let rec transpose list = match list with
| [] -> []
| [] :: xss -> transpose xss
| (x::xs) :: xss ->
(x :: List.map List.hd xss) :: transpose (xs :: List.map List.tl xss)
I found this code on this website and it transposes a list list or in my case a matrix from n × m to m × n.The only problem is I have no idea how it works. Can anyone please explain it step by step, line by line.
You should provide the link to the code as well, for context, or look on that question for an explanation.
Either way, I suspect other new OCaml programmers will come to this question looking for an answer.
-- After a quick Google search I believe this was the question you found the code from.
let rec transpose list = match list with
Create a function called transpose, make it recursive (rec) as we will make the function call itself. It will take an input parameter called list, where the type is determined by OCaml's type-system later.
Match the input parameter list with the following cases. That is, when the function finds a pattern matching those cases, it will execute that particular line of code. If code written does not cover all cases, OCaml will notify that not all cases are exhaustive and will cause a match_failure if it encounters cases that it doesn't cover.
First case:
| [] -> []
If we hit an empty list, return an empty list. This will end the recursion as the empty list is appended to the final result and returned. The empty list is a base case, where we meet a list with nothing inside, or it has recursed till there is nothing left. (the transpose of an empty list is ... and empty list)
| [] :: xss -> transpose xss
This is the case where there are still items in the list. Since it is a list of list, each element in the list we pass in is another list. i.e, [[1;2];[3;4]], so this function handles the case where we have [[];[1;2];[5;6]], it will skip the empty inner list, and then call transpose on the other [[1;2];[5;6]], since we don't need to transpose an empty list. Note that xss is the rest of the list (hence list of list), not just the next element only (which is just a list).
| (x::xs) :: xss ->
(x :: List.map List.hd xss) :: transpose (xs :: List.map List.tl xss)
This is the first case that will be matched when we pass in any regular list of list of length > 1, OCaml is able to match and also "unwrap" each element by breaking it down to fit the pattern, so we can immediately compute those elements without using nested match statements. In this case, we split the list of list into the (x::xs) list (first element) and the remaining list of list, xss. We could have also done it as x::xss but x will be a list that we need to break down into individual elements again.
For example, in [[1;2];[3;4]], (x::xs) will be [1;2] and xss will be [[3;4]]
(x :: List.map List.hd xss) will take the first element in the first list, in the given example, x will be 1. And apply that value to the List.hd of every item in xss list, (List.Map applies a function to every value in the list).
This is where it seems a little complicated. How you can look at it is: When you call List.hd [1;2;3] it will return 1 since it's the head. So when we do 3 :: List.map List.hd [[1;2]] it will return int list = [3; 1]. and it we did 3 :: List.map List.hd [[1;2];[4;5]] we get [3;1;4] as we are getting the head of every list inside the larger list, and appending 3 at the front of the resulting list.
If you did List.map List.tl [[1;2];[4;5]] instead your return type would be [2;5] since List.map applies the List.tl function to every list. You can use List.map to apply say a multiplication function (or anything complicated) to each item in the list in a quick and easy way.
The remaining transpose (xs :: List.map List.tl xss) just calls transpose to the remaining items in the list, which is the tail values of xss (all lists), appended with elements in xs. Try this out on paper if you feel you need a more concrete grasp on it.
Related
I am trying to understand this piece of code which returns the all possible combinations of [a] passed to it:
-- Infinite list of all combinations for a given value domain
allCombinations :: [a] -> [[a]]
allCombinations [] = [[]]
allCombinations values = [] : concatMap (\w -> map (:w) values)
(allCombinations values)
Here i tried this sample input:
ghci> take 7 (allCombinations [True,False])
[[],[True],[False],[True,True],[False,True],[True,False],[False,False]]
Here it doesn't seems understandable to me which is that how the recursion will eventually stops and will return [ [ ] ], because allCombinations function certainly doesn't have any pointer which moves through the list, on each call and when it meets the base case [ ] it returns [ [ ] ]. According to me It will call allCombinations function infinite and will never stop on its own. Or may be i am missing something?
On the other hand, take only returns the first 7 elements from the final list after all calculation is carried out by going back after completing recursive calls. So actually how recursion met the base case here?
Secondly what is the purpose of concatMap here, here we could also use Map function here just to apply function to the list and inside function we could arrange the list? What is actually concatMap doing here. From definition it concatMap tells us it first map the function then concatenate the lists where as i see we are already doing that inside the function here?
Any valuable input would be appreciated?
Short answer: it will never meet the base case.
However, it does not need to. The base case is most often needed to stop a recursion, however here you want to return an infinite list, so no need to stop it.
On the other hand, this function would break if you try to take more than 1 element of allCombination [] -- have a look at #robin's answer to understand better why. That is the only reason you see a base case here.
The way the main function works is that it starts with an empty list, and then append at the beginning each element in the argument list. (:w) does that recursively. However, this lambda alone would return an infinitely nested list. I.e: [],[[True],[False]],[[[True,True],[True,False] etc. Concatmap removes the outer list at each step, and as it is called recursively this only returns one list of lists at the end. This can be a complicated concept to grasp so look for other example of the use of concatMap and try to understand how they work and why map alone wouldn't be enough.
This obviously only works because of Haskell lazy evaluation. Similarly, you know in a foldr you need to pass it the base case, however when your function is supposed to only take infinite lists, you can have undefined as the base case to make it more clear that finite lists should not be used. For example, foldr f undefined could be used instead of foldr f []
#Lorenzo has already explained the key point - that the recursion in fact never ends, and therefore this generates an infinite list, which you can still take any finite number of elements from because of Haskell's laziness. But I think it will be helpful to give a bit more detail about this particular function and how it works.
Firstly, the [] : at the start of the definition tells you that the first element will always be []. That of course is the one and only way to make a 0-element list from elements of values. The rest of the list is concatMap (\w -> map (:w) values) (allCombinations values).
concatMap f is as you observe simply the composition concat . (map f): it applies the given function to every element of the list, and concatenates the results together. Here the function (\w -> map (:w) values) takes a list, and produces the list of lists given by prepending each element of values to that list. For example, if values == [1,2], then:
(\w -> map (:w) values) [1,2] == [[1,1,2], [2,1,2]]
if we map that function over a list of lists, such as
[[], [1], [2]]
then we get (still with values as [1,2]):
[[[1], [2]], [[1,1], [2,1]], [[1,2], [2,2]]]
That is of course a list of lists of lists - but then the concat part of concatMap comes to our rescue, flattening the outermost layer, and resulting in a list of lists as follows:
[[1], [2], [1,1], [2,1], [1,2], [2,2]]
One thing that I hope you might have noticed about this is that the list of lists I started with was not arbitrary. [[], [1], [2]] is the list of all combinations of size 0 or 1 from the starting list [1,2]. This is in fact the first three elements of allCombinations [1,2].
Recall that all we know "for sure" when looking at the definition is that the first element of this list will be []. And the rest of the list is concatMap (\w -> map (:w) [1,2]) (allCombinations [1,2]). The next step is to expand the recursive part of this as [] : concatMap (\w -> map (:w) [1,2]) (allCombinations [1,2]). The outer concatMap
then can see that the head of the list it's mapping over is [] - producing a list starting [1], [2] and continuing with the results of appending 1 and then 2 to the other elements - whatever they are. But we've just seen that the next 2 elements are in fact [1] and [2]. We end up with
allCombinations [1,2] == [] : [1] : [2] : concatMap (\w -> map (:w) values) [1,2] (tail (allCombinations [1,2]))
(tail isn't strictly called in the evaluation process, it's done by pattern-matching instead - I'm trying to explain more by words than explicit plodding through equalities).
And looking at that we know the tail is [1] : [2] : concatMap .... The key point is that, at each stage of the process, we know for sure what the first few elements of the list are - and they happen to be all 0-element lists with values taken from values, followed by all 1-element lists with these values, then all 2-element lists, and so on. Once you've got started, the process must continue, because the function passed to concatMap ensures that we just get the lists obtained from taking every list generated so far, and appending each element of values to the front of them.
If you're still confused by this, look up how to compute the Fibonacci numbers in Haskell. The classic way to get an infinite list of all Fibonacci numbers is:
fib = 1 : 1 : zipWith (+) fib (tail fib)
This is a bit easier to understand that the allCombinations example, but relies on essentially the same thing - defining a list purely in terms of itself, but using lazy evaluation to progressively generate as much of the list as you want, according to a simple rule.
It is not a base case but a special case, and this is not recursion but corecursion,(*) which never stops.
Maybe the following re-formulation will be easier to follow:
allCombs :: [t] -> [[t]]
-- [1,2] -> [[]] ++ [1:[],2:[]] ++ [1:[1],2:[1],1:[2],2:[2]] ++ ...
allCombs vals = concat . iterate (cons vals) $ [[]]
where
cons :: [t] -> [[t]] -> [[t]]
cons vals combs = concat [ [v : comb | v <- vals]
| comb <- combs ]
-- iterate :: (a -> a ) -> a -> [a]
-- cons vals :: [[t]] -> [[t]]
-- iterate (cons vals) :: [[t]] -> [[[t]]]
-- concat :: [[ a ]] -> [ a ]
-- concat . iterate (cons vals) :: [[t]]
Looks different, does the same thing. Not just produces the same results, but actually is doing the same thing to produce them.(*) The concat is the same concat, you just need to tilt your head a little to see it.
This also shows why the concat is needed here. Each step = cons vals is producing a new batch of combinations, with length increasing by 1 on each step application, and concat glues them all together into one list of results.
The length of each batch is the previous batch length multiplied by n where n is the length of vals. This also shows the need to special case the vals == [] case i.e. the n == 0 case: 0*x == 0 and so the length of each new batch is 0 and so an attempt to get one more value from the results would never produce a result, i.e. enter an infinite loop. The function is said to become non-productive, at that point.
Incidentally, cons is almost the same as
== concat [ [v : comb | comb <- combs]
| v <- vals ]
== liftA2 (:) vals combs
liftA2 :: Applicative f => (a -> b -> r) -> f a -> f b -> f r
So if the internal order of each step results is unimportant to you (but see an important caveat at the post bottom) this can just be coded as
allCombsA :: [t] -> [[t]]
-- [1,2] -> [[]] ++ [1:[],2:[]] ++ [1:[1],1:[2],2:[1],2:[2]] ++ ...
allCombsA [] = [[]]
allCombsA vals = concat . iterate (liftA2 (:) vals) $ [[]]
(*) well actually, this refers to a bit modified version of it,
allCombsRes vals = res
where res = [] : concatMap (\w -> map (: w) vals)
res
-- or:
allCombsRes vals = fix $ ([] :) . concatMap (\w -> map (: w) vals)
-- where
-- fix g = x where x = g x -- in Data.Function
Or in pseudocode:
Produce a sequence of values `res` by
FIRST producing `[]`, AND THEN
from each produced value `w` in `res`,
produce a batch of new values `[v : w | v <- vals]`
and splice them into the output sequence
(by using `concat`)
So the res list is produced corecursively, starting from its starting point, [], producing next elements of it based on previous one(s) -- either in batches, as in iterate-based version, or one-by-one as here, taking the input via a back pointer into the results previously produced (taking its output as its input, as a saying goes -- which is a bit deceptive of course, as we take it at a slower pace than we're producing it, or otherwise the process would stop being productive, as was already mentioned above).
But. Sometimes it can be advantageous to produce the input via recursive calls, creating at run time a sequence of functions, each passing its output up the chain, to its caller. Still, the dataflow is upwards, unlike regular recursion which first goes downward towards the base case.
The advantage just mentioned has to do with memory retention. The corecursive allCombsRes as if keeps a back-pointer into the sequence that it itself is producing, and so the sequence can not be garbage-collected on the fly.
But the chain of the stream-producers implicitly created by your original version at run time means each of them can be garbage-collected on the fly as n = length vals new elements are produced from each downstream element, so the overall process becomes equivalent to just k = ceiling $ logBase n i nested loops each with O(1) space state, to produce the ith element of the sequence.
This is much much better than the O(n) memory requirement of the corecursive/value-recursive allCombsRes which in effect keeps a back pointer into its output at the i/n position. And in practice a logarithmic space requirement is most likely to be seen as a more or less O(1) space requirement.
This advantage only happens with the order of generation as in your version, i.e. as in cons vals, not liftA2 (:) vals which has to go back to the start of its input sequence combs (for each new v in vals) which thus must be preserved, so we can safely say that the formulation in your question is rather ingenious.
And if we're after a pointfree re-formulation -- as pointfree can at times be illuminating -- it is
allCombsY values = _Y $ ([] :) . concatMap (\w -> map (: w) values)
where
_Y g = g (_Y g) -- no-sharing fixpoint combinator
So the code is much easier understood in a fix-using formulation, and then we just switch fix with the semantically equivalent _Y, for efficiency, getting the (equivalent of the) original code from the question.
The above claims about space requirements behavior are easily tested. I haven't done so, yet.
See also:
Why does GHC make fix so confounding?
Sharing vs. non-sharing fixed-point combinator
I want to generate all possible combinations of a 6 element list [x1,x2,x3,x4,x5,x6] with each xi being a number between 0 and 20.
I want to generate all possible combinations of such list, apply a function (takes the list as input and outputs a magical Int) to each list, then output the results to a list of tuples. So the list of tuples looks like
[([x11,x21,x31,x41,x51,x61],Int1), ([x12,x22,x32,x42,x52,x62],Int2), ...]
I tried to this by hand by quickly realised that there are too many combinations and it is practically impossible to do by hand.
The combinations are like [0,0,0,0,0,0], [1,7,0,10,11,6], [7,7,7,7,6,6], [20,20,20,20,20,20] and so on.
I know how to generate all combinations of a list and put them in a list of lists (because I asked this before)
foo [] = [[]]
foo (x:xs) = foo xs ++ map (x:) (foo xs)
What I want to achieve this time is different because I am not trying to generate the different combinations within a particular list, I am trying to generate all 6 element lists.
As far as I can tell, you want a cartesian product (denoted by × here) of 6 lists [0..20].
So essentially, something like this:
[0..20]×[0..20]×[0..20]×[0..20]×[0..20]×[0..20]
That's a lot of elements (85,766,121 to be exact). But, it could be done.
Perhaps an easier to understand version is as follows.
Let us define a function, cart, that would do something like a cartesian product over a list and a list of lists:
let cart xs ls = [x:l | x <- xs, l <- ls]
This function will take all elements from xs, and all elements from ls and build a list of lists with all possible concatenations.
Now, we need a base case. Suppose you wanted a list of lists of a single element instead of six. How would you apply our cart function? Well, since it adds each element from first argument to every element in second argument, we can pass a list of a single empty list as a second argument, and [0..20] as first argument:
cart [0..20] [[]]
We get
[[0],[1],[2],[3],[4],[5],[6],[7],[8],[9],[10],[11],[12],[13],[14],[15],[16],[17],[18],[19],[20]]
Great. Now we just apply it 6 times to its own result, starting with [[]]:
foldr ($) [[]] $ replicate 6 (cart [0..20])
($) is function application.
replicateM from Control.Monad (defined as sequence of n monadic actions: replicateM n x = sequence (replicate n x)) basically does the same when applied to lists (see e.g. Why does application of `sequence` on List of Lists lead to computation of its Cartesian Product? for why exactly that is) So shorter answer would be like this:
replicateM 6 [0..20]
You can map over those after this, e.g.
map (\x -> (x, magicFunction x)) $ replicateM 6 [0..20]
foo f xs = map (\ys -> (ys,f ys)) (replicateM (length xs) xs)
Here replicateM (length xs) xs will generate all combinations of the elements in xs. Then you just map over it.
Results in GHCI:
>import Control.Monad
>let foo f xs = map (\ys -> (ys,f ys)) (replicateM (length xs) xs)
>foo sum [1,2]
[([1,1],2),([1,2],3),([2,1],3),([2,2],4)]
I'm trying to add an int list list with another int list list using the append function, but I can't get it to work the way I want.
Say that I want to append [[1,2,3,4,5]] with [6,7] so that I get [[1,2,3,4,5,6,7]].
Here's my attempt: [1,2,3,4,5]::[]#[6,7]::[], but it just gives me the list I want to append as a list of its own instead of the two lists combined into one, like this: [[1,2,3,4,5],[6,7]].
How can I re-write the operation to make it return [[1,2,3,4,5,6,7]]?
Your question is too unspecific. You are dealing with nested lists. Do you want to append the second list to every inner list of the nested list, or only the first one? Your example doesn't tell.
For the former:
fun appendAll xss ys = List.map (fn xs => xs # ys) xss
For the latter:
fun appendHd [] ys = raise Empty
| appendHd (xs::xss) ys = (xs # ys)::xss
However, both of these should rarely be needed, and I somehow feel that you are trying to solve the wrong problem if you end up there.
I'm still fairly new to OCaml, and would like some assistance on optimizing code.
I'm trying to multiply each element of a given list by the list's last element.
Here's a snippet of my code:
(* Find the last element of a function *)
let rec lastE = function
| [] -> []
| [x] -> x
| _ :: t -> lastE t;;
(*multiply a list by the last element*)
let rec lmul list =
match list with
[] -> []
| hd::tl -> (hd *. (lastE tl)) :: lmul tl;;
When I run the code I get this error message:
Error: This expression has type float list but
an expression was expected of type 'a list list
I'm been studying it for a while, but any assistance on this problem will be much appreciated.
To rephrase differently what Dave Newman is telling you, your basic problem is that lastE needs to handle an empty list differently. If lastE is supposed to return a number, it has to return a number in all cases. As it stands, lastE returns a list when it receives an empty list.
If you don't want to use List.map (again as Dave Newman suggests), you might at least consider calling lastE just once rather than once for each element of the list. This will make a big difference for long lists.
how to get the value of the last element of a List? I've noted that List.hd (or .Head) return an item, while List.tl (or .Tail) returns a List.
Is rev the List and get the hd the only way around? Thanks.
Try this function. It uses recursion, though it gets optimised to iteration anyway since it's tail recursion. In any case, it is most likely quicker than reversing the entire list (using List.rev).
let rec last = function
| hd :: [] -> hd
| hd :: tl -> last tl
| _ -> failwith "Empty list."
The answer of Pavel Minaev is definitely worth taking into account, however. Nonetheless, the algorithm you have requested may be useful in some rare cases, and is the most efficient way to go about the task.
In general, if you need to do this, you're doing something wrong. Since F# lists are single-linked, accessing the last element is costly - O(N), where N is size of list. Try to rewrite your algorithm so that you always access the first element, not the last (which is O(1)). If you cannot do so, chances are good that your choice of list for a data structure wasn't correct in the first place.
A quick & dirty way of doing it is by using List.reduce. Assuming the list is called ls,
let lastElement ls = List.reduce (fun _ i -> i) ls
As for efficiency, I agree with Pavel.
A more concise version based on Mitch's answer:
let lastItem = myList |> List.rev |> List.head
The myList list is sent to List.rev function. The result is then processed by List.head
Agreed, not so efficient to get the last element of list, or any other "enumerable" sequence. That said, this function already exists in the Seq module, Seq.last.
As a novice F# developer, I don't see what the harm is in doing the following
let mylist = [1;2;3;4;5]
let lastValue = mylist.[mylist.Length - 1]
Imperative in nature? Yes but no need for recursion.
The regular way to work with lists in F# is to use recursion. The first item in a list is the head (obviously) and the rest of the list is the tail (as oppose to the last item). So when a function recieves a list it processes the head and then recursively processes the rest of the list (the tail).
let reversedList = List.rev originalList
let tailItem = List.hd reversedList
I think you can just write
list.[0..list.Length-1]
You can call List.Head to get the first element of a list, such that the below expression evaluates to true:
let lst = [1;2;3;4;5]
List.head lst = 1
However, calling List.Tail will return every element in the list after the first element, such that the below expression is true:
let lst = [1;2;3;4;5]
List.tail lst = [2;3;4;5]
Like some other people have mentioned, there isn't an efficient way in F# to get the tail end of a list, basic lists just aren't built with that functionality in mind. If you really want to get the last element you're going to have to reverse your list first, and then take the new head (which was the previous tail).
let lst = [1;2;3;4;5]
(List.head (List.rev lst) ) = 5
Below code worked fine with me, I've an array of integers, want to start from the 5th item, then take it minus the item number
Sum of [Array(xi) - Array(xi-5)] where i start at 5
The code used is:
series |> Array.windowed 5
|> Array.fold (fun s x ->
(x |> Array.rev |> Array.head) - (x |> Array.head) + s) 0
|> float
It's a very old question, but just in case someone comes here:
with FSharp 5, you can do x.[^index] where index will start at the end of the array / list.
let a = [1;2;3;4;5;6;7;8;9]
a.[^0] is 9
a.[^1] is 8
etc