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I don't understand the error message I'm getting or what's wrong with what I'm trying to do
I just want to use List.fold_left to apply my add1 function to this list [1,2,3]
My add1 function should just add 1 to each element, so I would get [2, 3, 4]
My main goal in doing this exercise is just to experiment with List.fold_left. I don't actually care about adding 1, I just choose that function because it seemed easy to write (I'm an ocaml beginner).
My ultimate goal is actually to populate the keys of a empty StringMap using List.fold_left and a function already written elsewhere, so if anyone has insight on that it would also be appreciated
Here's the 1st try (which I tried twice)
let rec add1 = function
| [] -> []
| h::t -> (h+1)::(add1 t) in List.fold_left add1 [1, 2, 3];;
Here's the 2nd try
let a(b) =
let rec add1 = function
| [] -> []
| h::t -> (h+1)::(add1 t)
in
let c = List.fold_left add1 b
in a [1,2,3];;
I think you should start with:
let add x = x + 1
And then build a function that applies a function to a list via List.fold_left:
let apply_f_to_list_elements fn lst = (*use List.fold_left here*)
Are you sure you want List.fold_left and not List.map?
It may help you to see how fold_left can be implemented.
let rec fold_left f init lst =
match lst with
| [] -> init
| x::xs -> fold_left f (f init x) xs
So consider what's happening when something like a sum function works, when implemented in term of fold_left.
let sum lst =
fold_left (+) 0 lst
If we evaluate sum [1; 2; 3; 4]:
sum [1; 2; 3; 4]
fold_left (+) 0 [1; 2; 3; 4]
fold_left (+) (0 + 1) [2; 3; 4]
fold_left (+) (1 + 2) [3; 4]
fold_left (+) (3 + 3) [4]
fold_left (+) (6 + 4) []
10
We can defined map in terms of fold_left:
let map f lst =
let f' init x = f x :: init in
fold_left f' [] lst
Let's evaluate map (fun x -> x + 1) [5; 2; 6]:
map (fun x -> x + 1) [5; 2; 6]
fold_left f' [] [5; 2; 6]
fold_left f' (5 + 1 :: []) [2; 6]
fold_left f' (2 + 1 :: [6]) [6]
fold_left f' (6 + 1 :: [3; 6]) []
[7; 3; 6]
Now, because of the way we destructure and create lists, the result is backwards. we can overcome this with fold_left by reversing the resulting list.
let map f lst =
let f' init x = f x :: init in
let lst' = fold_left f' [] lst in
List.rev lst'
Or with the |> operator:
let map f lst =
let f' init x = f x :: init in
fold_left f' [] lst |> List.rev
Taking this to the next level
At each iteration, fold_left transforms the first element in a list and an accumulator, into the accumulator for the next iteration. If you want to apply this concept to your StringMap module, consider StringMap.empty which generates an empty StringMap.t, and StringMap.add which take a key, an associated value, and an existing map, and returns a new map with that added mapping.
You can readily use fold_left to take an initially empty map and build it into a complete map step by step. The only question remaining will be what value you choose to associate with each string in your list.
As you seems to confuse map and fold_left I think this quote could help you to understand the difference:
Imagine you have a big dinner with numerous people. You are serving the dish: you go through all the people and replace their empty plates with plates containing food. This is a map operation: the number of plate on the table didn't change, but for each plate, you have done the same action (changing the content of the plate).
Once everything is done, you collect all the dirty plates: This is a fold operation, at the end, there are no more plates on the table, but you have done something for each plates (stacking them) and return the file result (a stack of dirty plates).
In both case, an action is applied systmatically. The difference is that Map preserves the current "structure" (the plates on the table) while Fold removes the structure, and build something else."
I'm playing around with Haskell, mostly trying to learn some new techniques to solve problems. Without any real application in mind I came to think about an interesting thing I can't find a satisfying solution to. Maybe someone has any better ideas?
The problem:
Let's say we want to generate a list of Ints using a starting value and a list of Ints, representing the pattern of numbers to be added in the specified order. So the first value is given, then second value should be the starting value plus the first value in the list, the third that value plus the second value of the pattern, and so on. When the pattern ends, it should start over.
For example: Say we have a starting value v and a pattern [x,y], we'd like the list [v,v+x,v+x+y,v+2x+y,v+2x+2y, ...]. In other words, with a two-valued pattern, next value is created by alternatingly adding x and y to the number last calculated.
If the pattern is short enough (2-3 values?), one could generate separate lists:
[v,v,v,...]
[0,x,x,2x,2x,3x, ...]
[0,0,y,y,2y,2y,...]
and then zip them together with addition. However, as soon as the pattern is longer this gets pretty tedious. My best attempt at a solution would be something like this:
generateLstByPattern :: Int -> [Int] -> [Int]
generateLstByPattern v pattern = v : (recGen v pattern)
where
recGen :: Int -> [Int] -> [Int]
recGen lastN (x:[]) = (lastN + x) : (recGen (lastN + x) pattern)
recGen lastN (x:xs) = (lastN + x) : (recGen (lastN + x) xs)
It works as intended - but I have a feeling there is a bit more elegant Haskell solution somewhere (there almost always is!). What do you think? Maybe a cool list-comprehension? A higher-order function I've forgotten about?
Separate the concerns. First look a just a list to process once. Get that working, test it. Hint: “going through the list elements with some accumulator” is in general a good fit for a fold.
Then all that's left to is to repeat the list of inputs and feed it into the pass-once function. Conveniently, there's a standard function for that purpose. Just make sure your once-processor is lazy enough to handle the infinite list input.
What you describe is
foo :: Num a => a -> [a] -> [a]
foo v pattern = scanl (+) v (cycle pattern)
which would normally be written even as just
foo :: Num a => a -> [a] -> [a]
foo v = scanl (+) v . cycle
scanl (+) v xs is the standard way to calculate the partial sums of (v:xs), and cycle is the standard way to repeat a given list cyclically. This is what you describe.
This works for a pattern list of any positive length, as you wanted.
Your way of generating it is inventive, but it's almost too clever for its own good (i.e. it seems overly complicated). It can be expressed with some list comprehensions, as
foo v pat =
let -- the lists, as you describe them:
lists = repeat v :
[ replicate i 0 ++
[ y | x <- [p, p+p ..]
, y <- map (const x) pat ]
| (p,i) <- zip pat [1..] ]
in
-- OK, so what do we do with that? How do we zipWith
-- over an arbitrary amount of lists?
-- with a fold!
foldr (zipWith (+)) (repeat 0) lists
map (const x) pat is a "clever" way of writing replicate (length pat) x. It can be further shortened to x <$ pat since (<$) x xs == map (const x) xs by definition. It might seem obfuscated, until you've become accustomed to it, and then it seems clear and obvious. :)
Surprised noone's mentioned the silly way yet.
mylist x xs = x : zipWith (+) (mylist x xs) (cycle xs)
(If you squint a bit you can see the connection to scanl answer).
When it is about generating series my first approach would be iterate or unfoldr. iterate is for simple series and unfoldr is for those who carry kind of state but without using any State monad.
In this particular case I think unfoldr is ideal.
series :: Int -> [Int] -> [Int]
series s [x,y] = unfoldr (\(f,s) -> Just (f*x + s*y, (s+1,f))) (s,0)
λ> take 10 $ series 1 [1,1]
[1,2,3,4,5,6,7,8,9,10]
λ> take 10 $ series 3 [1,1]
[3,4,5,6,7,8,9,10,11,12]
λ> take 10 $ series 0 [1,2]
[0,1,3,4,6,7,9,10,12,13]
It is probably better to implement the lists separately, for example the list with x can be implement with:
xseq :: (Enum a, Num a) => a -> [a]
xseq x = 0 : ([x, x+x ..] >>= replicate 2)
Whereas the sequence for y can be implemented as:
yseq :: (Enum a, Num a) => a -> [a]
yseq y = [0,y ..] >>= replicate 2
Then you can use zipWith :: (a -> b -> c) -> [a] -> [b] -> [c] to add the two lists together and add v to it:
mylist :: (Enum a, Num a) => a -> a -> a -> [a]
mylist v x y = zipWith ((+) . (v +)) (xseq x) (yseq y)
So for v = 1, x = 2, and y = 3, we obtain:
Prelude> take 10 (mylist 1 2 3)
[1,3,6,8,11,13,16,18,21,23]
An alternative is to see as pattern that we each time first add x and then y. We thus can make an infinite list [(x+), (y+)], and use scanl :: (b -> a -> b) -> b -> [a] -> [b] to each time apply one of the functions and yield the intermediate result:
mylist :: Num a => a -> a -> a -> [a]
mylist v x y = scanl (flip ($)) v (cycle [(x+), (y+)])
this yields the same result:
Prelude> take 10 $ mylist 1 2 3
[1,3,6,8,11,13,16,18,21,23]
Now the only thing left to do is to generalize this to a list. So for example if the list of additions is given, then you can impelement this as:
mylist :: Num a => [a] -> [a]
mylist v xs = scanl (flip ($)) v (cycle (map (+) xs))
or for a list of functions:
mylist :: Num a => [a -> a] -> [a]
mylist v xs = scanl (flip ($)) v (cycle (xs))
Implementing Haskell's take and drop functions using foldl.
Any suggestions on how to implement take and drop functions using foldl ??
take x ls = foldl ???
drop x ls = foldl ???
i've tried these but it's showing errors:
myFunc :: Int -> [a] -> [a]
myFunc n list = foldl func [] list
where
func x y | (length y) > n = x : y
| otherwise = y
ERROR PRODUCED :
*** Expression : foldl func [] list
*** Term : func
*** Type : a -> [a] -> [a]
*** Does not match : [a] -> [a] -> [a]
*** Because : unification would give infinite type
Can't be done.
Left fold necessarily diverges on infinite lists, but take n does not. This is so because left fold is tail recursive, so it must scan through the whole input list before it can start the processing.
With the right fold, it's
ntake :: Int -> [a] -> [a]
ntake 0 _ = []
ntake n xs = foldr g z xs 0
where
g x r i | i>=n = []
| otherwise = x : r (i+1)
z _ = []
ndrop :: Int -> [a] -> [a]
ndrop 0 xs = xs
ndrop n xs = foldr g z xs 0 xs
where
g x r i xs#(_:t) | i>=n = xs
| otherwise = r (i+1) t
z _ _ = []
ndrop implements a paramorphism nicely and faithfully, up to the order of arguments to the reducer function g, giving it access to both the current element x and the current list node xs (such that xs == (x:t)) as well as the recursive result r. A catamorphism's reducer has access only to x and r.
Folds usually encode catamorphisms, but this shows that right fold can be used to code up a paramorphism just as well. It's universal that way. I think it is beautiful.
As for the type error, to fix it just switch the arguments to your func:
func y x | ..... = .......
The accumulator in the left fold comes as the first argument to the reducer function.
If you really want it done with the left fold, and if you're really sure the lists are finite, two options:
ltake n xs = post $ foldl' g (0,id) xs
where
g (i,f) x | i < n = (i+1, f . (x:))
| otherwise = (i,f)
post (_,f) = f []
rltake n xs = foldl' g id xs r n
where
g acc x = acc . f x
f x r i | i > 0 = x : r (i-1)
| otherwise = []
r _ = []
The first counts from the left straight up, potentially stopping assembling the prefix in the middle of the full list traversal that it does carry to the end nevertheless, being a left fold.
The second also traverses the list in full turning it into a right fold which then gets to work counting down from the left again, being able to actually stop working as soon as the prefix is assembled.
Implementing drop this way is bound to be (?) even clunkier. Could be a nice exercise.
I note that you never specified the fold had to be over the supplied list. So, one approach that meets the letter of your question, though probably not the spirit, is:
sillytake :: Int -> [a] -> [a]
sillytake n xs = foldl go (const []) [1..n] xs
where go f _ (x:xs) = x : f xs
go _ _ [] = []
sillydrop :: Int -> [a] -> [a]
sillydrop n xs = foldl go id [1..n] xs
where go f _ (_:xs) = f xs
go _ _ [] = []
These each use left folds, but over the list of numbers [1..n] -- the numbers themselves are ignored, and the list is just used for its length to build a custom take n or drop n function for the given n. This function is then applied to the original supplied list xs.
These versions work fine on infinite lists:
> sillytake 5 $ sillydrop 5 $ [1..]
[6,7,8,9,10]
Will Ness showed a nice way to implement take with foldr. The least repulsive way to implement drop with foldr is this:
drop n0 xs0 = foldr go stop xs0 n0
where
stop _ = []
go x r n
| n <= 0 = x : r 0
| otherwise = r (n - 1)
Take the efficiency loss and rebuild the whole list if you have no choice! Better to drive a nail in with a screwdriver than drive a screw in with a hammer.
Both ways are horrible. But this one helps you understand how folds can be used to structure functions and what their limits are.
Folds just aren't the right tools for implementing drop; a paramorphism is the right tool.
You are not too far. Here are a pair of fixes.
First, note that func is passed the accumulator first (i.e. a list of a, in your case) and then the list element (an a). So, you need to swap the order of the arguments of func.
Then, if we want to mimic take, we need to add x when the length y is less than n, not greater!
So we get
myFunc :: Int -> [a] -> [a]
myFunc n list = foldl func [] list
where
func y x | (length y) < n = x : y
| otherwise = y
Test:
> myFunc 5 [1..10]
[5,4,3,2,1]
As you can see, this is reversing the string. This is because we add x at the front (x:y) instead of at the back (y++[x]). Or, alternatively, one could use reverse (foldl ....) to fix the order at the end.
Also, since foldl always scans the whole input list, myFunc 3 [1..1000000000] will take a lot of time, and myFunc 3 [1..] will fail to terminate. Using foldr would be much better.
drop is more tricky to do. I don't think you can easily do that without some post-processing like myFunc n xs = fst (foldl ...) or making foldl return a function which you immediately call (which is also a kind of post-processing).
I have list of lists of Int and I need to add an Int value to the last list from the list of lists. How can I do this? My attempt is below
f :: [[Int]] -> [Int] -> Int -> Int -> Int -> [[Int]]
f xs [] cur done total = [[]]
f xs xs2 cur done total = do
if total >= length xs2 then
xs
else
if done == fib cur then
f (xs ++ [[]]) xs2 (cur + 1) 0 total
else
f ((last xs) ++ [[xs2!!total]]) xs2 cur (done + 1) (total + 1)
The problem is:
We have a list A of Int
And we need to slpit it on N lists B_1 ,..., B_n , length of B_i is i-th Fibonacci number.
If we have list [1 , 2 , 3 , 4 , 5 , 6 , 7] (xs2 in my code)
The result should be [[1] , [2] , [3 , 4] , [5 , 6 , 7]]
The easy way to deal with problems like this is to separate the problem into sub-problems. In this case, you want to change the last item in a list. The way you want to change it is by adding an item to it.
First let's tackle changing the last item of a list. We'll do this by applying a function to the last item, but not to any other items.
onLast :: [a] -> (a -> a) -> [a]
onLast xs f = go xs
where
go [] = []
go [x] = [f x]
go (x:xs) = x:go xs
You want to change the last item in the list by adding an additional value, which you can do with (++ [value]).
Combining the two with the value you want to add (xs2!!total) we get
(onLast xs (++ [xs2!!total]))
f :: [[Int]] -> Int -> [[Int]]
f [] _ = []
f xs i = (take n xs) ++ [[x + i | x <- last xs]]
where n = (length xs) - 1
last = head . (drop n)
For example,
*Main> f [[1, 2, 3], [], [4, 5, 6]] 5
[[1,2,3],[],[9,10,11]]
*Main> f [[1, 2, 3]] 5
[[6,7,8]]
*Main> f [] 3
You approach uses a do block, this is kind of weird since do blocks are usually used for monads. Furthermore it is rather unclear what cur, done and total are doing. Furthermore you use (!!) :: [a] -> Int -> a and length :: [a] -> Int. The problem with these functions is that these run in O(n), so it makes the code inefficient as well.
Based on changed specifications, you want to split the list in buckets with length the Fibonacci numbers. In that case the signature should be:
f :: [a] -> [[a]]
because as input you give a list of numbers, and as output, you return a list of numbers. We can then implement that as:
f :: [a] -> [[a]]
f = g 0 1
where g _ _ [] = []
g a b xs = xa : g b (a+b) xb
where (xa,xb) = splitAt b xs
This generates:
*Main> f [1,2,3,4,5,6]
[[1],[2],[3,4],[5,6]]
*Main> f [1,2,3,4,5,6,7]
[[1],[2],[3,4],[5,6,7]]
*Main> f [1,2,3,4,5,6,7,8]
[[1],[2],[3,4],[5,6,7],[8]]
*Main> f [1,2,3,4,5,6,7,8,9]
[[1],[2],[3,4],[5,6,7],[8,9]]
The code works as follows: we state that f = g 0 1 so we pass the arguments of f to g, but g also gets an 0 and a 1 (the first Fibonacci numbers).
Each iteration g checks whether we reached the end of the list. If so, we return an empty list as well. Otherwise we determine the last Fibonacci number that far (b), and use a splitAt to obtain the first b elements of the list we process, as well as the remainder. We then emit the first part as head of the list, and for the tail we calculate the next Fibonacci number and pass that to g with the tail of splitAt.
Need increment every second item starting from the right in Haskell list but keeping origin order (e.g. reverse is not a case). For example:
f [1, 2, 3] -- [1, 3, 3]
f [1, 2, 3, 4] -- [2, 2, 4, 4]
I've tried something like a following:
fc ([]) = []
fc (x:[]) = [x]
fc (x:[y]) = [x+1,y]
fc( x:xs ) = fc [x] : ( fc xs ) -- this line is wrong
p.s. Obviously I could reverse (but prefer to understand original task) the list twice and apply something like:
helper (x:y:tail) = [x, y+1] ++ tail
fc x = reverse (helper (reverse x) )
The typical way to process a Haskell list from right to left would be to reverse it. Since you want to have the original order for the result, you would simply reverse again:
f1 = reverse . zipWith (+) (cycle [0,1]) . reverse
But if you really want to, you can have each recursive call return both the updated tail and a flag that indicates whether that position is even when counted from the end so you know whether to increase the element at that position or not:
f2 = snd . g
where
g [] = (False, [])
g (x:xs) = let (addOne, xs') = g xs
x' = if addOne then x + 1 else x
in (not addOne, x':xs')
We're basically mapping a function over the list, but this function requires an extra parameter that gets computed starting from the right end of the list. There's a standard function we can use:
import Data.List (mapAccumR)
f2' = snd . mapAccumR g False
where
g addOne x = (not addOne, if addOne then x + 1 else x)
I think a cleaner specification for what you want is that you increment even indicies if the length is even and odd indicies if the length is odd. For example, when indexing from zero, the list of length 3 resulted in index 1 being incremented. One way to do this is with the obvious two pass solution:
f xs = zipWith (+) (cycle sol) xs
where sol = map fromEnum [even len, odd len]
len = length xs
This can be done in one pass (without relying on the compiler fusion rules) by "tying the knot". For example (using manual recursive style as means of communication).
f2 xs = let (isEven, result) = go isEven xs in result
where
go _ [] = (True, [])
go e (x:xs) = let (ne,rest) = go (not e) xs
in (not ne, x+fromEnum e : rest)
This can be done efficiently using a left fold:
inc :: Num a => [a] -> [a]
inc xs = foldl go (\_ _ acc -> acc) xs id (+ 1) []
where go run x f g acc = run g f (f x: acc)
Note that even thought this is a left fold, the list is built using cons (:) operator; and it will perform linearly and not quadratic (similar construct as in difference lists).
\> inc [1, 2, 3]
[1,3,3]
\> inc [1, 2, 3, 4]
[2,2,4,4]
It can also be generalized to alternating functions other than id and (+ 1).
I like Thomas's solution. However, I think a simple foldr is enough here.
process = snd . foldr (\x (b,xs) -> (not b, x + fromEnum b:xs)) (False,[])