Which one is the appropriate morphism (recursion scheme) to use when the given item's position (index, or path) is required in the transformer function?
A simple example would be transforming a list ["foo", "bar", "qux"] into the string "foo, bar, and qux". The current element's position is needed to know when to insert the and.
You need to make the index part of the structure so it is available to the recursion scheme. An ad-hoc way to do this is to define a foldWithIndex function:
foldWithIndex :: (Foldable t, Num i) => (i -> a -> b -> b) -> b -> t a -> b
foldWithIndex f z t = snd $ foldr f' (0, z) t
where
f' z (i, x) = (i + 1, f i z x)
This function takes an operator for combining the elements which also considers the index:
foldWithIndex combine "" ["foo", "bar", "qux"]
where
combine 0 s1 s2 = s1 ++ s2
combine 1 s1 s2 = s1 ++ " and " ++ s2
combine _ s1 s2 = s1 ++ ", " ++ s2
results in "foo, bar and qux".
For a more general approach see Data.Foldable.WithIndex, which provides a foldable typeclass that also takes an index into account.
Related
So I want to partitision a List ItemModel in Elm into List (List ItemModel). List.partition only makes the list into two lists.
I wrote some code that makes the list into the parts I want (code below).
But it's not as nice of a solution as I'd like, and since it seems like an issue many people would have, I wonder are there better examples of doing this?
partition : List (ItemModel -> Bool) -> List ItemModel -> List (List ItemModel)
partition filters models =
let
filterMaybe =
List.head filters
in
case filterMaybe of
Just filter ->
let
part =
Tuple.first (List.partition filter models)
in
part :: (partition (List.drop 1 filters) models)
Nothing ->
[]
The returned list maps directly from the filters parameter, so it's actually pretty straightforward to do this using just List.map and List.filter (which is what you're really doing since you're discarding the remainder list returned from List.partition):
multifilter : List (a -> Bool) -> List a -> List (List a)
multifilter filters values =
filters |> List.map(\filter -> List.filter filter values)
Repeated partitioning needs to use the leftovers from each step as the input for the next step. This is different than simple repeated filtering of the same sequence by several filters.
In Haskell (which this question was initially tagged as, as well),
partitions :: [a -> Bool] -> [a] -> [[a]]
partitions preds xs = go preds xs
where
go [] xs = []
go (p:ps) xs = let { (a,b) = partition p xs } in (a : go ps b)
which is to say,
partitions preds xs = foldr g (const []) preds xs
where
g p r xs = let { (a,b) = partition p xs } in (a : r b)
or
-- mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])
partitions preds xs = snd $ mapAccumL (\xs p -> partition (not . p) xs) xs preds
Testing:
> partitions [ (<5), (<10), const True ] [1..15]
[[1,2,3,4],[5,6,7,8,9],[10,11,12,13,14,15]]
unlike the repeated filtering,
> [ filter p xs | let xs = [1..15], p <- [ (<5), (<10), const True ]]
[[1,2,3,4],[1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]]
In Elm, if I have an anonymous function
(\f x -> f x)
I can simplify it to
(<|)
Can the same be done for a two-parameter function where the parameters are arguments to another function?
(\x y -> f x y |> g)
I thought I could simply use
(f |> g)
but the compiler complains about the types.
Specifically, in one of the cases for my update function, I have something like this:
let
msgNames = [Foo, Bar]
values = ["f", "b"] // These values are actually derived
// by a more complicated operation
model_ = List.map2 (<|) msgNames values
|> List.foldl (\msg mod -> update msg mod |> Tuple.first)
model
in
( model_, Cmd.none )
I am trying to simplify the anonymous function argument to List.foldl to something like (update |> Tuple.first), but I get the following error from the compiler:
The right side of (|>) is causing a type mismatch.
159| update |> Tuple.first)
^^^^^^^^^^^
(|>) is expecting the right side to be a:
(Msg -> Model -> ( Model, Cmd Msg )) -> a
But the right side is:
(( Model, Cmd Msg )) -> Model
We can follow a few steps to simplify:
(\x y -> f x y |> g)
... can be written as
(\x y -> g (f x y))
... can be written as
(\x -> g << f x)
One more step and things get a little more confusing:
(((<<) g) << f)
This matches what you get from pointfree.io (which is Haskell where function composition is done using the . operator):
(g .) . f
If you are trying to improve readability, you might just want to make your own infix function:
infixr 9 <<<
(<<<) : (c -> d) -> (a -> b -> c) -> (a -> b -> d)
(<<<) g f x y =
g (f x y)
And now you can use it like this:
(g <<< f)
This is my first SML program. I am trying to write a function that returns the first number to the nth number of Hofstadter's Female or Male sequence in list form. What I have so far is:
val m = fn (n) => if n = 0 then 1 :: [] else m f (n - 1);
val f = fn (n) => if n = 0 then 0 :: [] else f m (n - 1);
You can learn about the sequence here:
https://en.wikipedia.org/wiki/Hofstadter_sequence#Hofstadter_Female_and_Male_sequences
The error that I am getting is:
[opening sequence.sml]
sequence.sml:1.49 Error: unbound variable or constructor: f
sequence.sml:1.47-1.58 Error: operator is not a function [tycon mismatch]
operator: int list
in expression:
(m <errorvar>) (n - 1)
val it = () : unit
How can I correct this?
I ended up taking this approach:
fun
m (n) = if n = 0 then 0 else n - (f (m (n - 1)))
and
f (n) = if n = 0 then 1 else n - (m (f (n - 1)));
val seq = fn n => List.tabulate((n), f);
It is quite slow. If anybody has a faster version, then I'd love to see it.
Although you have already fixed them, there were two problems with your original approach:
Function application is left-associative in SML so m f (n - 1) was being interpreted as (m f) (n - 1), not the desired m (f (n - 1)). You can fix this by explicitly specifying the bracketing m (f (n - 1)).
To be able to call f from m and m from f, you need to use the keyword fun instead of val on the first declaration (to make the function recursive), and the keyword and instead of fun or val on the second declaration (to make the function mutually recursive with the first function). This would look like
fun f n = ... (* I can call f or m from here! *)
and m n = ... (* I can call f or m from here! *)
To make it faster, you can memoize! The trick is to make f and m take as arguments memoized versions of themselves.
(* Convenience function: Update arr[i] to x, and return x. *)
fun updateAndReturn arr i x = (Array.update (arr, i, SOME x); x)
(*
* Look up result of f i in table; if it's not found, calculate f i and
* store in the table. The token is used so that deeper recursive calls
* to f can also try to store in the table.
*)
fun memo table f token i =
case Array.sub (table, i)
of NONE => updateAndReturn table i (f token i)
| SOME x => x
(*
* Given f, g, and n : int, returns a tuple (f', g') where f' and g' are memoized
* versions of f and g, respectively. f' and g' are defined only on the domain
* [0, n).
*)
fun memoizeMutual (f, g) n =
let
val fTable = Array.array (n, NONE)
val gTable = Array.array (n, NONE)
fun fMemo i = memo fTable f (fMemo, gMemo) i
and gMemo i = memo gTable g (gMemo, fMemo) i
in
(fMemo, gMemo)
end
fun female _ 0 = 1
| female (f, m) n = n - m (f (n - 1))
fun male _ 0 = 0
| male (m, f) n = n - f (m (n - 1))
fun hofstadter upTo =
let
val (male', female') = memoizeMutual (male, female) upTo
in
(List.tabulate (upTo, male'), List.tabulate (upTo, female'))
end
I renamed f and m to female and male. The memoized fMemo and gMemo are threaded through female and male by memoizeMutual. Interestingly, if we call male', then results for both male' and female' are memoized.
To confirm it's indeed faster, try evaluating hofstadter 10000. It's much faster than the forever that your version would take.
As a final note, the only recursive functions are fMemo and gMemo. Every other function I wrote could be written as an anonymous function (val memoizeMutual = fn ..., val female = fn ..., etc.), but I chose not to do so because the syntax for writing recursive functions is much more compact in SML.
To generalize this, you could replace the array version of memoizing with something like a hash table. Then we wouldn't have to specify the size of the memoization up front.
Lets say I have a list of points
[p1,p2,p3,p4,p5,p6, ...] or [[p1,p2,p3,...],[...]...]
were p1,p2,p3 are one stripe and p4,p5,p6 the other.
p1 - p4 - p7 ...
| / | / |
p2 - p5 - p8 ...
| / | / |
p3 - p6 - p9 ...
. . .
. . .
. . .
How can I transform this into a list of
[(p1,p2,p4), (p4,p5,p2), (p2,p3,p5), (p5,p6,p3), ...]
Is there a way without converting the list into an Array und use get and handle all the Maybes
First let's define how to split a square into two triangles:
squareToTriangles : a -> a -> a -> a -> List (a, a, a)
squareToTriangles topLeft botLeft topRight botRight =
[ (topLeft, botLeft, topRight)
, (topRight, botRight, botLeft)
]
Now, since squares are made of two lists, let's assume you can use a list of tuples as input. Now you can make triangles out of lists of left/right points:
triangles : List (a, a) -> List (a, a, a)
triangles list =
case list of
(tl, tr) :: ((bl, br) :: _ as rest) ->
List.append
(squareToTriangles tl bl tr br)
(triangles rest)
_ ->
[]
Of course, your input doesn't involve tuples, so let's define something that takes a list of lists as input:
triangleMesh : List (List a) -> List (a, a, a)
triangleMesh list =
case list of
left :: (right :: _ as rest) ->
List.append
(triangles <| List.map2 (,) left right)
(triangleMesh rest)
_ ->
[]
Now you can pass in your list of lists, such that:
triangleMesh [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
-- yields...
[(1,2,4),(4,5,2),(2,3,5),(5,6,3),(4,5,7),(7,8,5),(5,6,8),(8,9,6)]
Note that this can probably be optimized by using a better method than List.append, but the general algorithm holds.
You can simply pattern match on your list as follows:
toMesh: List Float -> List (Float, Float, Float)
toMesh list =
case list of
[ p1, p2, p3, p4, p5, p6] ->
Just [(p1,p2,p4), (p4,p5,p2), (p2,p3,p5), (p5,p6,p3)]
_ ->
[]
Say I have a List of records in elm:
[ { id = 1, magnitude = 100 }
, { id = 3, magnitude = 300 }
, { id = 2, magnitude = 200 } ]
and I want to get the record with the greatest magnitude value (300). What is a good way of doing this?
The docs gives an example of using the "maximum" -method, but it uses a simple list of integers. How is it done with records?
Update based on recommendation from #robertjlooby
There is a function called maximumBy which does exactly this in elm-community/list-extra. Example:
List.Extra.maximumBy .magnitude list
Original Answer
There are a few ways to achieve this.
This first way is more concise but it involves sorting the whole list, reversing it, then taking the head.
maxOfField : (a -> comparable) -> List a -> Maybe a
maxOfField field =
List.head << List.reverse << List.sortBy field
If you want something that's more efficient and only traverses the list once, here's a more efficient version:
maxOfField : (a -> comparable) -> List a -> Maybe a
maxOfField field =
let f x acc =
case acc of
Nothing -> Just x
Just y -> if field x > field y then Just x else Just y
in List.foldr f Nothing
An example of it in use:
list =
[ { id = 1, magnitude = 100 }
, { id = 3, magnitude = 300 }
, { id = 2, magnitude = 200 } ]
main =
text <| toString <| maxOfField .magnitude list
Here is a version that uses foldl and a default record:
bigger =
let
choose x y =
if x.magnitude > y.magnitude then
x
else
y
in
List.foldl choose {id = 0, magnitude = 0} items
Sebastian's answer add an arbitrary start value which could cause a problem if all your magnitudes were negative. I would adjust to
bigger items =
case items of
[] -> []
(h :: []) -> h
(h :: tail) ->
let
choose x y =
if x.magnitude > y.magnitude then
x
else
y
in
List.foldl choose h tail