fmap. So rather than liftFoo1 you would write
instance Functor Foo where
fmap f foo = ...
If you really want complete regularity you can then say
liftFoo1 = fmap
If you can make Foo into a functor, perhaps you can make it an applicative functor. In fact, if you can write liftFoo2 then the applicative instance looks like this:
import Control.Applicative
instance Applicative Foo where
pure x = Foo $ ... -- Wrap 'x' inside a Foo.
(<*>) = liftFoo2 ($)
The (<*>) operator for Foo has the type
(<*>) :: Foo (a -> b) -> Foo a -> Foo b
It applies the wrapped function to the wrapped value. So if you can implement liftFoo2 then you can write this in terms of it. Or you can implement it directly and not bother with liftFoo2, because the Control.Applicative module includes
liftA2 :: Applicative f => (a -> b -> c) -> f a -> f b -> f c
and likewise there are liftA and liftA3. But you don't actually use them very often because there is another operator
(<$>) = fmap
This lets you write:
result = myFunction <$> arg1 <*> arg2 <*> arg3 <*> arg4
The term myFunction <$> arg1 returns a new function wrapped in Foo. This in turn can be applied to the next argument using (<*>), and so on. So now instead of having a lift function for every arity, you just have a daisy chain of applicatives.
回答2:
Paul's and yairchu's are both good explanations.
I'd like to add that the function being lifted can have an arbitrary number of arguments and that they don't have to be of the same type. For example, you could also define a liftFoo1:
liftFoo1 :: (a -> b) -> Foo a -> Foo b
In general, the lifting of functions that take 1 argument is captured in the type class Functor, and the lifting operation is called fmap:
fmap :: Functor f => (a -> b) -> f a -> f b
Note the similarity with liftFoo1's type. In fact, if you have liftFoo1, you can make Foo an instance of Functor:
instance Functor Foo where
fmap = liftFoo1
Furthermore, the generalization of lifting to an arbitrary number of arguments is called applicative style. Don't bother diving into this until you grasp the lifting of functions with a fixed number of arguments. But when you do, Learn you a Haskell has a good chapter on this. The Typeclassopedia is another good document that describes Functor and Applicative (as well as other type classes; scroll down to the right chapter in that document).
Hope this helps!
回答3:
Let's start with an example (some white space is added for clearer presentation):
> import Control.Applicative
> replicate 3 'a'
"aaa"
> :t replicate
replicate :: Int -> b -> [b]
> :t liftA2
liftA2 :: (Applicative f) => (a -> b -> c) -> (f a -> f b -> f c)
> :t liftA2 replicate
liftA2 replicate :: (Applicative f) => f Int -> f b -> f [b]
> (liftA2 replicate) [1,2,3] ['a','b','c']
["a","b","c","aa","bb","cc","aaa","bbb","ccc"]
> ['a','b','c']
"abc"
liftA2 transforms a function of plain types to a function of same types wrapped in an Applicative, such as lists, IO, etc.
Another common lift is lift from Control.Monad.Trans. It transforms a monadic action of one monad to an action of a transformed monad.
In general, "lift" lifts a function/action into a "wrapped" type (so the original function gets to work "under the wraps").
The best way to understand this, and monads etc., and to understand why they are useful, is probably to code and use it. If there's anything you coded previously that you suspect can benefit from this (i.e. this will make that code shorter, etc.), just try it out and you'll easily grasp the concept.
回答4:
Lifting is a concept which allows you to transform a function into a corresponding function within another (usually more general) setting
take a look at http://haskell.org/haskellwiki/Lifting
回答5:
According to this shiny tutorial, a functor is some container (like Maybe<a>, List<a> or Tree<a> that can store elements of some another type, a). I have used Java generics notation, <a>, for element type a and think of the elements as berries on the tree Tree<a>. There is a function fmap, which takes an element conversion function, a->b and container functor<a>. It applies a->b to every element of the container effectively converting it into functor<b>. When only first argument is supplied, a->b, fmap waits for the functor<a>. That is, supplying a->b alone turns this element-level function into the function functor<a> -> functor<b> that operates over containers. This is called lifting of the function. Because the container is also called a functor, the Functors rather than Monads are a prerequisite for the lifting. Monads are sort of "parallel" to lifting. Both rely on the Functor notion and do f<a> -> f<b>. The difference is that lifting uses a->b for the conversion whereas Monad requires the user to define a -> f<b>.
来源：https://stackoverflow.com/questions/2395697/what-is-lifting-in-haskell