Memoizing multiplication - haskell

My application multiplies vectors after a (costly) conversion using an FFT. As a result, when I write
f :: (Num a) => a -> [a] -> [a]
f c xs = map (c*) xs
I only want to compute the FFT of c once, rather than for every element of xs. There really isn't any need to store the FFT of c for the entire program, just in the local scope.
I attempted to define my Num instance like:
data Foo = Scalar c
| Vec Bool v -- the bool indicates which domain v is in
instance Num Foo where
(*) (Scalar c) = \x -> case x of
Scalar d -> Scalar (c*d)
Vec b v-> Vec b $ map (c*) v
(*) v1 = let Vec True v = fft v1
in \x -> case x of
Scalar d -> Vec True $ map (c*) v
v2 -> Vec True $ zipWith (*) v (fft v2)
Then, in an application, I call a function similar to f (which works on arbitrary Nums) where c=Vec False v, and I expected that this would be just as fast as if I hack f to:
g :: Foo -> [Foo] -> [Foo]
g c xs = let c' = fft c
in map (c'*) xs
The function g makes the memoization of fft c occur, and is much faster than calling f (no matter how I define (*)). I don't understand what is going wrong with f. Is it my definition of (*) in the Num instance? Does it have something to do with f working over all Nums, and GHC therefore being unable to figure out how to partially compute (*)?
Note: I checked the core output for my Num instance, and (*) is indeed represented as nested lambdas with the FFT conversion in the top level lambda. So it looks like this is at least capable of being memoized. I have also tried both judicious and reckless use of bang patterns to attempt to force evaluation to no effect.
As a side note, even if I can figure out how to make (*) memoize its first argument, there is still another problem with how it is defined: A programmer wanting to use the Foo data type has to know about this memoization capability. If she wrote
map (*c) xs
no memoization would occur. (It must be written as (map (c*) xs)) Now that I think about it, I'm not entirely sure how GHC would rewrite the (*c) version since I have curried (*). But I did a quick test to verify that both (*c) and (c*) work as expected: (c*) makes c the first arg to *, while (*c) makes c the second arg to *. So the problem is that it is not obvious how one should write the multiplication to ensure memoization. Is this just an inherent downside to the infix notation (and the implicit assumption that the arguments to * are symmetric)?
The second, less pressing issue is that the case where we map (v*) onto a list of scalars. In this case, (hopefully) the fft of v would be computed and stored, even though it is unnecessary since the other multiplicand is a scalar. Is there any way around this?
Thanks

I believe stable-memo package could solve your problem. It memoizes values not using equality but by reference identity:
Whereas most memo combinators memoize based on equality, stable-memo does it based on whether the exact same argument has been passed to the function before (that is, is the same argument in memory).
And it automatically drops memoized values when their keys are garbage collected:
stable-memo doesn't retain the keys it has seen so far, which allows them to be garbage collected if they will no longer be used. Finalizers are put in place to remove the corresponding entries from the memo table if this happens.
So if you define something like
fft = memo fft'
where fft' = ... -- your old definition
you'll get pretty much what you need: Calling map (c *) xs will memoize the computation of fft inside the first call to (*) and it gets reused on subsequent calls to (c *). And if c is garbage collected, so is fft' c.
See also this answer to How to add fields that only cache something to ADT?

I can see two problems that might prevent memoization:
First, f has an overloaded type and works for all Num instances. So f cannot use memoization unless it is either specialized (which usually requires a SPECIALIZE pragma) or inlined (which may happen automatically, but is more reliable with an INLINE pragma).
Second, the definition of (*) for Foo performs pattern matching on the first argument, but f multiplies with an unknown c. So within f, even if specialized, no memoization can occur. Once again, it very much depends on f being inlined, and a concrete argument for c to be supplied, so that inlining can actually appear.
So I think it'd help to see how exactly you're calling f. Note that if f is defined using two arguments, it has to be given two arguments, otherwise it cannot be inlined. It would furthermore help to see the actual definition of Foo, as the one you are giving mentions c and v which aren't in scope.

Related

Why the pointfree style does not cause a problem?

I read about The Monomorphism Restriction from the page https://www.haskell.org/tutorial/pitfalls.html and could not understand the last point:
A common violation of the restriction happens with functions defined
in a higher-order manner, as in this definition of sum from the
Standard Prelude:
sum = foldl (+) 0
As is, this would cause a static type error. We can fix the problem by
adding the type signature:
sum :: (Num a) => [a] -> a
Also note that this problem would not have arisen if we had written:
sum xs = foldl (+) 0 xs
because the restriction only applies to pattern bindings.
Why the last point does not cause any error?
because the restriction only applies to pattern bindings.
Essentially, the MR does not apply when we are defining a function using a function binding of the form
f arg1 ... argN = ...
with N > 0.
The intuition is as follows. The purpose of the MR is to avoid turning Haskell non-functions into lower-level functions accidentally. For instance,
x = 3 + 4
is not a function. However, its type is Num a => a, which is usually implemented as a function from a Num dictionary to the result of 3+4 where + is a function defined by the dictionary. This can lead to a bad performance, since every time we use x the sum will need to be recomputed from scratch. This is unavoidable if we want to compute print (x :: Int) >> print (x :: Double), for instance. But actually using x at different types is rather uncommon.
So, the MR makes x monomorphic, preventing us to use it at more than a single type. In that way, recomputation can be avoided.
However, if x is already a function there is no harm in keeping that polymorphic, since we are "recomputing" function calls anyway. So, the MR does not apply to function bindings.

Apply function to all pairs efficiently

I need a second order function pairApply that applies a binary function f to all unique pairs of a list-like structure and then combines them somehow. An example / sketch:
pairApply (+) f [a, b, c] = f a b + f a c + f b c
Some research leads me to believe that Data.Vector.Unboxed probably will have good performance (I will also need fast access to specific elements); also it necessary for Statistics.Sample, which would come in handy further down the line.
With this in mind I have the following, which almost compiles:
import qualified Data.Vector.Unboxed as U      
pairElement :: (U.Unbox a, U.Unbox b)    
=> (U.Vector a)                    
  -> (a -> a -> b)                   
  -> Int                             
-> a                               
 -> (U.Vector b)                    
pairElement v f idx el =
U.map (f el) $ U.drop (idx + 1) v            
pairUp :: (U.Unbox a, U.Unbox b)   
=> (a -> a -> b)                        
 -> (U.Vector a)                         
-> (U.Vector (U.Vector b))
pairUp f v = U.imap (pairElement v f) v 
pairApply :: (U.Unbox a, U.Unbox b)
=> (b -> b -> b)                     
-> b                                 
 -> (a -> a -> b)                     
-> (U.Vector a)                      
 -> b
pairApply combine neutral f v =
folder $ U.map folder (pairUp f v) where
folder = U.foldl combine neutral
The reason this doesn't compile is that there is no Unboxed instance of a U.Vector (U.Vector a)). I have been able to create new unboxed instances in other cases using Data.Vector.Unboxed.Deriving, but I'm not sure it would be so easy in this case (transform it to a tuple pair where the first element is all the inner vectors concatenated and the second is the length of the vectors, to know how to unpack?)
My question can be stated in two parts:
Does the above implementation make sense at all or is there some quick library function magic etc that could do it much easier?
If so, is there a better way to make an unboxed vector of vectors than the one sketched above?
Note that I'm aware that foldl is probably not the best choice; once I've got the implementation sorted I plan to benchmark with a few different folds.
There is no way to define a classical instance for Unbox (U.Vector b), because that would require preallocating a memory area in which each element (i.e. each subvector!) has the same fixed amount of space. But in general, each of them may be arbitrarily big, so that's not feasible at all.
It might in principle be possible to define that instance by storing only a flattened form of the nested vector plus an extra array of indices (where each subvector starts). I once briefly gave this a try; it actually seems somewhat promising as far as immutable vectors are concerned, but a G.Vector instance also requires a mutable implementation, and that's hopeless for such an approach (because any mutation that changes the number of elements in one subvector would require shifting everything behind it).
Usually, it's just not worth it, because if the individual element vectors aren't very small the overhead of boxing them won't matter, i.e. often it makes sense to use B.Vector (U.Vector b).
For your application however, I would not do that at all – there's no need to ever wrap the upper element-choices in a single triangular array. (And it would be really bad for performance to do that, because it make the algorithm take O (n²) memory rather than O (n) which is all that's needed.)
I would just do the following:
pairApply combine neutral f v
= U.ifoldl' (\acc i p -> U.foldl' (\acc' q -> combine acc' $ f p q)
acc
(U.drop (i+1) v) )
neutral v
This corresponds pretty much to the obvious nested-loops imperative implementation
pairApply(combine, b, f, v):
for(i in 0..length(v)-1):
for(j in i+1..length(v)-1):
b = combine(b, f(v[i], v[j]);
return b;
My answer is basically the same as leftaroundabout's nested-loops imperative implementation:
pairApply :: (Int -> Int -> Int) -> Vector Int -> Int
pairApply f v = foldl' (+) 0 [f (v ! i) (v ! j) | i <- [0..(n-1)], j <- [(i+1)..(n-1)]]
where n = length v
As far as I know, I do not see any performance issue with this implementation.
Non-polymorphic for simplicity.

Can any recursive definition be rewritten using foldr?

Say I have a general recursive definition in haskell like this:
foo a0 a1 ... = base_case
foo b0 b1 ...
| cond1 = recursive_case_1
| cond2 = recursive_case_2
...
Can it always rewritten using foldr? Can it be proved?
If we interpret your question literally, we can write const value foldr to achieve any value, as #DanielWagner pointed out in a comment.
A more interesting question is whether we can instead forbid general recursion from Haskell, and "recurse" only through the eliminators/catamorphisms associated to each user-defined data type, which are the natural generalization of foldr to inductively defined data types. This is, essentially, (higher-order) primitive recursion.
When this restriction is performed, we can only compose terminating functions (the eliminators) together. This means that we can no longer define non terminating functions.
As a first example, we lose the trivial recursion
f x = f x
-- or even
a = a
since, as said, the language becomes total.
More interestingly, the general fixed point operator is lost.
fix :: (a -> a) -> a
fix f = f (fix f)
A more intriguing question is: what about the total functions we can express in Haskell? We do lose all the non-total functions, but do we lose any of the total ones?
Computability theory states that, since the language becomes total (no more non termination), we lose expressiveness even on the total fragment.
The proof is a standard diagonalization argument. Fix any enumeration of programs in the total fragment so that we can speak of "the i-th program".
Then, let eval i x be the result of running the i-th program on the natural x as input (for simplicity, assume this is well typed, and that the result is a natural). Note that, since the language is total, then a result must exist. Moreover, eval can be implemented in the unrestricted Haskell language, since we can write an interpreter of Haskell in Haskell (left as an exercise :-P), and that would work as fine for the fragment. Then, we simply take
f n = succ $ eval n n
The above is a total function (a composition of total functions) which can be expressed in Haskell, but not in the fragment. Indeed, otherwise there would be a program to compute it, say the i-th program. In such case we would have
eval i x = f x
for all x. But then,
eval i i = f i = succ $ eval i i
which is impossible -- contradiction. QED.
In type theory, it is indeed the case that you can elaborate all definitions by dependent pattern-matching into ones only using eliminators (a more strongly-typed version of folds, the generalisation of lists' foldr).
See e.g. Eliminating Dependent Pattern Matching (pdf)

What are the benefits of currying?

I don't think I quite understand currying, since I'm unable to see any massive benefit it could provide. Perhaps someone could enlighten me with an example demonstrating why it is so useful. Does it truly have benefits and applications, or is it just an over-appreciated concept?
(There is a slight difference between currying and partial application, although they're closely related; since they're often mixed together, I'll deal with both terms.)
The place where I realized the benefits first was when I saw sliced operators:
incElems = map (+1)
--non-curried equivalent: incElems = (\elems -> map (\i -> (+) 1 i) elems)
IMO, this is totally easy to read. Now, if the type of (+) was (Int,Int) -> Int *, which is the uncurried version, it would (counter-intuitively) result in an error -- but curryied, it works as expected, and has type [Int] -> [Int].
You mentioned C# lambdas in a comment. In C#, you could have written incElems like so, given a function plus:
var incElems = xs => xs.Select(x => plus(1,x))
If you're used to point-free style, you'll see that the x here is redundant. Logically, that code could be reduced to
var incElems = xs => xs.Select(curry(plus)(1))
which is awful due to the lack of automatic partial application with C# lambdas. And that's the crucial point to decide where currying is actually useful: mostly when it happens implicitly. For me, map (+1) is the easiest to read, then comes .Select(x => plus(1,x)), and the version with curry should probably be avoided, if there is no really good reason.
Now, if readable, the benefits sum up to shorter, more readable and less cluttered code -- unless there is some abuse of point-free style done is with it (I do love (.).(.), but it is... special)
Also, lambda calculus would get impossible without using curried functions, since it has only one-valued (but therefor higher-order) functions.
* Of course it actually in Num, but it's more readable like this for the moment.
Update: how currying actually works.
Look at the type of plus in C#:
int plus(int a, int b) {..}
You have to give it a tuple of values -- not in C# terms, but mathematically spoken; you can't just leave out the second value. In haskell terms, that's
plus :: (Int,Int) -> Int,
which could be used like
incElem = map (\x -> plus (1, x)) -- equal to .Select (x => plus (1, x))
That's way too much characters to type. Suppose you'd want to do this more often in the future. Here's a little helper:
curry f = \x -> (\y -> f (x,y))
plus' = curry plus
which gives
incElem = map (plus' 1)
Let's apply this to a concrete value.
incElem [1]
= (map (plus' 1)) [1]
= [plus' 1 1]
= [(curry plus) 1 1]
= [(\x -> (\y -> plus (x,y))) 1 1]
= [plus (1,1)]
= [2]
Here you can see curry at work. It turns a standard haskell style function application (plus' 1 1) into a call to a "tupled" function -- or, viewed at a higher level, transforms the "tupled" into the "untupled" version.
Fortunately, most of the time, you don't have to worry about this, as there is automatic partial application.
It's not the best thing since sliced bread, but if you're using lambdas anyway, it's easier to use higher-order functions without using lambda syntax. Compare:
map (max 4) [0,6,9,3] --[4,6,9,4]
map (\i -> max 4 i) [0,6,9,3] --[4,6,9,4]
These kinds of constructs come up often enough when you're using functional programming, that it's a nice shortcut to have and lets you think about the problem from a slightly higher level--you're mapping against the "max 4" function, not some random function that happens to be defined as (\i -> max 4 i). It lets you start to think in higher levels of indirection more easily:
let numOr4 = map $ max 4
let numOr4' = (\xs -> map (\i -> max 4 i) xs)
numOr4 [0,6,9,3] --ends up being [4,6,9,4] either way;
--which do you think is easier to understand?
That said, it's not a panacea; sometimes your function's parameters will be the wrong order for what you're trying to do with currying, so you'll have to resort to a lambda anyway. However, once you get used to this style, you start to learn how to design your functions to work well with it, and once those neurons starts to connect inside your brain, previously complicated constructs can start to seem obvious in comparison.
One benefit of currying is that it allows partial application of functions without the need of any special syntax/operator. A simple example:
mapLength = map length
mapLength ["ab", "cde", "f"]
>>> [2, 3, 1]
mapLength ["x", "yz", "www"]
>>> [1, 2, 3]
map :: (a -> b) -> [a] -> [b]
length :: [a] -> Int
mapLength :: [[a]] -> [Int]
The map function can be considered to have type (a -> b) -> ([a] -> [b]) because of currying, so when length is applied as its first argument, it yields the function mapLength of type [[a]] -> [Int].
Currying has the convenience features mentioned in other answers, but it also often serves to simplify reasoning about the language or to implement some code much easier than it could be otherwise. For example, currying means that any function at all has a type that's compatible with a ->b. If you write some code whose type involves a -> b, that code can be made work with any function at all, no matter how many arguments it takes.
The best known example of this is the Applicative class:
class Functor f => Applicative f where
pure :: a -> f a
(<*>) :: f (a -> b) -> f a -> f b
And an example use:
-- All possible products of numbers taken from [1..5] and [1..10]
example = pure (*) <*> [1..5] <*> [1..10]
In this context, pure and <*> adapt any function of type a -> b to work with lists of type [a]. Because of partial application, this means you can also adapt functions of type a -> b -> c to work with [a] and [b], or a -> b -> c -> d with [a], [b] and [c], and so on.
The reason this works is because a -> b -> c is the same thing as a -> (b -> c):
(+) :: Num a => a -> a -> a
pure (+) :: (Applicative f, Num a) => f (a -> a -> a)
[1..5], [1..10] :: Num a => [a]
pure (+) <*> [1..5] :: Num a => [a -> a]
pure (+) <*> [1..5] <*> [1..10] :: Num a => [a]
Another, different use of currying is that Haskell allows you to partially apply type constructors. E.g., if you have this type:
data Foo a b = Foo a b
...it actually makes sense to write Foo a in many contexts, for example:
instance Functor (Foo a) where
fmap f (Foo a b) = Foo a (f b)
I.e., Foo is a two-parameter type constructor with kind * -> * -> *; Foo a, the partial application of Foo to just one type, is a type constructor with kind * -> *. Functor is a type class that can only be instantiated for type constrcutors of kind * -> *. Since Foo a is of this kind, you can make a Functor instance for it.
The "no-currying" form of partial application works like this:
We have a function f : (A ✕ B) → C
We'd like to apply it partially to some a : A
To do this, we build a closure out of a and f (we don't evaluate f at all, for the time being)
Then some time later, we receive the second argument b : B
Now that we have both the A and B argument, we can evaluate f in its original form...
So we recall a from the closure, and evaluate f(a,b).
A bit complicated, isn't it?
When f is curried in the first place, it's rather simpler:
We have a function f : A → B → C
We'd like to apply it partially to some a : A – which we can just do: f a
Then some time later, we receive the second argument b : B
We apply the already evaluated f a to b.
So far so nice, but more important than being simple, this also gives us extra possibilities for implementing our function: we may be able to do some calculations as soon as the a argument is received, and these calculations won't need to be done later, even if the function is evaluated with multiple different b arguments!
To give an example, consider this audio filter, an infinite impulse response filter. It works like this: for each audio sample, you feed an "accumulator function" (f) with some state parameter (in this case, a simple number, 0 at the beginning) and the audio sample. The function then does some magic, and spits out the new internal state1 and the output sample.
Now here's the crucial bit – what kind of magic the function does depends on the coefficient2 λ, which is not quite a constant: it depends both on what cutoff frequency we'd like the filter to have (this governs "how the filter will sound") and on what sample rate we're processing in. Unfortunately, the calculation of λ is a bit more complicated (lp1stCoeff $ 2*pi * (νᵥ ~*% δs) than the rest of the magic, so we wouldn't like having to do this for every single sample, all over again. Quite annoying, because νᵥ and δs are almost constant: they change very seldom, certainly not at each audio sample.
But currying saves the day! We simply calculate λ as soon as we have the necessary parameters. Then, at each of the many many audio samples to come, we only need to perform the remaining, very easy magic: yⱼ = yⱼ₁ + λ ⋅ (xⱼ - yⱼ₁). So we're being efficient, and still keeping a nice safe referentially transparent purely-functional interface.
1 Note that this kind of state-passing can generally be done more nicely with the State or ST monad, that's just not particularly beneficial in this example
2 Yes, this is a lambda symbol. I hope I'm not confusing anybody – fortunately, in Haskell it's clear that lambda functions are written with \, not with λ.
It's somewhat dubious to ask what the benefits of currying are without specifying the context in which you're asking the question:
In some cases, like functional languages, currying will merely be seen as something that has a more local change, where you could replace things with explicit tupled domains. However, this isn't to say that currying is useless in these languages. In some sense, programming with curried functions make you "feel" like you're programming in a more functional style, because you more typically face situations where you're dealing with higher order functions. Certainly, most of the time, you will "fill in" all of the arguments to a function, but in the cases where you want to use the function in its partially applied form, this is a bit simpler to do in curried form. We typically tell our beginning programmers to use this when learning a functional language just because it feels like better style and reminds them they're programming in more than just C. Having things like curry and uncurry also help for certain conveniences within functional programming languages too, I can think of arrows within Haskell as a specific example of where you would use curry and uncurry a bit to apply things to different pieces of an arrow, etc...
In some cases, you want to think about more than functional programs, you can present currying / uncurrying as a way to state the elimination and introduction rules for and in constructive logic, which provides a connection to a more elegant motivation for why it exists.
In some cases, for example, in Coq, using curried functions versus tupled functions can produce different induction schemes, which may be easier or harder to work with, depending on your applications.
I used to think that currying was simple syntax sugar that saves you a bit of typing. For example, instead of writing
(\ x -> x + 1)
I can merely write
(+1)
The latter is instantly more readable, and less typing to boot.
So if it's just a convenient short cut, why all the fuss?
Well, it turns out that because function types are curried, you can write code which is polymorphic in the number of arguments a function has.
For example, the QuickCheck framework lets you test functions by feeding them randomly-generated test data. It works on any function who's input type can be auto-generated. But, because of currying, the authors were able to rig it so this works with any number of arguments. Were functions not curried, there would be a different testing function for each number of arguments - and that would just be tedious.

Evaluation strategy

How should one reason about function evaluation in examples like the following in Haskell:
let f x = ...
x = ...
in map (g (f x)) xs
In GHC, sometimes (f x) is evaluated only once, and sometimes once for each element in xs, depending on what exactly f and g are. This can be important when f x is an expensive computation. It has just tripped a Haskell beginner I was helping and I didn't know what to tell him other than that it is up to the compiler. Is there a better story?
Update
In the following example (f x) will be evaluated 4 times:
let f x = trace "!" $ zip x x
x = "abc"
in map (\i -> lookup i (f x)) "abcd"
With language extensions, we can create situations where f x must be evaluated repeatedly:
{-# LANGUAGE GADTs, Rank2Types #-}
module MultiEvG where
data BI where
B :: (Bounded b, Integral b) => b -> BI
foo :: [BI] -> [Integer]
foo xs = let f :: (Integral c, Bounded c) => c -> c
f x = maxBound - x
g :: (forall a. (Integral a, Bounded a) => a) -> BI -> Integer
g m (B y) = toInteger (m + y)
x :: (Integral i) => i
x = 3
in map (g (f x)) xs
The crux is to have f x polymorphic even as the argument of g, and we must create a situation where the type(s) at which it is needed can't be predicted (my first stab used an Either a b instead of BI, but when optimising, that of course led to only two evaluations of f x at most).
A polymorphic expression must be evaluated at least once for each type it is used at. That's one reason for the monomorphism restriction. However, when the range of types it can be needed at is restricted, it is possible to memoise the values at each type, and in some circumstances GHC does that (needs optimising, and I expect the number of types involved mustn't be too large). Here we confront it with what is basically an inhomogeneous list, so in each invocation of g (f x), it can be needed at an arbitrary type satisfying the constraints, so the computation cannot be lifted outside the map (technically, the compiler could still build a cache of the values at each used type, so it would be evaluated only once per type, but GHC doesn't, in all likelihood it wouldn't be worth the trouble).
Monomorphic expressions need only be evaluated once, they can be shared. Whether they are is up to the implementation; by purity, it doesn't change the semantics of the programme. If the expression is bound to a name, in practice you can rely on it being shared, since it's easy and obviously what the programmer wants. If it isn't bound to a name, it's a question of optimisation. With the bytecode generator or without optimisations, the expression will often be evaluated repeatedly, but with optimisations repeated evaluation would indicate a compiler bug.
Polymorphic expressions must be evaluated at least once for every type they're used at, but with optimisations, when GHC can see that it may be used multiple times at the same type, it will (usually) still be shared for that type during a larger computation.
Bottom line: Always compile with optimisations, help the compiler by binding expressions you want shared to a name, and give monomorphic type signatures where possible.
Your examples are indeed quite different.
In the first example, the argument to map is g (f x) and is passed once to map most likely as partially applied function.
Should g (f x), when applied to an argument within map evaluate its first argument, then this will be done only once and then the thunk (f x) will be updated with the result.
Hence, in your first example, f xwill be evaluated at most 1 time.
Your second example requires a deeper analysis before the compiler can arrive at the conclusion that (f x) is always constant in the lambda expression. Perhaps it will never optimize it at all, because it may have knowledge that trace is not quite kosher. So, this may evaluate 4 times when tracing, and 4 times or 1 time when not tracing.
This is really dependent on GHC's optimizations, as you've been able to tell.
The best thing to do is to study the GHC core that you get after optimizing the program. I would look at the generated Core and examine whether f x had its own let statement outside the map or not.
If you want to be sure, then you should factor f x out into its own variable assigned in a let, but there's not really a guaranteed way to figure it out other than reading through Core.
All that said, with the exception of things like trace that use unsafePerformIO, this will never change the semantics of your program: how it actually behaves.
In GHC without optimizations, the body of a function is evaluated every time the function is called. (A "call" means the function is applied to arguments and the result is evaluated.) In the following example, f x is inside a function, so it will execute each time the function is called.
(GHC may optimize this expression as discussed in the FAQ [1].)
let f x = trace "!" $ zip x x
x = "abc"
in map (\i -> lookup i (f x)) "abcd"
However, if we move f x out of the function, it will execute only once.
let f x = trace "!" $ zip x x
x = "abc"
in map ((\f_x i -> lookup i f_x) (f x)) "abcd"
This can be rewritten more readably as
let f x = trace "!" $ zip x x
x = "abc"
g f_x i = lookup i f_x
in map (g (f x)) "abcd"
The general rule is that, each time a function is applied to an argument, a new "copy" of the function body is created. Function application is the only thing that may cause an expression to re-execute. However, be warned that some functions and function calls do not look like functions syntactically.
[1] http://www.haskell.org/haskellwiki/GHC/FAQ#Subexpression_Elimination

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