I'm trying to write in Haskell a function that in Java would be something like this: char find_match (char[] l1, char[] l2, char e){ //l1 and l2 are not empty int i = 0; while (l2){ char aux = l2[i]; char[n] laux = l2; while(laux){ int j = 0; if(laux[j] = aux) laux[j] = e; j++; } if compare (l1, laux) return aux; else i++; } return ''; } compare function just compares the two lists and return true if they are equal, or false if they are not. it is really a simple function, but I've been thinking about it a lot of time and I can't get the goal. It works like this: find_match "4*h&a" "4*5&a" 'h' ----> returns '5' (5 matches with the h) find_match "4*n&s" "4dhnn" "k" ----> returns '' (no match at all - lists are different anyway) I'm trying to use map for scanning the list l2, then using some function for substitute the char, and finally something like zipWith x == y laux l1 for comparing the lists, but I can't find a way for putting all this together!! -- View this message in context: http://www.nabble.com/Java-or-C-to-Haskell-tf2303820.html#a6403589 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.
crespi.albert:
I'm trying to write in Haskell a function that in Java would be something like this:
char find_match (char[] l1, char[] l2, char e){ //l1 and l2 are not empty int i = 0; while (l2){ char aux = l2[i]; char[n] laux = l2; while(laux){ int j = 0; if(laux[j] = aux) laux[j] = e; j++; } if compare (l1, laux) return aux; else i++; } return ''; }
Yikes!
compare function just compares the two lists and return true if they are equal, or false if they are not. it is really a simple function, but I've been thinking about it a lot of time and I can't get the goal. It works like this:
find_match "4*h&a" "4*5&a" 'h' ----> returns '5' (5 matches with the h) find_match "4*n&s" "4dhnn" "k" ----> returns '' (no match at all - lists are different anyway)
That's almost a spec there :) How about: import Data.List findMatch s t c | Just n <- elemIndex c s = Just (t !! n) | otherwise = Nothing Using it in GHCi:
findMatch "4*h&a" "4*5&a" 'h' Just '5'
findMatch "4*n&s" "4dhnn" 'k' Nothing
-- Don
dons:
crespi.albert:
I'm trying to write in Haskell a function that in Java would be something like this:
char find_match (char[] l1, char[] l2, char e){ //l1 and l2 are not empty int i = 0; while (l2){ char aux = l2[i]; char[n] laux = l2; while(laux){ int j = 0; if(laux[j] = aux) laux[j] = e; j++; } if compare (l1, laux) return aux; else i++; } return ''; }
Yikes!
compare function just compares the two lists and return true if they are equal, or false if they are not. it is really a simple function, but I've been thinking about it a lot of time and I can't get the goal. It works like this:
find_match "4*h&a" "4*5&a" 'h' ----> returns '5' (5 matches with the h) find_match "4*n&s" "4dhnn" "k" ----> returns '' (no match at all - lists are different anyway)
That's almost a spec there :)
Ah, I see I misread the spec :) Time for some tea. -- Don
That works good, but I have a problem with the return type, I forgot to mention... can it be a [char]?? Donald Bruce Stewart wrote:
crespi.albert:
I'm trying to write in Haskell a function that in Java would be something like this:
char find_match (char[] l1, char[] l2, char e){ //l1 and l2 are not empty int i = 0; while (l2){ char aux = l2[i]; char[n] laux = l2; while(laux){ int j = 0; if(laux[j] = aux) laux[j] = e; j++; } if compare (l1, laux) return aux; else i++; } return ''; }
Yikes!
compare function just compares the two lists and return true if they are equal, or false if they are not. it is really a simple function, but I've been thinking about it a lot of time and I can't get the goal. It works like this:
find_match "4*h&a" "4*5&a" 'h' ----> returns '5' (5 matches with the h) find_match "4*n&s" "4dhnn" "k" ----> returns '' (no match at all - lists are different anyway)
That's almost a spec there :)
How about:
import Data.List
findMatch s t c | Just n <- elemIndex c s = Just (t !! n) | otherwise = Nothing
Using it in GHCi:
findMatch "4*h&a" "4*5&a" 'h' Just '5'
findMatch "4*n&s" "4dhnn" 'k' Nothing
-- Don _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
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Carajillu <crespi.albert@gmail.com> writes:
That works good, but I have a problem with the return type, I forgot to mention... can it be a [char]??
If that's what you want, how about this: import Maybe find_match l1 l2 c = fmap catMaybes . sequence $ zipWith match l1 l2 where match a b | a == c = Just (Just b) | a == b = Just Nothing | otherwise = Nothing although part of the reason for writing it like that is to make you work hard to understand it ;-) ... and it returns Maybe [Char] since I can't bring myself to use "" to indicate failure... but judicious use of catMaybes . concat. maybeToList might help with that. -- Jón Fairbairn Jon.Fairbairn@cl.cam.ac.uk
On Wed, Sep 20, 2006 at 01:31:22AM -0700, Carajillu wrote:
I'm trying to write in Haskell a function that in Java would be something like this:
char find_match (char[] l1, char[] l2, char e){ //l1 and l2 are not empty int i = 0; while (l2){ char aux = l2[i]; char[n] laux = l2; while(laux){ int j = 0; if(laux[j] = aux) laux[j] = e; j++; } if compare (l1, laux) return aux; else i++; } return ''; }
compare function just compares the two lists and return true if they are equal, or false if they are not.
I know that this is far too simple. But I'm simple minded: comp [] [] = True comp (x:xs) (y:ys) = if x == y then comp xs ys else False andrea
On Wed, Sep 20, 2006 at 01:31:22AM -0700, Carajillu wrote:
compare function just compares the two lists and return true if they are equal, or false if they are not. it is really a simple function, but I've been thinking about it a lot of time and I can't get the goal.
I forgot, obviously, that lists are an instance of the Eq class... so, this is enough: comp l1 l2 = if l1 == l2 then True else False You never stop learning! andrea
wow, the simpliest ever! Andrea Rossato wrote:
On Wed, Sep 20, 2006 at 01:31:22AM -0700, Carajillu wrote:
compare function just compares the two lists and return true if they are equal, or false if they are not. it is really a simple function, but I've been thinking about it a lot of time and I can't get the goal.
I forgot, obviously, that lists are an instance of the Eq class... so, this is enough: comp l1 l2 = if l1 == l2 then True else False
You never stop learning! andrea _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
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mailing_list:
On Wed, Sep 20, 2006 at 01:31:22AM -0700, Carajillu wrote:
compare function just compares the two lists and return true if they are equal, or false if they are not. it is really a simple function, but I've been thinking about it a lot of time and I can't get the goal.
I forgot, obviously, that lists are an instance of the Eq class... so, this is enough: comp l1 l2 = if l1 == l2 then True else False
You never stop learning! andrea
which you would just write as: comp = (==) and then you'd just use == anyway :) -- Don
On Wed, Sep 20, 2006 at 07:20:23PM +1000, Donald Bruce Stewart wrote:
comp l1 l2 = if l1 == l2 then True else False
You never stop learning! andrea
which you would just write as: comp = (==)
and then you'd just use == anyway :)
this is why I came to love haskell: it remembers me when I was doing mathematics at the high school. The most challenging stuff, for me, was finding a way to simplifying expressions... It's just an endless chess game. Sending mails to this mailing list, I mean. Andrea
Andrea Rossato <mailing_list@istitutocolli.org> writes:
I forgot, obviously, that lists are an instance of the Eq class... so, this is enough:
comp l1 l2 = if l1 == l2 then True else False
Or why not:
comp l1 l2 = l1 == l2
Or simply:
comp = (==)
:-) -k -- If I haven't seen further, it is by standing in the footprints of giants
Carajillu <crespi.albert@gmail.com> writes:
compare function just compares the two lists and return true if they are equal, or false if they are not.
find_match "4*h&a" "4*5&a" 'h' ----> returns '5' (5 matches with the h) find_match "4*n&s" "4dhnn" "k" ----> returns '' (no match at all - lists are different anyway)
Must they be equal the whole way, or just up to the occurrence of the searched-for character? find_match (x:xs) (y:ys) c | x==c = Just y | x/=y = Nothing | True = find_match xs ys c find_match [] [] _ = Nothing Or, to check the whole list: find_match (x:xs) (y:ys) c | x==c && xs == ys = Just y | x/=y = Nothing | True = find_match xs ys c find_match [] [] _ = Nothing -k -- If I haven't seen further, it is by standing in the footprints of giants
Yes, they must be equal the whole way, I like this recursive solution :) Ketil Malde-3 wrote:
Carajillu <crespi.albert@gmail.com> writes:
compare function just compares the two lists and return true if they are equal, or false if they are not.
find_match "4*h&a" "4*5&a" 'h' ----> returns '5' (5 matches with the h) find_match "4*n&s" "4dhnn" "k" ----> returns '' (no match at all - lists are different anyway)
Must they be equal the whole way, or just up to the occurrence of the searched-for character?
find_match (x:xs) (y:ys) c | x==c = Just y | x/=y = Nothing | True = find_match xs ys c find_match [] [] _ = Nothing
Or, to check the whole list:
find_match (x:xs) (y:ys) c | x==c && xs == ys = Just y | x/=y = Nothing | True = find_match xs ys c find_match [] [] _ = Nothing
-k -- If I haven't seen further, it is by standing in the footprints of giants
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... and if you want to search strings not single characters: findmatch s t e = take m . drop n $ t where m' = length e (n, m) = f 0 s f i s | take m' s == e = (i, m') | null s = (0, 0) | otherwise = f (i+1) (tail s) findmatch "asdfasdf" "asdfxvdf" "fas" == "fxv" (this one skips equality checks before *and* after the match. feel free post the necessary modifications. :) matthias On Wed, Sep 20, 2006 at 02:22:29AM -0700, Carajillu wrote:
To: haskell-cafe@haskell.org From: Carajillu <crespi.albert@gmail.com> Date: Wed, 20 Sep 2006 02:22:29 -0700 (PDT) Subject: Re: [Haskell-cafe] Java or C to Haskell
Yes, they must be equal the whole way, I like this recursive solution :)
Ketil Malde-3 wrote:
Carajillu <crespi.albert@gmail.com> writes:
compare function just compares the two lists and return true if they are equal, or false if they are not.
find_match "4*h&a" "4*5&a" 'h' ----> returns '5' (5 matches with the h) find_match "4*n&s" "4dhnn" "k" ----> returns '' (no match at all - lists are different anyway)
Must they be equal the whole way, or just up to the occurrence of the searched-for character?
find_match (x:xs) (y:ys) c | x==c = Just y | x/=y = Nothing | True = find_match xs ys c find_match [] [] _ = Nothing
Or, to check the whole list:
find_match (x:xs) (y:ys) c | x==c && xs == ys = Just y | x/=y = Nothing | True = find_match xs ys c find_match [] [] _ = Nothing
-k
How about something like this? import Data.List findMatch xs ys k = lookup k . concat $ zipWith zip (substrings xs) (substrings ys) where substrings = nonempty . map (nonempty . inits) . tails where nonempty = filter (not . null) On 20/09/06, Matthias Fischmann <fis@wiwi.hu-berlin.de> wrote:
... and if you want to search strings not single characters:
findmatch s t e = take m . drop n $ t where m' = length e (n, m) = f 0 s f i s | take m' s == e = (i, m') | null s = (0, 0) | otherwise = f (i+1) (tail s)
findmatch "asdfasdf" "asdfxvdf" "fas" == "fxv"
(this one skips equality checks before *and* after the match. feel free post the necessary modifications. :)
matthias
On Wed, Sep 20, 2006 at 02:22:29AM -0700, Carajillu wrote:
To: haskell-cafe@haskell.org From: Carajillu <crespi.albert@gmail.com> Date: Wed, 20 Sep 2006 02:22:29 -0700 (PDT) Subject: Re: [Haskell-cafe] Java or C to Haskell
Yes, they must be equal the whole way, I like this recursive solution :)
Ketil Malde-3 wrote:
Carajillu <crespi.albert@gmail.com> writes:
compare function just compares the two lists and return true if they are equal, or false if they are not.
find_match "4*h&a" "4*5&a" 'h' ----> returns '5' (5 matches with the h) find_match "4*n&s" "4dhnn" "k" ----> returns '' (no match at all - lists are different anyway)
Must they be equal the whole way, or just up to the occurrence of the searched-for character?
find_match (x:xs) (y:ys) c | x==c = Just y | x/=y = Nothing | True = find_match xs ys c find_match [] [] _ = Nothing
Or, to check the whole list:
find_match (x:xs) (y:ys) c | x==c && xs == ys = Just y | x/=y = Nothing | True = find_match xs ys c find_match [] [] _ = Nothing
-k
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participants (7)
-
Andrea Rossato -
Cale Gibbard -
Carajillu -
dons＠cse.unsw.edu.au -
Jón Fairbairn -
Ketil Malde -
Matthias Fischmann