I'm a Haskell beginner and I am doing a small file for a project that should take the input of interaction data for groups of two people and save it to a list to be output at the end. I have done my best to implement this but it seems like the program hits the "stop" case no matter what is input. Any help or advice would be appreciated.
import Data.List
import Text.Read
main :: IO ()
main = do
putStrLn "This program is a means to record interactions between individuals during the COVID-19 pandemic."
putStrLn "Please enter your interactions in this format: 'x interacted with y'"
inputs <- getUserInputs
putStr "input: "
putStrLn ("list sequence " ++ show (inputs))
parseInput :: String -> Maybe String
parseInput input = if input == "stop" then Nothing else (readMaybe input):: Maybe String
getUserInputs :: IO [String]
getUserInputs = do
input <- getLine
case parseInput input of
Nothing -> return []
Just aString -> do
moreinputs <- getUserInputs
return (aString : moreinputs)
Show and Read are intended to produce and consume the representation of a value as a Haskell expression. That’s why when you call show on a String, it produces a quoted string:
> show "beans"
"\"beans\""
Therefore Read expects the string to be quoted as well, so readMaybe is always returning Nothing in your code because you’re not supplying quotes:
> readMaybe "beans" :: Maybe String
Nothing
> readMaybe "\"beans\"" :: Maybe String
Just "beans"
Therefore the fix is simple: remove the call to readMaybe and just return the string directly:
parseInput1 :: String -> Maybe String
parseInput1 input = if input == "stop"
then Nothing
else Just input
Which, as a matter of style preference, you could also write with guards, pattern matching, or the Maybe monad instead of if:
parseInput2 input
| input == "stop" = Nothing
| otherwise = Just input
parseInput3 "stop" = Nothing
parseInput3 input = Just input
import Control.Monad (guard)
parseInput4 input = do
-- ‘guard’ returns ‘Nothing’,
-- short-circuiting the ‘do’ block,
-- if its condition is ‘False’.
guard (input /= "stop")
pure input
Read and Show are fine for simple programs, particularly when you’re learning Haskell, but in larger applications it’s helpful to use them mostly for debug input and output and reading input you’ve already validated. Parsing and pretty-printing libraries are preferable for more involved parsing and producing human-readable output, respectively; megaparsec and prettyprinter are good default choices in that area.
Related
I have been trying to a read file creating data in haskell. i have the following data type constructed:
data Participant = Participant {name:: String, age:: Int,
country:: String, }
The text file i have is in this format
Jack 21 England
Natalie 20 France
Sophie 24 France
Each word corresponds to name age and country respectivly. I want to read the file and create a list of participants. But IO:String seem to be a pain in the neck. Do you have a suitable soliton for this problem.
IO isolation is not a defect, but a feature of Haskell, and one of its clear success.
If you want to serialize data to file, and load data back into your program, you don't have to do that by hand. Use any serializing library, there are many of them. You can also use the great Aeson library, that can convert your data in JSON and load them back.
If you want to do that by hand for any reason, you must first define your own file format, and have it unambiguous. For example, what happens on your example if name contains a space ?
Then you should define a function that can parse one line of your file format and produce a Participant readParticipant :: String -> Maybe Participant. Notice the maybe, because if the string is ill-formated, your function won't be able to create a participant out of its hat, so it will produce a Nothing.
Then you can have a function that parse a list of participants (notice the plural) readParticipants :: String -> [Participants]. No Maybe here because the list itself allows for failure.
Now you can have a tiny IO function that will read the file content, and run readParticipants on it.
readParticipantsIO :: IO [Participants]
readParticipantsIO = readParticipants <$> readFile "participants.data"
-- alternative definition, exact same behaviour in this case
readParticipantsIO :: IO [Participants]
readParticipantsIO = do
content <- readFile "participants.data"
return $ readParticipants content
The lines and words functions can be used to split the string into a list of strings, and then you can package these strings into a list of Participant.
import Text.Read (readMaybe)
data Participant = Participant { name :: String, age :: Int, country :: String }
parseParticipants :: String -> [Participant]
parseParticipants fileContents = do
[name', age', country'] <- words <$> lines fileContents
Just age'' <- return (readMaybe age')
return (Participant { name = name', age = age'', country = country' })
main :: IO ()
main = do
participants <- parseParticipants <$> readFile "yourfile.txt"
-- do things with the participants
return ()
I have the following ocaml code:
let rec c_write =
"printf(\" %d \");\n"
On calling this function in the interpreter, I expect to get the output
printf("%d"); followed by a new line, but instead I get
printf(\" %d \");\n
How can I get my expected output when I'm calling the function without using any other I/O functions?
The expression let rec c_write = "printf(\" %d \");\n" is not a function. It is a value of type string which is bound to a variable named c_write. So you're not using any I/O functions in your code.
When entered in the interactive toplevel, this value is printed by the interpreter evaluation loop for user convenience. The same as when a Python interpreter will print for you the value that you've just entered.
The representation, chosen by the OCaml toplevel interpreter, in general, has nothing to do with the representation which is used to store a value in a file or to print it. Moreover, in OCaml, there is no canonical representations.
If you want to write a function that prints a C printf statement then this is how it will look like in OCaml
let print_printf () =
print_endline {|printf("%d");|}
In the example above, I've used {||} to denote a sting literal instead of more common "", since in this literal there is no need to escape special characters and they are interpreted literally (i.e., the don't have any special meaning).
You can achieve the same result using the regular "" quotes for denoting it
let print_printf () =
print_endline "printf(\"%d\");"
Here is an example of the toplevel interaction using these definitions:
# let print_printf () =
print_endline {|printf("%d");|};;
val print_printf : unit -> unit = <fun>
# print_printf ();;
printf("%d");
- : unit = ()
# let print_printf () =
print_endline "printf(\"%d\");";;
val print_printf : unit -> unit = <fun>
# print_printf ();;
printf("%d");
- : unit = ()
If you will put this code in a file, compile, and execute and redirect into a C file it will be a well-formed C file (modulo the absence of the function body).
Since you are somehow using the toplevel printer for printing, and that you somehow needs a very specific format, you need to install a custom printer.
The following would work:
# #install_printer Format.pp_print_string;;
# " This \" is not escaped " ;;
- : string = This " is not escaped
However, it seems very likely that this is not really the problem that you are trying to solve.
I'm writing a lexer in Alex with the monad wrapper. It's not behaving as I expect, and I would like to write some unit tests for it. I can write unit tests for lexing a single token by doing:
runAlex "foo" alexMonadScan `shouldBe` Right TokenFoo
but I don't know how to test that the string "foo bar" gets lexed to [TokenFoo, TokenBar].
Given that Token is my token type, I'd need a function like runAlex that has the type String -> Alex [Token] -> Either String [Token], but I don't know how to transform alexMonadScan so that it has the type Alex [Token] rather than Alex Token.
I tried
runAlex "foo bar" (liftM (:[]) alexMonadScan) `shouldBe` [TokenFoo, TokenBar]
which seems to have the right type, but it returns Right [TokenEOF], apparently dropping the tokens it saw along the way.
How can I achieve this?
There is a function alexScanTokens :: String -> [token] which you can use.
It's defined in the file templates/wrappers.hs
Here's a monadic version I found here:
alexScanTokens :: String -> Either String [Keyword]
alexScanTokens inp = runAlex inp gather
where
gather = do
t <- alexMonadScan
case trace (show t) t of
EOF -> return [EOF]
_ -> (t:) `liftM` gather
I need to scan through a document and accumulate the output of different functions for each string in the file. The function run on any given line of the file depends on what is in that line.
I could do this very inefficiently by making a complete pass through the file for every list I wanted to collect. Example pseudo-code:
at :: B.ByteString -> Maybe Atom
at line
| line == ATOM record = do stuff to return Just Atom
| otherwise = Nothing
ot :: B.ByteString -> Maybe Sheet
ot line
| line == SHEET record = do other stuff to return Just Sheet
| otherwise = Nothing
Then, I would map each of these functions over the entire list of lines in the file to get a complete list of Atoms and Sheets:
mapper :: [B.ByteString] -> IO ()
mapper lines = do
let atoms = mapMaybe at lines
let sheets = mapMaybe to lines
-- Do stuff with my atoms and sheets
However, this is inefficient because I am maping through the entire list of strings for every list I am trying to create. Instead, I want to map through the list of line strings only once, identify each line as I am moving through it, and then apply the appropriate function and store these values in different lists.
My C mentality wants to do this (pseudo code):
mapper' :: [B.ByteString] -> IO ()
mapper' lines = do
let atoms = []
let sheets = []
for line in lines:
| line == ATOM record = (atoms = atoms ++ at line)
| line == SHEET record = (sheets = sheets ++ ot line)
-- Now 'atoms' is a complete list of all the ATOM records
-- and 'sheets' is a complete list of all the SHEET records
What is the Haskell way of doing this? I simply can't get my functional-programming mindset to come up with a solution.
First of all, I think that the answers others have supplied will work at least 95% of the time. It's always good practice to code for the problem at hand by using appropriate data types (or tuples in some cases). However, sometimes you really don't know in advance what you're looking for in the list, and in these cases trying to enumerate all possibilities is difficult/time-consuming/error-prone. Or, you're writing multiple variants of the same sort of thing (manually inlining multiple folds into one) and you'd like to capture the abstraction.
Fortunately, there are a few techniques that can help.
The framework solution
(somewhat self-evangelizing)
First, the various "iteratee/enumerator" packages often provide functions to deal with this sort of problem. I'm most familiar with iteratee, which would let you do the following:
import Data.Iteratee as I
import Data.Iteratee.Char
import Data.Maybe
-- first, you'll need some way to process the Atoms/Sheets/etc. you're getting
-- if you want to just return them as a list, you can use the built-in
-- stream2list function
-- next, create stream transformers
-- given at :: B.ByteString -> Maybe Atom
-- create a stream transformer from ByteString lines to Atoms
atIter :: Enumeratee [B.ByteString] [Atom] m a
atIter = I.mapChunks (catMaybes . map at)
otIter :: Enumeratee [B.ByteString] [Sheet] m a
otIter = I.mapChunks (catMaybes . map ot)
-- finally, combine multiple processors into one
-- if you have more than one processor, you can use zip3, zip4, etc.
procFile :: Iteratee [B.ByteString] m ([Atom],[Sheet])
procFile = I.zip (atIter =$ stream2list) (otIter =$ stream2list)
-- and run it on some data
runner :: FilePath -> IO ([Atom],[Sheet])
runner filename = do
resultIter <- enumFile defaultBufSize filename $= enumLinesBS $ procFile
run resultIter
One benefit this gives you is extra composability. You can create transformers as you like, and just combine them with zip. You can even run the consumers in parallel if you like (although only if you're working in the IO monad, and probably not worth it unless the consumers do a lot of work) by changing to this:
import Data.Iteratee.Parallel
parProcFile = I.zip (parI $ atIter =$ stream2list) (parI $ otIter =$ stream2list)
The result of doing so isn't the same as a single for-loop - this will still perform multiple traversals of the data. However, the traversal pattern has changed. This will load a certain amount of data at once (defaultBufSize bytes) and traverse that chunk multiple times, storing partial results as necessary. After a chunk has been entirely consumed, the next chunk is loaded and the old one can be garbage collected.
Hopefully this will demonstrate the difference:
Data.List.zip:
x1 x2 x3 .. x_n
x1 x2 x3 .. x_n
Data.Iteratee.zip:
x1 x2 x3 x4 x_n-1 x_n
x1 x2 x3 x4 x_n-1 x_n
If you're doing enough work that parallelism makes sense this isn't a problem at all. Due to memory locality, the performance is much better than multiple traversals over the entire input as Data.List.zip would make.
The beautiful solution
If a single-traversal solution really does make the most sense, you might be interested in Max Rabkin's Beautiful Folding post, and Conal Elliott's followup work (this too). The essential idea is that you can create data structures to represent folds and zips, and combining these lets you create a new, combined fold/zip function that only needs one traversal. It's maybe a little advanced for a Haskell beginner, but since you're thinking about the problem you may find it interesting or useful. Max's post is probably the best starting point.
I show a solution for two types of line, but it is easily extended to five types of line by using a five-tuple instead of a two-tuple.
import Data.Monoid
eachLine :: B.ByteString -> ([Atom], [Sheet])
eachLine bs | isAnAtom bs = ([ {- calculate an Atom -} ], [])
| isASheet bs = ([], [ {- calculate a Sheet -} ])
| otherwise = error "eachLine"
allLines :: [B.ByteString] -> ([Atom], [Sheet])
allLines bss = mconcat (map eachLine bss)
The magic is done by mconcat from Data.Monoid (included with GHC).
(On a point of style: personally I would define a Line type, a parseLine :: B.ByteString -> Line function and write eachLine bs = case parseLine bs of .... But this is peripheral to your question.)
It is a good idea to introduce a new ADT, e.g. "Summary" instead of tuples.
Then, since you want to accumulate the values of Summary you came make it an istance of Data.Monoid. Then you classify each of your lines with the help of classifier functions (e.g. isAtom, isSheet, etc.) and concatenate them together using Monoid's mconcat function (as suggested by #dave4420).
Here is the code (it uses String instead of ByteString, but it is quite easy to change):
module Classifier where
import Data.List
import Data.Monoid
data Summary = Summary
{ atoms :: [String]
, sheets :: [String]
, digits :: [String]
} deriving (Show)
instance Monoid Summary where
mempty = Summary [] [] []
Summary as1 ss1 ds1 `mappend` Summary as2 ss2 ds2 =
Summary (as1 `mappend` as2)
(ss1 `mappend` ss2)
(ds1 `mappend` ds2)
classify :: [String] -> Summary
classify = mconcat . map classifyLine
classifyLine :: String -> Summary
classifyLine line
| isAtom line = Summary [line] [] [] -- or "mempty { atoms = [line] }"
| isSheet line = Summary [] [line] []
| isDigit line = Summary [] [] [line]
| otherwise = mempty -- or "error" if you need this
isAtom, isSheet, isDigit :: String -> Bool
isAtom = isPrefixOf "atom"
isSheet = isPrefixOf "sheet"
isDigit = isPrefixOf "digits"
input :: [String]
input = ["atom1", "sheet1", "sheet2", "digits1"]
test :: Summary
test = classify input
If you have only 2 alternatives, using Either might be a good idea. In that case combine your functions, map the list, and use lefts and rights to get the results:
import Data.Either
-- first sample function, returning String
f1 x = show $ x `div` 2
-- second sample function, returning Int
f2 x = 3*x+1
-- combined function returning Either String Int
hotpo x = if even x then Left (f1 x) else Right (f2 x)
xs = map hotpo [1..10]
-- [Right 4,Left "1",Right 10,Left "2",Right 16,Left "3",Right 22,Left "4",Right 28,Left "5"]
lefts xs
-- ["1","2","3","4","5"]
rights xs
-- [4,10,16,22,28]
I have a question about tuples and lists in Haskell. I know how to add input into a tuple a specific number of times. Now I want to add tuples into a list an unknown number of times; it's up to the user to decide how many tuples they want to add.
How do I add tuples into a list x number of times when I don't know X beforehand?
There's a lot of things you could possibly mean. For example, if you want a few copies of a single value, you can use replicate, defined in the Prelude:
replicate :: Int -> a -> [a]
replicate 0 x = []
replicate n | n < 0 = undefined
| otherwise = x : replicate (n-1) x
In ghci:
Prelude> replicate 4 ("Haskell", 2)
[("Haskell",2),("Haskell",2),("Haskell",2),("Haskell",2)]
Alternately, perhaps you actually want to do some IO to determine the list. Then a simple loop will do:
getListFromUser = do
putStrLn "keep going?"
s <- getLine
case s of
'y':_ -> do
putStrLn "enter a value"
v <- readLn
vs <- getListFromUser
return (v:vs)
_ -> return []
In ghci:
*Main> getListFromUser :: IO [(String, Int)]
keep going?
y
enter a value
("Haskell",2)
keep going?
y
enter a value
("Prolog",4)
keep going?
n
[("Haskell",2),("Prolog",4)]
Of course, this is a particularly crappy user interface -- I'm sure you can come up with a dozen ways to improve it! But the pattern, at least, should shine through: you can use values like [] and functions like : to construct lists. There are many, many other higher-level functions for constructing and manipulating lists, as well.
P.S. There's nothing particularly special about lists of tuples (as compared to lists of other things); the above functions display that by never mentioning them. =)
Sorry, you can't1. There are fundamental differences between tuples and lists:
A tuple always have a finite amount of elements, that is known at compile time. Tuples with different amounts of elements are actually different types.
List an have as many elements as they want. The amount of elements in a list doesn't need to be known at compile time.
A tuple can have elements of arbitrary types. Since the way you can use tuples always ensures that there is no type mismatch, this is safe.
On the other hand, all elements of a list have to have the same type. Haskell is a statically-typed language; that basically means that all types are known at compile time.
Because of these reasons, you can't. If it's not known, how many elements will fit into the tuple, you can't give it a type.
I guess that the input you get from your user is actually a string like "(1,2,3)". Try to make this directly a list, whithout making it a tuple before. You can use pattern matching for this, but here is a slightly sneaky approach. I just remove the opening and closing paranthesis from the string and replace them with brackets -- and voila it becomes a list.
tuplishToList :: String -> [Int]
tuplishToList str = read ('[' : tail (init str) ++ "]")
Edit
Sorry, I did not see your latest comment. What you try to do is not that difficult. I use these simple functions for my task:
words str splits str into a list of words that where separated by whitespace before. The output is a list of Strings. Caution: This only works if the string inside your tuple contains no whitespace. Implementing a better solution is left as an excercise to the reader.
map f lst applies f to each element of lst
read is a magic function that makes a a data type from a String. It only works if you know before, what the output is supposed to be. If you really want to understand how that works, consider implementing read for your specific usecase.
And here you go:
tuplish2List :: String -> [(String,Int)]
tuplish2List str = map read (words str)
1 As some others may point out, it may be possible using templates and other hacks, but I don't consider that a real solution.
When doing functional programming, it is often better to think about composition of operations instead of individual steps. So instead of thinking about it like adding tuples one at a time to a list, we can approach it by first dividing the input into a list of strings, and then converting each string into a tuple.
Assuming the tuples are written each on one line, we can split the input using lines, and then use read to parse each tuple. To make it work on the entire list, we use map.
main = do input <- getContents
let tuples = map read (lines input) :: [(String, Integer)]
print tuples
Let's try it.
$ runghc Tuples.hs
("Hello", 2)
("Haskell", 4)
Here, I press Ctrl+D to send EOF to the program, (or Ctrl+Z on Windows) and it prints the result.
[("Hello",2),("Haskell",4)]
If you want something more interactive, you will probably have to do your own recursion. See Daniel Wagner's answer for an example of that.
One simple solution to this would be to use a list comprehension, as so (done in GHCi):
Prelude> let fstMap tuplist = [fst x | x <- tuplist]
Prelude> fstMap [("String1",1),("String2",2),("String3",3)]
["String1","String2","String3"]
Prelude> :t fstMap
fstMap :: [(t, b)] -> [t]
This will work for an arbitrary number of tuples - as many as the user wants to use.
To use this in your code, you would just write:
fstMap :: Eq a => [(a,b)] -> [a]
fstMap tuplist = [fst x | x <- tuplist]
The example I gave is just one possible solution. As the name implies, of course, you can just write:
fstMap' :: Eq a => [(a,b)] -> [a]
fstMap' = map fst
This is an even simpler solution.
I'm guessing that, since this is for a class, and you've been studying Haskell for < 1 week, you don't actually need to do any input/output. That's a bit more advanced than you probably are, yet. So:
As others have said, map fst will take a list of tuples, of arbitrary length, and return the first elements. You say you know how to do that. Fine.
But how do the tuples get into the list in the first place? Well, if you have a list of tuples and want to add another, (:) does the trick. Like so:
oldList = [("first", 1), ("second", 2)]
newList = ("third", 2) : oldList
You can do that as many times as you like. And if you don't have a list of tuples yet, your list is [].
Does that do everything that you need? If not, what specifically is it missing?
Edit: With the corrected type:
Eq a => [(a, b)]
That's not the type of a function. It's the type of a list of tuples. Just have the user type yourFunctionName followed by [ ("String1", val1), ("String2", val2), ... ("LastString", lastVal)] at the prompt.