Does any one know of a function/idiom (in any language) that takes a set and returns two or more subsets, determined by one or more predicates?
It is easy to do this in an imperative style e.g:
a = b = []
for x in range(10):
if even(x):
a.append(x)
else:
b.append(x)
or slightly better:
[even(x) and a.append(x) or b.append(x) for x in range(10)]
Since a set comprehension returns a single list based upon a single predicate (and it effectively just a map) I think there ought to be something that splits the input into 2 or more bins based on either a binary predicate or multiple predicates.
The neatest syntax I can come up with is:
>> def partition(iterable, *functions):
>> return [filter(f,iterable) for f in functions]
>> partition(range(10), lambda x: bool(x%2), lambda x: x == 2)
[[1, 3, 5, 7, 9], [2]]
Searching for (a -> Bool) -> [a] -> ([a], [a]) on Hoogle yields Data.List.partition.
The partition function takes a predicate a list and returns the pair of lists of elements which do and do not satisfy the predicate, respectively; i.e.,
partition p xs == (filter p xs, filter (not . p) xs)
If you look at its source and translate to Python,
def partition(predicate, sequence):
def select((yes, no), value):
if predicate(value):
return (yes + [value], no)
else:
return (yes, no + [value])
return reduce(select, sequence, ([], []))
which is pretty nicely functional. Unlike the original, it's not lazy, but that's a bit trickier to pull off in Python.
Ruby's Enumerable mixin has a partition method that does what you describe.
Related
I have the following question "Given a list of integer pairs, write a function to return a list of even numbers in that list in sml".
this is what I've achieved so far
val x = [(6, 2), (3, 4), (5, 6), (7, 8), (9, 10)];
fun isEven(num : int) =
if num mod 2 = 0 then num else 0;
fun evenNumbers(list : (int * int) list) =
if null list then [] else
if isEven(#1 (hd list)) <> 0
then if isEven(#2 (hd list)) <> 0
then #1 (hd list) :: #1 (hd list) :: evenNumbers(tl list)
else []
else if isEven(#2 (hd list)) <> 0
then #1 (hd list) :: evenNumbers(tl list)
else [];
evenNumbers(x);
the result should be like this [6,2,4,6,8,10]
any help would be appreciated.
I see two obvious problems.
If both the first and second number are even, you do
#1 (hd list) :: #1 (hd list) :: evenNumbers(tl list)
which adds the first number twice and ignores the second.
If the first number is odd and the second even, you do
#1 (hd list) :: evenNumbers(tl list)
which adds the number that you know is odd and ignores the one you know is even.
Programming with selectors and conditionals gets complicated very quickly (as you've noticed).
With pattern matching, you could write
fun evenNumbers [] = []
| evenNumber ((x,y)::xys) = ...
and reduce the risk of using the wrong selector.
However, this still makes for complicated logic, and there is a better way.
Consider the simpler problem of filtering the odd numbers out of a list of numbers, not pairs.
If you transform the input into such a list, you only need to solve that simpler problem (and there's a fair chance that you've already solved something very similar in a previous exercise).
Exercise: implement this transformation. Its type will be ('a * 'a) list -> 'a list.
Also, your isEven is more useful if it produces a truth value (if you ask someone, "is 36 even?", "36" is a very strange answer).
fun isEven x = x mod 2 = 0
Now, evenNumbers can be implemented as "just" a combination of other, more general, functions.
So running your current code,
- evenNumbers [(6, 2), (3, 4), (5, 6), (7, 8), (9, 10)];
val it = [6,6,3,5,7,9] : int list
suggests that you're not catching all even numbers, and that you're catching some odd numbers.
The function isEven sounds very much like you want to have the type int -> bool like so:
fun isEven n =
n mod 2 = 0
Instead of addressing the logic error of your current solution, I would like to propose a syntactically much simpler approach which is to use pattern matching and fewer explicit type annotations. One basis for such a solution could look like:
fun evenNumbers [] = ...
| evenNumbers ((x,y)::pairs) = ...
Using pattern matching is an alternative to if-then-else: the [] pattern is equivalent to if null list ... and the (x,y)::pairs pattern matches when the input list is non-empty (holds at least one element, being (x,y). At the same time, it deconstructs this one element into its parts, x and y. So in the second function body you can express isEven x and isEven y.
As there is a total of four combinations of whether x and y are even or not, this could easily end up with a similarly complicated nest of if-then-else's. For this I might do either one of two things:
Use case-of (and call evenNumbers recursively on pairs):
fun evenNumbers [] = ...
| evenNumbers ((x,y)::pairs) =
case (isEven x, isEven y) of
... => ...
| ... => ...
Flatten the list of pairs into a list of integers and filter it:
fun flatten [] = ...
| flatten ((x,y)::pairs) = ...
val evenNumbers pairs = ...
I'm a beginner in functional programming but I'm famaliar with imperative programming. I'm having trouble translating a piece of cpp code involving updatating two objects at the same time (context is n-body simulation).
It's roughly like this in c++:
for (Particle &i: particles) {
for (Particle &j: particles) {
collide(i, j) // function that mutates particles i and j
}
}
I'm translating this to Ocaml, with immutable objects and immutable Lists. The difficult part is that I need to replace two objects at the same time. So far I have this:
List.map (fun i ->
List.map (fun j ->
let (new_i, new_j) = collide(i, j) in // function that returns new particles i, j
// how do i update particles with new i, j?
) particles
) particles
How do I replace both objects in the List at the same time?
The functional equivalent of the imperative code is just as simple as,
let nbody f xs =
List.map (fun x -> List.fold_left f x xs) xs
It is a bit more generic, as a I abstracted the collide function and made it a parameter. The function f takes two bodies and returns the state of the first body as affected by the second body. For example, we can implement the following symbolic collide function,
let symbolic x y = "f(" ^ x ^ "," ^ y ^ ")"
so that we can see the result and associativity of the the collide function application,
# nbody symbolic [
"x"; "y"; "z"
];;
- : string list =
["f(f(f(x,x),y),z)"; "f(f(f(y,x),y),z)"; "f(f(f(z,x),y),z)"]
So, the first element of the output list is the result of collision of x with x itself, then with y, then with z. The second element is the result of collision of y with x, and y, and z. And so on.
Obviously the body shall not collide with itself, but this could be easily fixed by either modifying the collide function or by filtering the input list to List.fold and removing the currently being computed element. This is left as an exercise.
List.map returns a new list. The function you supply to List.map may transform the elements from one type to another or just apply some operation on the same type.
For example, let's assume you start with a list of integer tuples
let int_tuples = [(1, 3); (4, 3); (8, 2)];;
and let's assume that your update function takes an integer tuple and doubles the integers:
let update (i, j) = (i * 2, j * 2) (* update maybe your collide function *)
If you now do:
let new_int_tuples = List.map update int_tuples
You'll get
(* [(2, 6); (8, 6); (16, 4)] *)
Hope this helps
The function should take a list of tuples, and return the ones that have sum > 5
Let's say I have the following code:
fn :: [(Int, Int)] -> [(Int, Int)]
fn tuples = map (\(x,y) -> if (x + y) > 5 then (x,y) else (0,0)) tuples
fn [(3,4), (4,4), (0,1)] returns [(3,4),(4,4),(0,0)] but really I just want it to return [(3,4),(4,4)]
Is this possible in haskell while still following the type signature?
What you're asking for is mapMaybe:
mapMaybe :: (a -> Maybe b) -> [a] -> [b]
base Data.Maybe
The mapMaybe function is a version of map which can throw out elements. In particular, the functional argument returns something of type Maybe b. If this is Nothing, no element is added on to the result list. If it is Just b, then b is included in the result list.
The smallest change to use it in your code would be:
import Data.Maybe
fn :: [(Int, Int)] -> [(Int, Int)]
fn tuples = mapMaybe (\(x,y) -> if (x + y) > 5 then Just (x,y) else Nothing) tuples
However, in this specific case, you don't actually transform, you just remove. If you don't plan on adding transformation later, filter is more suitable:
fn = filter (\(x,y) -> x+y > 5)
Bit of a weird question, but after the comment
I dont see a concept of “empty tuple” in haskell
I guess I see where you're coming from. Actually Haskell does have “empty tuples”: the unit type () is the type of “tuples with zero elements”. So what you're thinking of seems to be
fn tuples = map (\(x,y) -> if x + y > 5 then (x,y) else ()) tuples
But that doesn't work because () is a different type from (Int,Int). The elements of a list must all have the same type. Even if it did work, à la dynamic-types, the result of fn [(3,4), (4,4), (0,1)] would then actually be [(3,4), (4,4), ()]. I.e. you'd still get three elements, just one of them would be “boring”.
map does in fact by design guarantee to never change the number of elements in the list, only the values of their elements. So if that's what you want, you need to use a different function. The closest to your approach would be concatMap:
fn tuples = concatMap (\(x,y) -> if x + y > 5 then [(x,y)] else []) tuples
What happens here can also be described in two steps:
You map a function that generates a list for each element. The result is thus a list of lists.
You flatten that list.
So [(3,4), (4,4), (0,1)] -> [[(3,4)], [(4,4)], []] -> [(3,4), (4,4)].
Really though, there's no need to use a mapping step at all here – the elements are kept as they are eventually, so filter is the tool to use.
I know how to do the equivalent of Scheme's (or Python's) map and filter functions with the list monad using only the "bind" operation.
Here's some Scala to illustrate:
scala> // map
scala> List(1,2,3,4,5,6).flatMap {x => List(x * x)}
res20: List[Int] = List(1, 4, 9, 16, 25, 36)
scala> // filter
scala> List(1,2,3,4,5,6).flatMap {x => if (x % 2 == 0) List() else List(x)}
res21: List[Int] = List(1, 3, 5)
and the same thing in Haskell:
Prelude> -- map
Prelude> [1, 2, 3, 4, 5, 6] >>= (\x -> [x * x])
[1,4,9,16,25,36]
Prelude> -- filter
Prelude> [1, 2, 3, 4, 5, 6] >>= (\x -> if (mod x 2 == 0) then [] else [x])
[1,3,5]
Scheme and Python also have a reduce function that's often grouped with map and filter. The reduce function combines the first two elements of a list using the supplied binary function, and then combines that result the the next element, and then so on. A common use to to compute the sum or product of a list of values. Here's some Python to illustrate:
>>> reduce(lambda x, y: x + y, [1,2,3,4,5,6])
21
>>> (((((1+2)+3)+4)+5)+6)
21
Is there any way to do the equivalent of this reduce using just the bind operation on a list monad? If bind can't do this on its own, what's the most "monadic" way to perform this operation?
If possible, please limit/avoid the use of syntactic sugar (ie: do notation in Haskell or sequence comprehensions in Scala) when answering.
One of the defining properties of the bind operation is that the result is still "inside" the monad¹. So when you perform bind on a list, the result will again be a list. Since the reduce operation² often results in something other than a list, it can't be expressed in terms of the bind operation.
In addition to that the bind operation on lists (i.e. concatMap/flatMap) only looks at one element at a time and offers no way of reusing the result of previous steps. So even if we're okay with getting the result wrapped in a single-element list, there's no way to do it just with monad operations.
¹ So if you have a type that allows you to perform no operations on it except the ones defined by the monad type class, you can never "break out" of the monad. That's what makes the IO monad works.
² Which is called fold in Haskell and Scala by the way.
If bind can't do this on its own, what's the most "monadic" way to perform this operation?
While the answer given by #sepp2k is correct, there is a way to do a reduce-like operation on a list monadically, but using the product or "writer" monad and an operation which corresponds to distributing the product monad over the list functor.
The definition is:
import Control.Monad.Writer.Lazy
import Data.Monoid
reduce :: Monoid a => [a] -> a
reduce xs = snd . runWriter . sequence $ map tell xs
Let me unpack:
The Writer monad has a data type Writer w a which is basically a tuple (product) of a value a and "written" value w. The type of written values w must be a monoid where the bind operation of the Writer monad is defined something like:
(w, a) >>= f = let (w', b) = f a in (mappend w w', b)
i.e. take the incoming written value, and the result written value, and combine them using the binary operation of the monoid.
The tell operation writes a value, tell :: w -> Writer w (). Thus map tell has type [a] -> [Writer a ()] i.e. a list of monadic values where each element of the original list has been "written" in the monad.
sequence :: Monad m => [m a] -> m [a] corresponds to a distributive law between lists and monads i.e. distribute the monad type over the list type; sequence can be defined in terms of bind as:
sequence [] = return []
sequnece (x:xs) = x >>= (\x' -> (sequence xs) >>= (\xs' -> return $ x':xs'))
(actually the implementation in Prelude uses foldr, a clue to the reduction-like usage)
Thus, sequence $ map tell xs has type Writer a [()]
The runWriter operation unpacks the Writer type, runWriter :: Writer w a -> (a, w),
which is composed here with snd to project out the accumulated value.
An example usage on lists of Ints would be to use the monoid instance:
instance Monoid Int where
mappend = (+)
mempty = 0
then:
> reduce ([1,2,3,4]::[Int])
10
I'm new to ocaml and tryin to write a continuation passing style function but quite confused what value i need to pass into additional argument on k
for example, I can write a recursive function that returns true if all elements of the list is even, otherwise false.
so its like
let rec even list = ....
on CPS, i know i need to add one argument to pass function
so like
let rec evenk list k = ....
but I have no clue how to deal with this k and how does this exactly work
for example for this even function, environment looks like
val evenk : int list -> (bool -> ’a) -> ’a = <fun>
evenk [4; 2; 12; 5; 6] (fun x -> x) (* output should give false *)
The continuation k is a function that takes the result from evenk and performs "the rest of the computation" and produces the "answer". What type the answer has and what you mean by "the rest of the computation" depends on what you are using CPS for. CPS is generally not an end in itself but is done with some purpose in mind. For example, in CPS form it is very easy to implement control operators or to optimize tail calls. Without knowing what you are trying to accomplish, it's hard to answer your question.
For what it is worth, if you are simply trying to convert from direct style to continuation-passing style, and all you care about is the value of the answer, passing the identity function as the continuation is about right.
A good next step would be to implement evenk using CPS. I'll do a simpler example.
If I have the direct-style function
let muladd x i n = x + i * n
and if I assume CPS primitives mulk and addk, I can write
let muladdk x i n k =
let k' product = addk x product k in
mulk i n k'
And you'll see that the mulptiplication is done first, then it "continues" with k', which does the add, and finally that continues with k, which returns to the caller. The key idea is that within the body of muladdk I allocated a fresh continuation k' which stands for an intermediate point in the multiply-add function. To make your evenk work you will have to allocate at least one such continuation.
I hope this helps.
Whenever I've played with CPS, the thing passed to the continuation is just the thing you would normally return to the caller. In this simple case, a nice "intuition lubricant" is to name the continuation "return".
let rec even list return =
if List.length list = 0
then return true
else if List.hd list mod 2 = 1
then return false
else even (List.tl list) return;;
let id = fun x -> x;;
Example usage: "even [2; 4; 6; 8] id;;".
Since you have the invocation of evenk correct (with the identity function - effectively converting the continuation-passing-style back to normal style), I assume that the difficulty is in defining evenk.
k is the continuation function representing the rest of the computation and producing a final value, as Norman said. So, what you need to do is compute the result of v of even and pass that result to k, returning k v rather than just v.
You want to give as input the result of your function as if it were not written with continuation passing style.
Here is your function which tests whether a list has only even integers:
(* val even_list : int list -> bool *)
let even_list input = List.for_all (fun x -> x mod 2=0) input
Now let's write it with a continuation cont:
(* val evenk : int list -> (bool -> 'a) -> 'a *)
let evenk input cont =
let result = even_list input in
(cont result)
You compute the result your function, and pass resultto the continuation ...