Hey guys, I need some help, how can I sum a list from a file using the sum builtin function?
example: text.txt contains [1,2,3,4]
and i want to get this list from the file and sum these numbers with the sum builtin. Is it possible?
Thanks!
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1See also this FAQ in the haskell+io tag.Don Stewart– Don Stewart2011-05-11 22:07:32 +00:00Commented May 11, 2011 at 22:07
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3 Answers
So, you won't be able to actually get a value :: [Int], because that would be unsafe. But we can get you to IO [Int], and then pass that through >>= into print:
main = (sum . read) `fmap` readFile "myFile" >>= print
If you're using Control.Applicative, you can make it even prettier:
main = sum . read <$> readFile "myFile" >>= print
Voilà! As you can see, this way is much more succinct than using do-notation. do-notation is awesome for getting rid of nested lambdas, but in a lot of cases, it's really not necessary.
Edit: @augustss suggests the following superior alternative:
main = print . sum . read =<< readFile "myFile"
This is awesome, in that it's simpler, and doesn't depend on Control.Applicative; even more importantly, it doesn't have to read be read "inside-out", as @hammar points out.
answered May 11, 2011 at 21:48
Jonathan Sterling
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2 Comments
Jonathan Sterling
@augustss Wow, that's way better! Thanks.
hammar
It also reads better, from right to left, whereas you have to read the applicative one "inside out".
Sure.
main = do
contents <- readFile "text.txt"
print . sum . read $ contents
This uses some standard functions from the Prelude:
readFile :: FilePath -> IO Stringreads the file into a string.read :: Read a => String -> aconverts from aStringto any type in theReadtypeclass. You might have to add a type annotation if the compiler couldn't figure out which type you wanted. In this case the compiler can figure it out since we usedsumon the result, so it infers thatamust be[Integer]. (It actually infersNum a => [a], but this defaults to[Integer]due to type defaulting).readexpects the same format generated byshowon the same type.
Note that I had to use do-notation to extract the String from IO String in order to apply read to it.
1 Comment
Robin Green
Technically the compiler guesses which type you wanted (by using a default), since
sum can operate on multiple numeric types.If the list is always in this kind of format, you want to use the read function. This function deserializes data, i.e. turns a string into a program data type. For example,
Prelude> read "[1, 2, 3, 4]" :: [Int]
[1,2,3,4]
Now, you can combine this with file IO, but you'll need to read up on Monads if you're not familiar. Some good resources are [ http://www.haskell.org/haskellwiki/Monad ] and Wikipedia. Essentially, each monad will represent a step in a sequential execution -- which is necessary since IO modifies the environment, hence its execution order must be determined.
The code I came up with for reading and then summing is:
main = do
fc <- readFile "file1"
parsed <- return (read fc :: [Int])
putStrLn (printf "sum: %d" (sum parsed))
2 Comments
Alehar
Thanks, though it's not nice it doesn't understand space delimited numbers without "[" chars around it.
gatoatigrado
@Alehar if you don't have braces, you can just try the
words function, the strip off trailing commas. If you need something more advanced, you might try a Parsec parser.