I would like to foldl a string so that any occurrence of zero which is preceded by # is replaced with "k". So "a.#0.1.2.0" becomes, "a.#k.1.2.0". How do I do that? So far my attempt is
test = foldl(\x acc-> if((last acc) == "#" && x == "0" then acc ++ "k" else acc ++ x)) "" "a.#0.1.2.0"
However, it doesn't work. Foldl is expecting list of string and what I'm providing is just a string. How do I overcome that?
Taking chi's advice,
rep "" = ""
rep ('#' : '0' : xs) = "#k" ++ rep xs
rep (x : xs) = x : rep xs
If we want to get fancier,
rep = snd . mapAccumL go False
where
go True '0' = (False, 'k')
go _ '#' = (True, '#')
go _ x = (False, x)
or even
rep = snd . mapAccumL go 'x'
where
go '#' '0' = ('0', 'k')
go _ y = (y, y)
To use foldr with this second approach (just because it's shorter; the first will work fine too, and allows generalization),
rep xs = foldr go (const "") xs 'x'
where
go '0' r '#' = 'k' : r '0'
go x r _ = x : r x
To use zipWith (which is more awkward to generalize):
rep xs = zipWith go ('x' : xs) xs where
go '#' '0' = 'k'
go _ x = x
As others have commented, foldl would not fit well for this task. However, using the idea of difference list, you may still do this efficiently using foldl:
repl :: String -> String
repl a = foldl loop id a ""
where
loop :: (String -> String) -> Char -> (String -> String)
loop f '#' ('0':rest) = f ('#':'k':rest)
loop f x xs = f (x:xs)
Just for the purpose of demonstration and the fact that the question has asked for a foldl solution. (not that I would recommend doing it this way).
Related
I could understand if the question doesn't really clarify my problem, so here is some more explanation:
I am trying to add the string "+" at the start of my string, which I get like this:
printLine :: [Int] -> String --Type of the function
printLine [] = "" --Base case
printLine (x:xs) = "+" ++ foldr (++) "+" f ++ printLine xs
where f = replicate x "-"
The result I get from the above:
+-----++------++------++------+
The result I would like to get:
+-----+------+------+------+
Basically my question is: How do I add "+" only at the start?
I can understand that this might be a silly question, but I am stuck for a while now and I can't find the answer on SO or elsewhere.
Proposal: don't detect when you're in the first iteration, which is hard; instead detect when you're in the last iteration, which is easy because it's the [] case in the first line.
printLine :: [Int] -> String
-- final iteration; add an extra + at the end
printLine [] = "+"
-- not the final iteration; don't include a + at the end of the -s
printLine (x:xs) = "+" ++ replicate x '-' ++ printLine xs
If an empty list must map to an empty string, one option is to fold with a special case for an empty list.
printLine :: [Int] -> String
printLine [] = ""
printLine xs = foldr (\x res -> '+' : replicate x '-' ++ res) "+" xs
So that
λ> map printLine [[], [1..4], [5]]
["","+-+--+---+----+","+-----+"]
Alternatively, since the original question asked for control during the first iteration, one option is to use a helper function. Here are two alternatives.
printLine' :: [Int] -> String
printLine' [] = ""
printLine' xs = '+' : go xs
where go :: [Int] -> String
go [] = ""
go (n:ns) = replicate n '-' ++ "+" ++ go ns
printLine'' :: [Int] -> String
printLine'' xs = go True xs
where go :: Bool -> [Int] -> String
go _ [] = ""
go isFirst (n:ns) = (if isFirst then "+" else "")
++ replicate n '-' ++ "+" ++ go False ns
With these definitions
λ> map printLine' [[], [1..4], [5]]
["","+-+--+---+----+","+-----+"]
λ> map printLine'' [[], [1..4], [5]]
["","+-+--+---+----+","+-----+"]
If I was given a string like skhfbvqa, how would I generate the next string? For this example, it would be skhfbvqb, and the next string of that would be skhfbvqc, and so on. The given string (and the answer) will always be N characters long (in this case, N=8).
What I tried:
I tried to generate the entire (infinite) list of possible combinations, and get the required (next) string of the given string, but unsurprisingly, it's so slow, that I don't even get the answer for N=6.
I used list comprehension:
allStrings = [ c : s | s <- "" : allStrings, c <- ['a'..'z'] ]
main = do
input <- readFile "k.in"
putStrLn . head . tail . dropWhile (not . (==) input) . map reverse $ allStrings
(Please excuse my incredibly bad Haskell-ing :) Still a noob)
So my question is, how can I do this? If there are multiple methods, a comparison between them is much appreciated. Thanks!
Here's a version with base conversion (this way you could add and subtract arbitrarily if you like):
encode x base = encode' x [] where
encode' x' z | x' == 0 = z
| otherwise = encode' (div x' base) ((mod x' base):z)
decode num base =
fst $ foldr (\a (b,i) -> (b + a * base^i,i + 1)) (0,0) num
Output:
*Main> map (\x -> toEnum (x + 97)::Char)
$ encode (decode (map (\x -> fromEnum x - 97) "skhfbvqa") 26 + 1) 26
"skhfbvqb"
I would go and make a helper function f :: Integer -> String and one g :: String -> Integer, where f 1 = "a", ... f 27 = "aa", f 28 = "ab" and so on and the inverse g.
Then incrementString = f . succ . g
Note: I omitted the implementation of f on purpose for learning
Update
for a different approach you could define a increment with carry function inc' :: Char -> (Char, Bool), and then
incString :: String -> String
incString = reverse . incString'
where incString' [] = []
incString' (x:xs) = case inc' x of (x',True) -> x': incString' xs
(x',False) -> x':xs
Note: this function is not tail recursive!
I found this to work. It just uses pattern matching to see if the string begins with a z and adds an additional a accordingly.
incrementString' :: String -> String
incrementString' [] = ['a']
incrementString' ('z':xs) = 'a' : incrementString' xs
incrementString' (x:xs) = succ x : xs
incrementString :: String -> String
incrementString = reverse . incrementString' . reverse
I can remove the vowels without a recursive function like this:
NoVowels:: String -> String
NoVowels xs = filter f xs where f x = not (x == ’a’ || x == ’e’ || x == ’i’ || x == ’o’ || x == ’u’)
But how may I do that with a recursive function ?
I tried something like this but of course did not work (parse Error):
NoVowels :: String -> String
NoVowels "" = error "Empty String!!"
NoVowels (x:xs)
| x in (x == 'a'|| x == 'e' || x == 'i' || x == 'o' || x == 'u') = NoVowels (tail x)
If the head is a vowel then I CUT it from the string and pass recursively the tail, if its not a vowel how may I verify the other string without removing it.
Update obs: I want to return the function without the vowels.
I thins it makes sense to define a function isVowel :: Char->Bool and after that write something like this :
noVowels :: String -> String
noVowels [] = []
noVowels (x:xs)
|isVowel x = noVowels xs
|otherwise = x : noVowels xs
If you don't want to define one more function you can try next code :
noVowels :: String ->String
noVowels [] = []
noVowels (x:xs)
|not( x `elem` "aeiou") = x: noVowels xs
|otherwise = noVowels xs
Here's a bit of an improvement on the code. Just in case you missed out on preserving the cases involved.
module Disemvowel where
disemvowel :: String -> String
disemvowel [] = []
disemvowel " " = " "
disemvowel (x:xs)
| x `elem` "aeiouAEIOU" = disemvowel xs
| otherwise = [x] ++ disemvowel xs
Supposing I had the string "HELLO WORLD" is there a way I can call a function that replaces the character 'O' in the string with the character 'X' so that the new string would look like "HELLX WXRLD"?
How about:
let
repl 'o' = 'x'
repl c = c
in map repl "Hello World"
If you need to replace additional characters later, just add clauses to the repl function.
Sorry for picking up this old thread but why not use lambda expressions?
λ> let replaceO = map (\c -> if c=='O' then 'X'; else c)
λ> replaceO "HELLO WORLD"
"HELLX WXRLD"`
Alternative 1 - Using MissingH
First:
import Data.List.Utils (replace)
Then use:
replace "O" "X" "HELLO WORLD"
Alternative 2 - Using Control.Monad
One funny bastard:
import Control.Monad (mfilter)
replace a b = map $ maybe b id . mfilter (/= a) . Just
Example:
λ> replace 'O' 'X' "HELLO WORLD"
"HELLX WXRLD"
Alternative 3 - Using if
Amon's suggestions was probably the finest I believe! No imports and easy to read and understand!
But to be picky - there's no need for semicolon:
replace :: Eq a => a -> a -> [a] -> [a]
replace a b = map $ \c -> if c == a then b else c
If you depend on the text package (like 99.99% of Haskell applications), you can use T.replace:
>>> replace "ofo" "bar" "ofofo"
"barfo"
Here's another possible solution using divide and conquer:
replaceO [] = []
replaceO (x:xs) =
if x == 'O'
then 'X' : replaceO xs
else x : replaceO xs
First, you set the edge condition "replaceO [] = []".
If the list is empty, there is nothing to replace, returning an empty list.
Next, we take the string and divide it into head and tail. in this case 'H':"ELLOWORLD"
If the head is equal to 'O', it will replace it with 'X'. and apply the replaceO function to the rest of the string.
If the head is not equal to 'O', then it will put the head back where it is and apply the replaceO function to the rest of the string.
replace :: Char -> Char -> String -> String
replace _ _ [] = []
replace a b (x : xs)
| x == a = [b] ++ replace a b xs
| otherwise = [x] ++ replace a b xs
I'm new to Haskell and I've tried to make it simpler for others like me.
I guess this could be useful.
main = print $ charRemap "Hello WOrld" ['O','o'] ['X','x']
charRemap :: [Char] -> [Char] -> [Char] -> [Char]
charRemap [] _ _ = []
charRemap (w:word) mapFrom mapTo =
if snd state
then mapTo !! fst state : charRemap word mapFrom mapTo
else w : charRemap word mapFrom mapTo
where
state = hasChar w mapFrom 0
hasChar :: Char -> [Char] -> Int -> (Int,Bool)
hasChar _ [] _ = (0,False)
hasChar c (x:xs) i | c == x = (i,True)
| otherwise = hasChar c xs (i+1)
I have to write two functions converting decimal numers into a (-2)adian number system (similar to binary only with -2) and vice versa.
I already have managed to get the decimal -> (-2)adian running.
But with (-2)adian -> decimal I have a problem and just don't know where to begin.
Hope you can Help me
type NegaBinary = String
-- Function (-2)adisch --> decimal
negbin_dezi :: NegaBinary -> Integer -> Integer
negbin_dezi (xs:x) n
| (x == 0) = if ([xs] == "") then 0 else (negbin_dezi [xs] (n+1))
| (x == 1) = if ([xs] == "") then (-2)**n else (-2)**n + (negbin_dezi [xs] (n+1))
It always throws:
"Instances of (Num [Char], Floating Integer) required for definition of negbin_dezi.
Anyone an idea why it wont work?
Please please please :)
You have your list pattern-matching syntax backwards. In _ : _ the first argument is the head of the list (one element), and the second is the tail of the list (another list). e.g. x:xs matched with "abc" gives x = 'a' xs = "bc". So xs:x should be x:xs. The reason for GHC asking for an instance of Num [Char], is the comparison x == 0 (and x == 1). In this, it is trying to match the type of x (String == [Char]) with the type of 0 (Num a => a), and to do this, it requires a Num instance for String.
The fix is: negbin_dezi (x:xs) n
The problem asking for an Floating Integer instance is because (**) has type Floating a => a -> a -> a, where as you want (^) which has type (Num a, Integral b) => a -> b -> a (i.e. it is restricted to integer powers.)
Once you've done this, you'll find that your algorithm doesn't work for a few reasons:
The number 0 is different to the character '0', you should be comparing x with the characters '0' and '1' rather than the numbers 0 and 1.
xs is already a string, so [xs] is a list containing a string, which isn't what you want. This is fixed by removing the square brackets.
Possibly the ordering of the reduction is wrong.
On a different note, the duplicated if statement suggests that there is some optimisations that could happen with your code. Specifically, if you handle the empty string as part of negbin_dezi then you won't have to special case it. You could write it something like
negbin_dezi "" _ = 0
negbin_dezi (x:xs) n
| n == '0' = negbin_dezi xs (n+1)
| n == '1' = (-2)^n + negbin_dezi
(This has the bonus of meaning that the function is "more total", i.e. it is defined on more inputs.)
A few more things:
The code is "stringly-typed": your data is being represented as a string, despite having more structure. A list of booleans ([Bool]) would be much better.
The algorithm can be adapted to be cleaner. For the following, I'm assuming you are storing it like "01" = -2 "001" = 4, etc. If so, then we know that number = a + (-2) * b + (-2)^2 * c ... = a + (-2) * (b + (-2) * (c + ...)) where a,b,c,... are the digits. Looking at this, we can see the stuff inside the brackets is actually the same as the whole expression, just starting at the second digit. This is easy to express in Haskell (I'm using the list-of-bools idea.):
negbin [] = 0
negbin (x:xs) = (if x then 1 else 0) + (-2) * negbin xs
And that's the whole thing. If you aren't storing it in that order, then a call to reverse fixes that! (Being really tricky, one could write
negbin = foldr (\x n -> (if x then 1 else 0) + (-2)*n) 0
)
Some problems:
x == 0 or x == 1, but x is a Char, so you mean x == '0'.
You write (xs:x). There's no pattern for matching at the end of a list. Perhaps use a helper function that reverses the list first.
[xs] has one element, and will never be "". Use a base case instead.
Pattern matching is more helpful than equality checking.
** is for floating point powers, ^ is for integer powers
You often use [xs] where you mean xs. You don't need to put square brackets to make a list.
Here's a rewrite that works:
negbin_dezi1 :: NegaBinary -> Integer
negbin_dezi1 xs = negbin (reverse xs) 0
negbin [] _ = 0
negbin (x:xs) n
| x == '0' = negbin xs (n+1)
| x == '1' = (-2)^n + (negbin xs (n+1))
It would be nicer to use pattern matching:
negbin_dezi2 :: NegaBinary -> Integer
negbin_dezi2 xs = negbin (reverse xs) 0 where
negbin [] _ = 0
negbin ('0':xs) n = negbin xs (n+1)
negbin ('1':xs) n = (-2)^n + negbin xs (n+1)
But maybe it would be nicer to convert '0' to 0 and '1' to 1 and just multiply by that:
val :: Char -> Int
val '0' = 0
val '1' = 1
negbin_dezi3 :: NegaBinary -> Integer
negbin_dezi3 xs = negbin (reverse xs) 0 where
negbin [] _ = 0
negbin (x:xs) n = val x * (-2)^n + negbin xs (n+1)
I'd not write it that way, though:
A completely different approach is to think about the whole thing at once.
"10010" -rev> [0,1,0,0,1] -means> [ 0, 1, 0, 0, 1 ]
[(-2)^0, (-2)^1, (-2)^2, (-2)^3, (-2)^4]
so let's make both lists
powers = [(-2)^n | n <- [0..]]
coefficients = reverse.map val $ xs
and multiply them
zipWith (*) powers coefficients
then add up, giving:
negbin_dezi4 xs = sum $ zipWith (*) powers coefficients
where powers = [(-2)^n | n <- [0..]]
coefficients = reverse.map val $ xs
You could rewrite powers as map ((-2)^) [0..],
or even nicer: powers = 1:map ((-2)*) powers.
(It's nicer because it reuses previous calculations and is pleasantly clean.)
this
convB2D::NegaBinary->Integer
convB2D xs|(length xs)==0 =0
|b=='0' = convB2D(drop 1 xs)
|b=='1' = val+convB2D(drop 1 xs)
|otherwise= error "invalid character "
where b=head xs
val=(-2)^((length xs)-1)
worked for me.
I on the other hand have problems to convert dec->nbin :D