2 weeks new to haskell and functional programming. In the process of covering foldl and foldr in class, I found that I was quite new to tail recursion and never actually tried writing a tail recursive function before (also new to how foldl traverses a list which is where it came up).
As an exercise, I tried to rewrite the following function to be tail recursive.
--replace replaces all instances of value x in a list with y
replace :: (Eq a) => a -> a -> [a] -> [a]
replace _ _ [] = []
replace x y (h:t) =
(if x==h then y: (replace x y t) else h: (replace x y t))
--tail recursive version of the same function
replacet _ _ [] = []
replacet x y (h:t) =
(if x==h then (replacet x y (y:t)) else (replacet x y (h:t)))
...But the output is just the following in ghci:
Nothing seems to be running at all, let alone getting an overflow.
Any ideas?
Your replacet never makes any progress. Observe that at each step, your input list stays the same size: you replace (h:t) with either (y:t) or (h:t), and then call replacet on the result.
Related
I am a newbie to Haskell and I wrote the following code
module RememberMap (rememberMap) where
rememberMap :: (a -> b -> (a, c)) -> a -> [b] -> [c]
rememberMap f acc xs = go acc xs []
where
go acc [x] carry = carry <> [step]
where
(_, step) = f acc x
go acc (x:xs) carry = go accStep xs (carry <> [step])
where
(accStep, step) = f acc x
I wrote this contaminator with the Intent to help me with the most Common difficulty that i have when writing my Haskell code, That is that I recurrently find myself willing to map something (specially in CodeWarrior's Katas) like to map something, but that something required knowledge of the elements before it. But it had the problem of being non-streaming, ergo, it does no allow me to use lazy proprieties of Haskell with it, thus I would like to know if (a) there is already a solution to this problem (preferably Arrows) or (b) how to make it lazy.
To make the function stream you need to have the cons operator outside the recursive call, so a caller can see the first element without the whole recursion needing to happen. So you expect it to look something like:
rememberMap f acc (x:xs) = element : ... recursion ...
Once you understand this there is not much more to do:
rememberMap _ _ [] = []
rememberMap f acc (x:xs) = y : rememberMap f acc' xs
where
(acc', y) = f acc x
You can make an auxiliary function to avoid passing f around if you want, but there's no reason for it to have the extra list that you called carry.
There are mapAccumL and traverse with the lazy State monad.
What is the difference between this two, in terms of evaluation?
Why this "obeys" (how to say?) non-strictness
recFilter :: (a -> Bool) -> [a] -> [a]
recFilter _ [] = []
recFilter p (h:tl) = if (p h)
then h : recFilter p tl
else recFilter p tl
while this doesn't?
recFilter :: (a -> Bool) -> [a] -> Int -> [a]
recFilter _ xs 0 = xs
recFilter p (h:tl) len
| p(h) = recFilter p (tl ++ [h]) (len-1)
| otherwise = recFilter p tl (len-1)
Is it possible to write a tail-recursive function non-strictly?
To be honest I also don't understand the call stack of the first example, because I can't see where that h: goes. Is there a way to see this in ghci?
The non-tail recursive function roughly consumes a portion of the input (the first element) to produce a portion of the output (well, if it's not filtered out at least). Then recursion handles the next portion of the input, and so on.
Your tail recursive function will recurse until len becomes zero, and only at that point it will output the whole result.
Consider this pseudocode:
def rec1(p,xs):
case xs:
[] -> []
(y:ys) -> if p(y): print y
rec1(p,ys)
and compare it with this accumulator-based variant. I'm not using len since I use a separate accumulator argument, which I assume to be initially empty.
def rec2(p,xs,acc):
case xs:
[] -> print acc
(y:ys) -> if p(y):
rec2(p,ys,acc++[y])
else:
rec2(p,ys,acc)
rec1 prints before recursing: it does not need to inspect the whole input list to start printing its output. It works in a "steraming" fashion, in a sense. Instead, rec2 will only start to print at the very end, after the input list was completely scanned.
In your Haskell code there are no prints, of course, but you can thing of returning x : function call as "printing x", since x is made available to the caller of our function before function call is actually made. (Well, to be pedantic this depends on how the caller will consume the output list, but I'll neglect this.)
Hence the non-tail recursive code can also work on infinite lists. Even on finite inputs, performance is improved: if we call head (rec1 p xs), we only evaluate xs until the first non-discarded element. By contrast head (rec2 p xs) would fully filter the whole list xs, even we don't need that.
The second implementation does not make much sense: a variable named len will not contain the length of the list. You thus need to pass this, for infinite lists, this would not work, since there is no length at all.
You likely want to produce something like:
recFilter :: (a -> Bool) -> [a] -> [a]
recFilter p = go []
where go ys [] = ys -- (1)
go ys (x:xs) | p x = go (ys ++ [x]) xs
| otherwise = go ys xs
where we thus have an accumulator to which we append the items in the list, and then eventually return the accumulator.
The problem with the second approach is that as long as the accumulator is not returned, Haskell will need to keep recursing until at least we reach weak head normal form (WHNF). This means that if we pattern match the result with [] or (_:_), we will need at least have to recurse until case one, since the other cases only produce a new expression, and it will thus not yield a data constructor on which we can pattern match.
This in contrast to the first filter where if we pattern match on [] or (_:_) it is sufficient to stop at the first case (1), or the third case 93) where the expression produces an object with a list data constructor. Only if we require extra elements to pattern match, for example (_:_:_), it will require to evaluate the recFilter p tl in case (2) of the first implementation:
recFilter :: (a -> Bool) -> [a] -> [a]
recFilter _ [] = [] -- (1)
recFilter p (h:tl) = if (p h)
then h : recFilter p tl -- (2)
else recFilter p tl
For more information, see the Laziness section of the Wikibook on Haskell that describes how laziness works with thunks.
I do not understand a sample solution for the following problem: given a list of elements, remove the duplicates. Then count the unique digits of a number. No explicit recursion may be used for either problem.
My code:
removeDuplicates :: Eq a => [a] -> [a]
removeDuplicates = foldr (\x ys -> x:(filter (x /=) ys)) []
differentDigits :: Int -> Int
differentDigits xs = length (removeDuplicates (show xs))
The solution I am trying to understand has a different definition for differentDigits, namely
differentDigits xs = foldr (\ _ x -> x + 1) 0 ( removeDuplicates ( filter (/= '_') ( show xs )))
Both approaches work, but I cannot grasp the sample solution. To break my question down into subquestions,
How does the first argument to filter work? I mean
(/= '_')
How does the lambda for foldr work? In
foldr (\ _ x -> x + 1)
^
the variable x should still be the Char list? How does Haskell figure out that actually 0 should be incremented?
filter (/= '_') is, I'm pretty sure, redundant. It filters out underscore characters, which shouldn't be present in the result of show xs, assuming xs is a number of some sort.
foldr (\ _ x -> x + 1) 0 is equivalent to length. The way foldr works, it takes the second argument (which in your example is zero) as the starting point, then applies the first argument (in your example, lambda) to it over and over for every element of the input list. The element of the input list is passed into the lambda as first argument (denoted _ in your example), and the running sum is passed as second argument (denoted x). Since the lambda just returns a "plus one" number on every pass, the result will be a number representing how many times the lambda was called - which is the length of the list.
First, note that (2) is written in so called point free style, leaving out the third argument of foldr.
https://en.wikipedia.org/wiki/Tacit_programming#Functional_programming
Also, the underscore in \_ x -> x + 1 is a wild card, that simply marks the place of a parameter but that does not give it a name (a wild card works as a nameless parameter).
Second, (2) is a really nothing else than a simple recursive function that folds to the right. foldr is a compact way to write such recursive functions (in your case length):
foldr :: (a -> b -> b) -> b -> [a]
foldr f z [] = z
foldr f z (x:xs) = f x (foldr f z xs)
If we write
foldr f c ls
ls is the list over which our recursive function should recur (a is the type of the elements).
c is the result in the base case (when the recursive recursive function is applied on an empty list).
f computes the result in the general case (when the recursive function is applied on a non-empty list). f takes two arguments:
The head of the list and
the result of the recursive call on the tail of the list.
So, given f and c, foldr will go through the list ls recursively.
A first example
The Wikipedia page about point free style gives the example of how we can compute the sum of all elements in a list using foldr:
Instead of writing
sum [] = 0
sum (x:xs) = x + sum xs
we can write
sum = foldr (+) 0
The operator section (+) is a 2-argument function that adds its arguments. The expression
sum [1,2,3,4]
is computed as
1 + (2 + (3 + (4)))
(hence "folding to the right").
Example: Multiplying all elements.
Instead of
prod [] = 1
prod (x:xs) = x * prod xs
we can write
prod = foldr (*) 1
Example: Remove all occurrences of a value from a list.
Instead of
remove _ [] = []
remove v (x:xs) = if x==v then remove v xs else x:remove v xs
we can write
remove v = foldr (\x r -> if x==v then r else x:r) []
Your case, (2)
We can now fully understand that
length = foldr (\ _ x -> x + 1) 0
in fact is the same as
length [] = 0
length (x:xs) = length xs + 1
that is, the length function.
Hope this recursive view on foldr helped you understand the code.
This question already has answers here:
Eq => function in Haskell
(2 answers)
Closed 3 years ago.
I am new to Haskell and am trying to write quite an easy function which gathers each repeated consecutive elements under separate sub-lists, For example:
f :: Eq a => [a] -> [[a]]
So:
f [] = []
f [3] = [[3]]
f [1,1,1,3,2,2,1,1,1,1] = [[1,1,1],[3],[2,2],[1,1,1,1]]
I thought about this function:
f :: Eq a => [a] -> [[a]]
f [] = []
f (x:[]) = [[x]]
f (x:x':xs) = if x == x' then [[x, x']] ++ (f (xs))
else [[x]] ++ (f (xs))
It seems to not work well since when it arrives to the last element, it wants to compare it to its consecutive, which clearly does not exist.
I would like to receive a simple answer (beginner level) that will not be too different than mine, correcting my code will be the best.
Thanks in advance.
The problem isn't really what you said, it's just that you only hard-coded the cases that either one or two consecutive elements are equal. Actually you want to ground together an arbitrary number of equal consecutives. IOW, for every element, you pop off as many following ones as are equal.
Generally, splitting of the head-part of a list which fulfills some condition is what the span function does. In this case, the condition it's supposed to check is being equal to the element you already removed. That's written thus:
f [] = []
f (x:xs) = (x:xCopies) : f others
where (xCopies,others) = span (==x) xs
Here, x:xCopies puts together the chunk of elements equal to x (with x itself on front), use that as the heading chunk-list of the result, and then you recurse over all the elements that remain.
Your problem is that both halves of your if have the same structure: they cons exactly one element onto the front of the recursive call. This can't be right: sometimes you want to add an element to the front of the list, and other times you want to combine your element with what's already in the recursive call.
Instead, you need to pattern-match on the recursive call to get the first item in the recursive result, and then prepend to that when the first two items match.
f :: Eq a => [a] -> [[a]]
f [] = []
f [x] = [[x]]
f (x:xs#(y:_)) | x == y = (x:head):more
| otherwise = [x]:result
where result#(head:more) = f xs
Currently I am using
takeWhile (\x -> x /= 1 && x /= 89) l
to get the elements from a list up to either a 1 or 89. However, the result doesn't include these sentinel values. Does Haskell have a standard function that provides this variation on takeWhile that includes the sentinel in the result? My searches with Hoogle have been unfruitful so far.
Since you were asking about standard functions, no. But also there isn't a package containing a takeWhileInclusive, but that's really simple:
takeWhileInclusive :: (a -> Bool) -> [a] -> [a]
takeWhileInclusive _ [] = []
takeWhileInclusive p (x:xs) = x : if p x then takeWhileInclusive p xs
else []
The only thing you need to do is to take the value regardless whether the predicate returns True and only use the predicate as a continuation factor:
*Main> takeWhileInclusive (\x -> x /= 20) [10..]
[10,11,12,13,14,15,16,17,18,19,20]
Is span what you want?
matching, rest = span (\x -> x /= 1 && x /= 89) l
then look at the head of rest.
The shortest way I found to achieve that is using span and adding a function before it that takes the result of span and merges the first element of the resulting tuple with the head of the second element of the resulting tuple.
The whole expression would look something like this:
(\(f,s) -> f ++ [head s]) $ span (\x -> x /= 1 && x /= 89) [82..140]
The result of this expression is
[82,83,84,85,86,87,88,89]
The first element of the tuple returned by span is the list that takeWhile would return for those parameters, and the second element is the list with the remaining values, so we just add the head from the second list to our first list.