On Mon, 2007-09-10 at 15:47 +1000, Stuart Cook wrote: > On 9/10/07, PR Stanley <[EMAIL PROTECTED]> wrote: > > Hi > > Any comments and/or criticisms would be most appreciated: > > --count the occurrences of char in string > > countC :: Char -> [Char] -> Int > > countC x xs = sum [1 | c <- xs, c == x] > > That's a clever implementation, but I think there are clearer ways of > achieving the same goal. You could replace sum with length and get the > same answer, at which point the list comprehension isn't buying you > much. > > How about this? > > countC ch xs = length $ filter (ch ==) xs > > > --count occurrences of chars in string > > countCS :: [Char] -> [(Char, Int)] > > countCS xs = [(x, (countC x xs)) | x <- [' '..'z'], (countC x xs) > 0] > > A few things to note: > > * Those extra parens around countC are unnecessary; function > application has highest precedence. > * [' '..'z'] has a few issues: it misses a few ASCII characters > (including all the control codes), and won't catch non-ASCII > characters. > * Duplicating the call to countC will probably result in redundant > evaluation, so it's best to avoid that if possible -- difficult to > avoid cleanly inside a comprehension though. > * Scanning the string repeatedly for each potential character [' > '..'z'] seems excessive, especially if you want to expand it to > include non-ASCII. Better just to deal with the characters actually in > the string as they appear, I think.
You can use 'let' in a list comprehension, [(x,count) | x <- [' '..'z'], let count = countC x xs, count > 0] _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
