I have just tried rewriting some code, originally a short Javascript function, in Haskell. The original has 2 nested loops and the inner loop contains a check for equality against both loop counters:
function f(x, points){
var i, j;
var n = points.length;
var result = 0;
for(i=0; i<n; i++){
var xprod = 1;
for(j=0; j<n; j++){
if(j != i){
xprod *= (x - points[j][0]);
}
}
result += points[i][1] * xprod;
}
return result;
}
I was hoping to be able to simplify it in Haskell, but I couldn't figure out how get hold of the i and j values without effectively writing out every step of the original recursively. In Javascript Array.map passes the list position into the callback function as the second parameter, but it seems that map in Haskell doesn't do this. My current Haskell version looks awful to me as I'm passing in 2 copies of the array (one for each loop):
xproduct :: Int -> Int -> Double -> [(Double,Double)] -> Double
xproduct _ _ _ [] = 1
xproduct i j x (pt:todo)
| i == j = (xproduct i (j+1) x todo)
| otherwise = (xproduct i (j+1) x todo) * (x - (fst pt))
solvestep :: Int -> Double -> [(Double,Double)] -> [(Double,Double)] -> Double
solvestep _ _ _ [] = 0
solvestep i x pts (pt:todo) = ((snd pt) * xprod) + (solvestep (i+1) x pts todo)
where xprod = xproduct i 0 x pts
solve :: Double -> [(Double,Double)] -> Double
solve x points = solvestep 0 x points points
Is there a better way to do this?
I generally avoid using any indices at all, if possible. In this case, what you're really working with is: any one element of the list with all the other elements. No need to express that with index comparison, instead write a function that will give you a suitable look into the list:
pickouts :: [a] -> [(a,[a])]
pickouts [] = []
pickouts (x:xs) = (x,xs) : (second (x:) <$> pickouts xs)
Then, the actual computation becomes just
f :: Double -> [(Double,Double)] -> Double
f x points = sum [q * product [x-p | (p,_)<-ps] | ((_,q),ps) <- pickouts points]
Related
I want to create a function as mentioned in the title. The specific is that it adds the digits in reversed order, you can see that in the test cases: 12 -> 1; 852369 -> 628; 1714 -> 11; 12345 -> 42; 891 -> 9; 448575 -> 784; 4214 -> 14
The main idea is that when the number is bigger than 99 it enters the helper function which has i - indicator if the the digit is on an even position, and res which stores the result. Helper begins to cycle n as it checks whether or not the current digit is on even position and adds it to the result.
So far I've tried the following code:
everyOther :: Int -> Int
everyOther n
| n < 10 = error "n must be bigger than 10 or equal"
| n < 100 = div n 10
| otherwise = helper n 0 0
where
helper :: Int -> Int -> Int -> Int
helper n i res
| n < 100 = res
| i == 1 = helper (div n 10) (i - 1) (res + (mod n 10)*10)
| otherwise = helper (div n 10) i res
Any help would be appreciated!
You can obtain the one but last digit of x with mod (div x 10) 10. You can use this with an accumulator that accumulates the value by each time multiplying with 10, so:
everyOther :: Int -> Int
everyOther = go 0
where go a v
| v < 10 = a
| otherwise = go (10*a + mod (div v 10) 10) (div v 100)
If v is thus less than 10, we can return the accumulator, since there is no "other digit" anymore. If that is not the case, we multiply a with 10, and add mod (div v 10) 10 to add the other digit to it, and recurse with the value divided by 100 to move it two places to the right.
We can improve this, as #Daniel Wagner says, by making use of quotRem :: Integral a => a -> a -> (a, a):
everyOther :: Int -> Int
everyOther = go 0
where go a v
| v < 10 = a
| otherwise = let (q, r) = v `quotRem` 100 in go (10*a + r `quot` 10) q
here we thus work with the remainder of a division by 100, and this thus avoids an extra modulo.
Can anybody help me with this function?
getCell :: [[Int]] -> Int -> Int -> Int
Where m i j are the indexes of the lines and the columns of the list of lists m.
the indexes start from zero and every line is the same size.
The function should return -1 if i or j are not valid.
I'm having an exam on Haskell, and despite the fact that this might show up, i still want to know how can i do it, and because i've never worked with lists of lists in Haskell, i have no idea how to start solving this problem. Can you give me a hand ?
here's what i've done so far:
getCell :: [[Int]] -> Int -> Int -> Int
getCell [] _ _ = "the list is empty!"
getCell zs x y =
if x > length zs || y > length (z:zs) then -1 else
let row = [x| x == !! head z <- zs]
column = ...
I don't know how to find the rows and the columns
This should work using the (!!) operator. First check if index is in the list, then access the element at that index using (!!).
getCell m i j = if i >= length m then -1
else let
m0 = m !! i
in if j >= length m0 then -1
else m0 !! j
Just for fun - one liner
getCell l i j = (((l ++ repeat []) !! i) ++ repeat (-1)) !! j
I need to count values inbetween values in a list i.e. [135,136,138,140] would count all the numbers between 135-136,136-138,138-140. with the input list [135.2,135.3,137,139] would out put[2,1,1] using type [Float] [Float] [Int]. So far I have:
heightbetween :: Float -> Float -> [Float] -> Int
heightbetween _ _ [] = 0
heightbetween n s (x:xs)
| (n < x) && (s > x) = 1 + (heightbetween n s xs)
| otherwise = heightbetween n s xs
count :: [Float] -> [Float] -> [Int]
count [] [] = []
count [x,y] = [(x,y)]
count (x:y:ys) = (x,y):count (y:ys)
forEach fun lst = heightbetween op ([],lst)
where
op (start,[]) = Nothing
op (start,a:as) = Just (start++(fun a):as
,(start++[a],as))
forPairs fun lst lst2 = map (map fst)
$ forEach (\(a,b)->(fun a b,b))
$ zip lst lst2
Your count looks strange. It should be like this:
-- count -> ranges -> data -> [counts]
count :: [Float] -> [Float] -> [Int]
count [] _ = [] -- no ranges given -> empty list
count [_] _ = [] -- no ranges, but single number -> empty list
count _ [] = [] -- no data given -> empty list
count (x:y:xs) d =
(heightbetween x y d) : count (y:xs) d
heightbetween :: Float -> Float -> [Float] -> Int
heightbetween _ _ [] = 0
heightbetween n s (x:xs)
| (n < x) && (s > x) = 1 + (heightbetween n s xs)
| otherwise = heightbetween n s xs
The other lines are obsolete.
Then invoking
count [135,136,138,140] [135.2,135.3,137,139]
gives
[2,1,1]
First, make sure that your range list is in order....
rangePoints = [135,136,138,140]
orderedRangePoints = sort rangePoints
Next, you will find it much easier to work with actual ranges (which you can represent using a 2-tuple (low,high))
ranges = zip orderedRangePoints $ tail orderedRangePoints
You will need an inRange function (one already exists in Data.Ix, but unfortunately it includes the upperbound, so you can't use it)
inRange (low,high) val | val >= low && val < high = True
inRange _ _ = False
You will also want to order your input points
theData = sort [135.2,135.3,137,139]
With all of this out of the way, the binCount function is easy to write.
binCount'::[(Float, Float)]->[Float]->[Int]
binCount' [] [] = []
binCount' (range:rest) vals =
length valsInRange:binCount' rest valsAboveRange
where
(valsInRange, valsAboveRange) = span (`inRange` range) vals
Notice, that I defined a function called binCount', not binCount. I did this, because I consider this an unsafe function, because it only works on ordered ranges and values.... You should finalize this by writing a safer binCount function, which puts all of the stuff above in its where clause. You should probably add all the types and some error checking also (what happens if a value is outside of all ranges?).
Can anybody help me with this function?
getCell :: [[Int]] -> Int -> Int -> Int
Where m i j are the indexes of the lines and the columns of the list of lists m.
the indexes start from zero and every line is the same size.
The function should return -1 if i or j are not valid.
I'm having an exam on Haskell, and despite the fact that this might show up, i still want to know how can i do it, and because i've never worked with lists of lists in Haskell, i have no idea how to start solving this problem. Can you give me a hand ?
here's what i've done so far:
getCell :: [[Int]] -> Int -> Int -> Int
getCell [] _ _ = "the list is empty!"
getCell zs x y =
if x > length zs || y > length (z:zs) then -1 else
let row = [x| x == !! head z <- zs]
column = ...
I don't know how to find the rows and the columns
This should work using the (!!) operator. First check if index is in the list, then access the element at that index using (!!).
getCell m i j = if i >= length m then -1
else let
m0 = m !! i
in if j >= length m0 then -1
else m0 !! j
Just for fun - one liner
getCell l i j = (((l ++ repeat []) !! i) ++ repeat (-1)) !! j
I'm stuck with my homework task, somebody help, please..
Here is the task:
Find all possible partitions of string into words of some dictionary
And here is how I'm trying to do it:
I use dynamical programming concept to fill matrix and then I'm stuck with how to retrieve data from it
-- Task5_2
retrieve :: [[Int]] -> [String] -> Int -> Int -> Int -> [[String]]
retrieve matrix dict i j size
| i >= size || j >= size = []
| index /= 0 = [(dict !! index)]:(retrieve matrix dict (i + sizeOfWord) (i + sizeOfWord) size) ++ retrieve matrix dict i (next matrix i j) size
where index = (matrix !! i !! j) - 1; sizeOfWord = length (dict !! index)
next matrix i j
| j >= (length matrix) = j
| matrix !! i !! j > 0 = j
| otherwise = next matrix i (j + 1)
getPartitionMatrix :: String -> [String] -> [[Int]]
getPartitionMatrix text dict = [[ indiceOfWord (getWord text i j) dict 1 | j <- [1..(length text)]] | i <- [1..(length text)]]
--------------------------
getWord :: String -> Int -> Int -> String
getWord text from to = map fst $ filter (\a -> (snd a) >= from && (snd a) <= to) $ zip text [1..]
indiceOfWord :: String -> [String] -> Int -> Int
indiceOfWord _ [] _ = 0
indiceOfWord word (x:xs) n
| word == x = n
| otherwise = indiceOfWord word xs (n + 1)
-- TESTS
dictionary = ["la", "a", "laa", "l"]
string = "laa"
matr = getPartitionMatrix string dictionary
test = retrieve matr dictionary 0 0 (length string)
Here is a code that do what you ask for. It doesn't work exactly like your solution but should work as fast if (and only if) both our dictionary lookup were improved to use tries as would be reasonable. As it is I think it may be a bit faster than your solution :
module Partitions (partitions) where
import Data.Array
import Data.List
data Branches a = Empty | B [([a],Branches a)] deriving (Show)
isEmpty Empty = True
isEmpty _ = False
flatten :: Branches a -> [ [ [a] ] ]
flatten Empty = []
flatten (B []) = [[]]
flatten (B ps) = concatMap (\(word, bs) -> ...) ps
type Dictionary a = [[a]]
partitions :: (Ord a) => Dictionary a -> [a] -> [ [ [a] ] ]
partitions dict xs = flatten (parts ! 0)
where
parts = listArray (0,length xs) $ zipWith (\i ys -> starting i ys) [0..] (tails xs)
starting _ [] = B []
starting i ys
| null words = ...
| otherwise = ...
where
words = filter (`isPrefixOf` ys) $ dict
go word = (word, parts ! (i + length word))
It works like this : At each position of the string, it search all possible words starting from there in the dictionary and evaluates to a Branches, that is either a dead-end (Empty) or a list of pairs of a word and all possible continuations after it, discarding those words that can't be continued.
Dynamic programming enter the picture to record every possibilities starting from a given index in a lazy array. Note that the knot is tied : we compute parts by using starting, which uses parts to lookup which continuations are possible from a given index. This only works because we only lookup indices after the one starting is computing and starting don't use parts for the last index.
To retrieve the list of partitions from this Branches datatype is analogous to the listing of all path in a tree.
EDIT : I removed some crucial parts of the solution in order to let the questioner search for himself. Though that shouldn't be too hard to complete with some thinking. I'll probably put them back with a somewhat cleaned up version later.