Confused over behavior of List.mapi in F# - collections

I am building some equations in F#, and when working on my polynomial class I found some odd behavior using List.mapi
Basically, each polynomial has an array, so 3*x^2 + 5*x + 6 would be [|6, 5, 3|] in the array, so, when adding polynomials, if one array is longer than the other, then I just need to append the extra elements to the result, and that is where I ran into a problem.
Later I want to generalize it to not always use a float, but that will be after I get more working.
So, the problem is that I expected List.mapi to return a List not individual elements, but, in order to put the lists together I had to put [] around my use of mapi, and I am curious why that is the case.
This is more complicated than I expected, I thought I should be able to just tell it to make a new List starting at a certain index, but I can't find any function for that.
type Polynomial() =
let mutable coefficients:float [] = Array.empty
member self.Coefficients with get() = coefficients
static member (+) (v1:Polynomial, v2:Polynomial) =
let ret = List.map2(fun c p -> c + p) (List.ofArray v1.Coefficients) (List.ofArray v2.Coefficients)
let a = List.mapi(fun i x -> x)
match v1.Coefficients.Length - v2.Coefficients.Length with
| x when x < 0 ->
ret :: [((List.ofArray v1.Coefficients) |> a)]
| x when x > 0 ->
ret :: [((List.ofArray v2.Coefficients) |> a)]
| _ -> [ret]

I think that a straightforward implementation using lists and recursion would be simpler in this case. An alternative implementation of the Polynomial class might look roughly like this:
// The type is immutable and takes initial list as constructor argument
type Polynomial(coeffs:float list) =
// Local recursive function implementing the addition using lists
let rec add l1 l2 =
match l1, l2 with
| x::xs, y::ys -> (x+y) :: (add xs ys)
| rest, [] | [], rest -> rest
member self.Coefficients = coeffs
static member (+) (v1:Polynomial, v2:Polynomial) =
// Add lists using local function
let newList = add v1.Coefficients v2.Coefficients
// Wrap result into new polynomial
Polynomial(newList)
It is worth noting that you don't really need mutable field in the class, since the + operator creates and returns a new instance of the type, so the type is fully immutable (as you'd usually want in F#).
The nice thing in the add function is that after processing all elements that are available in both lists, you can simply return the tail of the non-empty list as the rest.
If you wanted to implement the same functionality using arrays, then it may be better to use a simple for loop (since arrays are, in principle, imperative, the usual imperative patterns are usually the best option for dealing with them). However, I don't think there is any particular reason for preferring arrays (maybe performance, but that would have to be evaluated later during the development).
As Pavel points out, :: operator appends a single element to the front of a list (see the add function above, which demonstrates that). You could write what you wanted using # which concatenates lists, or using Array.concat (which concatenates a sequence of arrays).
An implementation using higher-order functions and arrays is also possible - the best version I can come up with would look like this:
let add (a1:_[]) (a2:_[]) =
// Add parts where both arrays have elements
let l = min a1.Length a2.Length
let both = Array.map2 (+) a1.[0 .. l-1] a2.[0 .. l-1]
// Take the rest of the longer array
let rest =
if a1.Length > a2.Length
then a1.[l .. a1.Length - 1]
else a2.[l .. a2.Length - 1]
// Concatenate them
Array.concat [ both; rest ]
add [| 6; 5; 3 |] [| 7 |]
This uses slices (e.g. a.[0 .. l]) which give you a part of an array - you can use these to take the parts where both arrays have elements and the remaining part of the longer array.

I think you're misunderstanding what operator :: does. It's not used to concatenate two lists. It's used to prepend a single element to the list. Consequently, it's type is:
'a -> 'a list -> 'a list
In your case, you're giving ret as a first argument, and ret is itself a float list. Consequently, it expects the second argument to be of type float list list - hence why you need to add an extra [] to the second argument to make it to compile - and that will also be the result type of your operator +, which is probably not what you want.
You can use List.concat to concatenate two (or more) lists, but that is inefficient. In your example, I don't see the point of using lists at all - all this converting back & forth is going to be costly. For arrays, you can use Array.append, which is better.
By the way, it's not clear what is the purpose of mapi in your code at all. It's exactly the same as map, except for the index argument, but you're not using it, and your mapping is the identity function, so it's effectively a no-op. What's it about?

Related

Recursive functions and pattern matching in ocaml

The following code snippet comes from the official OCaml website:
# let rec compress = function
| a :: (b :: _ as t) -> if a = b then compress t else a :: compress t
| smaller -> smaller;;
val compress : 'a list -> 'a list = <fun>
The above function 'compresses' a list with consecutive, duplicative elements, e.g. :
# compress ["a";"a";"a";"a";"b";"c";"c";"a";"a";"d";"e";"e";"e";"e"];;
- : string list = ["a"; "b"; "c"; "a"; "d"; "e"]
I'm having a devil of a time understanding the logic of the above code. I'm used to coding imperatively, so this recursive, functional approach, combined with OCamls laconic - but obscure - syntax is causing me to struggle.
For example, where is the base case? Is it smaller -> smaller? I know smaller is a variable, or an identifier, but what is it returning (is returning even the right term in OCaml for what's happening here)?
I know that lists in OCaml are singly linked, so I'm also wondering if a new list is being generated, or if elements of the existed list are being cut? Since OCaml is functional, I'm inclined to think that lists are not mutable - is that correct? If you want to change a list, you essentially need to generate a new list with the elements you're seeking to add (or with the elements you're seeking to excise absent). Is this a correct understanding?
Yes, the base case is this:
| smaller -> smaller
The first pattern of the match expression matches any list of length 2 or greater. (It would be good to make sure you see why this is the case.)
Since OCaml matches patterns in order, the base case matches lists of lengths 0 and 1. That's why the programmer chose the name smaller. They were thinking "this is some smaller list".
The parts of a match statement look like this in general:
| pattern -> result
Any names in the pattern are bound to parts of the value matched against the pattern (as you say). So smaller is bound to the whole list. So in sum, the second part of the match says that if the list is of length 0 or 1, the result should just be the list itself.
Lists in OCaml are immutable, so it's not possible for the result of the function to be a modified version of the list. The result is a new list, unless the list is already a short list (of length 0 or 1).
So, what you say about the immutability of OCaml lists is exactly correct.

How do I create a function which reverses an int list in OCaml?

I have a type intlist:
type intlist = Nil | Cons of int * intlist
and I'd now like to write a function that reverses the list. I have no real idea how to go about this. in class, our prof defined a function within a function, I believe, but I don't have the solution with me. How would one do this, simply? I would appreciate if we can keep the coding rather rudimentary, seeing as I'm relatively new to this.
So far, I only have
let reverse (l : intlist) : intlist =
match l with
Nil -> Nil
| Cons(a, Nil) -> Cons(a, Nil)
This is how I tend to create these kinds of functions, so I've written the trivial part (which granted, may not actually be what I need to start with). Any help is appreciated, thanks!
You indeed need a helper function, since for the reversing you need to build another list, so you need a function that will recurse into one list, while building another list, i.e., that has two arguments. In fact, this helper function is called rev_append and it is appending a reversed contents of one list to another. But let's try to make it using a helper function defined in the scope of the rev function:
let rev xs =
let rec loop xs ys = match xs with
| Nil -> ys
| Cons (x,xs) -> loop xs (Cons (x,ys)) in
loop xs Nil
So, to reverse a list we just take each element of a list and put it into another list. Since list behaves like a stack we are getting the reversed list for free. It is like Hanoi towers, when you pick elements from one list (tower) and put them to another, they will end up in a reversed order.

Function calling for each element in a list

Using only recursion (ie. no loops of any sort), given a list of elements, how can I call a function each time for every element of the list using that element as an argument each time in OCaml? Fold and map would not work because although they are applying a function to each element, it returns a list of whatever function I called on each element, which is not what I want.
To better illustrate what I'm essentially trying to do in OCaml, here's the idea of what I want in Ruby code:
arr.each {|x| some_function x}
but I must do this using only recursion and no iter functions
The correct recursive function is described as:
if the list is empty, do nothing;
else, process the first element and then the tail of the list.
The corresponding code is:
let rec do_all f lst =
match lst with
| [] -> ()
| x :: xs -> f x; do_all f xs
A fairly general template for a recursive function would be this:
let rec f x =
if x is trival to handle then
handle x
else
let (part, rest) = division of x into smaller parts in
let part_result = handle_part part in
let recursive_result = f rest in
combine part_result recursive_result
Since you don't need a result, you can skip a lot of this.
Which parts of this template seem most difficult to do for your problem?
Update
(As #EduardoLeĆ³n points out, when working with lists you can test for a trivial list and break down the list into smaller parts using pattern matching. Pattern matching is cool.)
Update 2
My question is sincere. Which part are you having trouble with? Otherwise we don't know what to suggest.

Choosing unique items from a list, using recursion

As follow up to yesterday's question Erlang: choosing unique items from a list, using recursion
In Erlang, say I wanted choose all unique items from a given list, e.g.
List = [foo, bar, buzz, foo].
and I had used your code examples resulting in
NewList = [bar, buzz].
How would I further manipulate NewList in Erlang?
For example, say I not only wanted to choose all unique items from List, but also count the total number of characters of all resulting items from NewList?
In functional programming we have patterns that occur so frequently they deserve their own names and support functions. Two of the most widely used ones are map and fold (sometimes reduce). These two form basic building blocks for list manipulation, often obviating the need to write dedicated recursive functions.
Map
The map function iterates over a list in order, generating a new list where each element is the result of applying a function to the corresponding element in the original list. Here's how a typical map might be implemented:
map(Fun, [H|T]) -> % recursive case
[Fun(H)|map(Fun, T)];
map(_Fun, []) -> % base case
[].
This is a perfect introductory example to recursive functions; roughly speaking, the function clauses are either recursive cases (result in a call to iself with a smaller problem instance) or base cases (no recursive calls made).
So how do you use map? Notice that the first argument, Fun, is supposed to be a function. In Erlang, it's possible to declare anonymous functions (sometimes called lambdas) inline. For example, to square each number in a list, generating a list of squares:
map(fun(X) -> X*X end, [1,2,3]). % => [1,4,9]
This is an example of Higher-order programming.
Note that map is part of the Erlang standard library as lists:map/2.
Fold
Whereas map creates a 1:1 element mapping between one list and another, the purpose of fold is to apply some function to each element of a list while accumulating a single result, such as a sum. The right fold (it helps to think of it as "going to the right") might look like so:
foldr(Fun, Acc, [H|T]) -> % recursive case
foldr(Fun, Fun(H, Acc), T);
foldr(_Fun, Acc, []) -> % base case
Acc.
Using this function, we can sum the elements of a list:
foldr(fun(X, Sum) -> Sum + X, 0, [1,2,3,4,5]). %% => 15
Note that foldr and foldl are both part of the Erlang standard library, in the lists module.
While it may not be immediately obvious, a very large class of common list-manipulation problems can be solved using map and fold alone.
Thinking recursively
Writing recursive algorithms might seem daunting at first, but as you get used to it, it turns out to be quite natural. When encountering a problem, you should identify two things:
How can I decompose the problem into smaller instances? In order for recursion to be useful, the recursive call must take a smaller problem as its argument, or the function will never terminate.
What's the base case, i.e. the termination criterion?
As for 1), consider the problem of counting the elements of a list. How could this possibly be decomposed into smaller subproblems? Well, think of it this way: Given a non-empty list whose first element (head) is X and whose remainder (tail) is Y, its length is 1 + the length of Y. Since Y is smaller than the list [X|Y], we've successfully reduced the problem.
Continuing the list example, when do we stop? Well, eventually, the tail will be empty. We fall back to the base case, which is the definition that the length of the empty list is zero. You'll find that writing function clauses for the various cases is very much like writing definitions for a dictionary:
%% Definition:
%% The length of a list whose head is H and whose tail is T is
%% 1 + the length of T.
length([H|T]) ->
1 + length(T);
%% Definition: The length of the empty list ([]) is zero.
length([]) ->
0.
You could use a fold to recurse over the resulting list. For simplicity I turned your atoms into strings (you could do this with list_to_atom/1):
1> NewList = ["bar", "buzz"].
["bar","buzz"]
2> L = lists:foldl(fun (W, Acc) -> [{W, length(W)}|Acc] end, [], NewList).
[{"buzz",4},{"bar",3}]
This returns a proplist you can access like so:
3> proplists:get_value("buzz", L).
4
If you want to build the recursion yourself for didactic purposes instead of using lists:
count_char_in_list([], Count) ->
Count;
count_char_in_list([Head | Tail], Count) ->
count_char_in_list(Tail, Count + length(Head)). % a string is just a list of numbers
And then:
1> test:count_char_in_list(["bar", "buzz"], 0).
7

Miranda going through lists

is there an easy way to go through a list?
lets say i wanted to access the 5th data on the list not knowing it was a B
["A","A","A","A","B","A","A","A","A"]
is there a way i can do it without having to sort through the list?
I do not know Miranda that well, but I expect the functions skip and take are available.
you can address the 5th element by making a function out of skip and take. When skip and take are not available, it is easy to create them yourself.
skip: skips the y number of elements in a list, when y is greater than the number of items in the list, it will return an empty list
take: takes the first y number of elements in a list, when y is greater than the number of items in the list, the full list will be returned.
skip y [] = []
skip 0 xs = xs
skip y (x:xs) = skip xs (y-1)
take y [] = []
take 0 xs = []
take y (x:xs) = x : take (y-1) xs
elementAt x xs = take 1 (skip x xs)
Lists are inductive datatypes. This means that functions defined over lists - for instance, accessing the nth element - are defined by recursion. The data structure you are looking for appears to be an array, which allows constant time lookup. The easiest way to find the element at an index in a list is directly:
lookup :: Int -> [a] -> Maybe [a]
lookup n [] = Nothing
lookup 0 (x:xs) = Just x
lookup n (x:xs) = lookup (n - 1) xs
Another way to do this would be to use the ! operator. Let's say you have a program with defined data in the list, such as:
plist = [A,A,A,A,B,A,A,A,A]
then executing plist!4 will give you the 5th element of that list. (4 being the 5th unit if you include 0,1,2,3,4)
So plist!4 returns B.
Lists are not arrays.
You can only access elements beginning from first. Think of lists as streams (like a song playing in radio). Lists may be of infinite length (as radio never stops).
Most programmers uses "syntactic" sugar, which hides the nature of lists behind an easier syntax.
Miranda automatically loads a default library named stdenv.m, which you can study.
Now, let's think about your problem:
You want to ignore ("drop") all elements before the 5th and then get the first element from the rest of the ramaining list.
This is expressed in Miranda as:
nth :: num -> [*] -> *
nth n = hd . drop (n-1)
This is a function with explicit type declaration to see, that function works with every list (elements are of wildcard type *).
Sample:
plist :: [[char]]
plist = ["A","A","A","A","B","A","A","A","A"]
result :: [char]
result = nth 5 plist
If you want to code your functions with error handling, you need techniques to catch that there is no 5th element in your list.
As seen above, one technique is "Maybe". Another is continuations.
A bad technique is to check the length of list first, because this will crash with infinite lists.

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