getChar :: Int -> IO Char
getChar n = do
c <- getLine
return (c !! n)
The program must needs a number and a line and it will return char, but how do I catch exception, if the number is too big?
I tried like this but it doesnt seem to work
getChar n
= do
c <-getLine
| n>=0 && n < b
= return c !! n
| otherwise
= error "Too big number"
where
b = length c
This is not a homework, im trying to involve myself. Google didint give me useful answers
Couldn't implement catch in there. Examples?
You probably want to restructure things a bit as you've got IO mixed up in something it doesn't have to be. What about changing the signature to something like this?
getChar :: Int -> String -> Maybe Char
getChar n x | n < length x = Just (x !! n)
| otherwise = Nothing
Data.Maybe allows you to indicate that you are either going to return something (e.g. the length is within range) or Nothing (the length isn't within range). The function that calls getChar can then decide what to do with things. Data.Either provides a way of returning an error message with an error instead. From what I've seen (and I'm by no means an expert) exceptions are rarely used in Haskell, and choice types such as Either or Maybe are much more commonly used.
Now in the code that calls this, you can use pattern matching to see what happened e.g.
main :: IO ()
main = do
x <- getLine
let z = getChar' 5 x
case z of
(Just z) -> print $ "The 5th character is " ++ show z
Nothing -> print $ "The 5th character is out of range"
You can use the drop function to drop the first n characters of the line (drop will just give an empty result if there are fewer than n chars), and the listToMaybe function to turn a list to a Maybe (either Just c where c is the first element of the list, or Nothing if the list is empty):
import Data.Maybe (listToMaybe)
getchar :: Int -> IO (Maybe Char)
getchar n = do
line <- getLine
return . listToMaybe . drop n $ line
getChar' :: Int -> IO Char
getChar' n =
do
c <- getLine
if (n < length c)
then
return (c !! n)
else
getChar' n
You can do something like above. This is just an example though. But, since you are a beginner, it is strongly recommended not to play with IO and Monads. You can come to it after you get familiarized with pure functional concepts.
Related
I am trying to write a simple function which reads in one line at a time (which I know will be integers) and then stores them in to a list. For the life of me however, it seems like the list comes out to empty.
import System.IO
import Control.Monad
type Int2 = [Int]
valueTab = [] :: [Int]
app [ ] list = list
app (h:t) list = h:(app t list)
main :: IO ()
main = do
hSetBuffering stdout NoBuffering -- DO NOT REMOVE
-- Auto-generated code below aims at helping you parse
-- the standard input according to the problem statement.
input_line <- getLine
let n = read input_line :: Int
let value = [] :: [Int]
replicateM n $ do
input_line <- getLine
let pi = read input_line :: Int
hPutStrLn stderr (show input_line)
hPutStrLn stderr (show valueTab)
return $ app valueTab [pi]
-- hPutStrLn stderr "Debug messages..."
-- Write answer to stdout
--putStrLn input_line
return ()
So when I run this with
8
6
4
3 all on their own lines,
It prints 6, [], 4, [], 3 [].
Is this a problem with my printing, with my list declaration, with the way that I store them or? I have both value and valueTab to check whether it was a scope problem.
Note: The code is boilerplate code on a coding website that tests it on it's platform. Assume replicateM is just a loop that runs through the code x number of times.
It looks like you're doing a problem on codinggame.com. Other coding sites seem to do a better job with their Haskell templates, particularly for beginner exercises -- the template usually takes care of all the input and output, and you just need to supply the missing pure function. In contrast, codinggame.com's Haskell templates seem to assume the beginner has a pretty firm grasp of the IO monad, and leave out a lot of essential details (e.g., the fact that the result of the replicateM action should actually be bound to a variable) which you're expected to fill in.
The replicateM call in the original template probably looked something like:
replicateM n $ do
input_line <- getLine
let pi = read input_line :: Int -- maybe this line was there, maybe not
return ()
This expression creates a composite IO action that, when executed, will repeat the following subaction n times:
read a line, binding the string read to input_line
prepare to convert that line to an integer value pi (though this is never done because pi isn't used)
regardless of the string read, return "unit" (the value ()) as the result of the subaction
The value of the composite action is then a list of the values returned by the subactions. Since those were all units, the final value of the composite action is a list [(),(),()...()], with one () for each line read. However, because the value of this composite action is never bound to a variable (i.e., because there's no result <- ... before the replicateM expression), this list is thrown away.
So, this template provides a needlessly verbose method of reading n lines of input and doing nothing with them.
Fortunately, all you need to do to make this template do something useful is to have the subaction return a value that's useful (e.g., the integer pi) which will cause the composite action to return a list of the integers read, and then make sure you bind the resulting list to a variable using the <- notation.
In other words, you want to write:
main = do
...
pis <- replicateM n $ do
input_line <- getLine
let pi = read input_line :: Int
return pi
hPutStrLn stderr (show pis)
You won't need the helper function app, and you won't need to pre-declare a list valueTab to contain the result. The result is automatically produced by replicateM, and you just need to name it so you can use it.
The complete working program will look like:
import System.IO
import Control.Monad
type Int2 = [Int]
main :: IO ()
main = do
hSetBuffering stdout NoBuffering -- DO NOT REMOVE
-- Auto-generated code below aims at helping you parse
-- the standard input according to the problem statement.
input_line <- getLine
let n = read input_line :: Int
let value = [] :: [Int]
pis <- replicateM n $ do
input_line <- getLine
let pi = read input_line :: Int
return pi
hPutStrLn stderr (show pis)
-- hPutStrLn stderr "Debug messages..."
-- Write answer to stdout
return ()
I'm trying to build a list in OCaml that takes its end range variable from a function which returns an int cast from a float:
#require "batteries"
#require "pa_comprehension"
#require "core_kernel"
open Batteries
open Core_kernel
let factor_of num fact = num mod fact == 0 ;;
let limit num = Float.to_int (floor (sqrt num) ) ;;
the population of this list is not performed due to:
Error: This expression has type int but an expression was expected of type [<Downto | To ]
in both the (yummy) batteries list comprehension:
[? List: x | x <- 0--(limit num) ; factor_of num x ?] ;;
and the (slightly less but still quite readable) core_kernel List constructor:
List.(range 0 (limit num) |> filter ~f:( fun x -> factor_of num x) );;
I'm thinking that the return from limit num is trapped inside the (evil) monad that provides the Float.to_int function. This also happens when using Float.int_of_float, the type signatures are the same:
utop # Float.to_int;;
- : float -> int = <fun>
So... how do I get my int 'out' of the monad, or, if this is not the problem, what is going on and how do I cast to an actual int that is usable in this way?
Also, could someone point me to a decent 'What the hell are these things: Monads' tutorial for tiny, tiny brains? I am at my wits end with them.
UPDATE: The error was not caused by any monadic behaviour (or use at all, in fact) it was due to the incorrect use of an infix operator (mod) and a couple of other quirks in functional thinking that I have not completely understood. I'm not sure this post should still exist but maybe it is an example of the mistakes you can make when moving into the functional paradigm...?
I have managed to get to the source of the problem. Over thinking and a late night:
The error was not caused by any monadic behaviour (or use of monads at all, in fact) it was due to the incorrect use of an infix operator (mod) and a couple of other quirks in functional thinking that I have not completely understood. I'm not sure this post should still exist but maybe it is an example of the mistakes you can make when moving into the functional paradigm...?
Credit to helping me fix the issue to Str for helping me bend my mind.
Joe Gob has also pointed out the main issue in the code. More casting between float and int based of the types required by my functions.
This is the corrected code (deps.ml is a requirements file):
#use "deps.ml";;
open Batteries
open Core_kernel
let factor_of num fact = num mod fact == 0 ;;
let limit num = 2 * Float.to_int (floor (sqrt num)) ;;
let factor_list_of num =
[? List: x | x <- 1--(limit (Int.to_float num)) ; factor_of num x ?]
let sum_factors num = ( List.fold_right (+) (factor_list_of num) 0 )
let is_perfect num = sum_factors num == num
let is_abundant num = sum_factors num > num
let is_deficient num = sum_factors num < num
Basically I have been spoiled by Ruby, but I strongly advise rubyists to take up ocaml, it's has the closest style to ruby code of the functional languages.
Suppose I am writing an OCaml program and my input will be a large stream of integers separated by spaces i.e.
let string = input_line stdin;;
will return a string which looks like e.g. "2 4 34 765 5 ..." Now, the program itself will take a further two values i and j which specify a small subsequence of this input on which the main procedure will take place (let's say that the main procedure is the find the maximum of this sublist). In other words, the whole stream will be inputted into the program but the program will only end up acting on a small subset of the input.
My question is: what is the best way to translate the relevant part of the input stream into something usable i.e. a string of ints? One option would be to convert the whole input string into a list of ints using
let list = List.map int_of_string(Str.split (Str.regexp_string " ") string;;
and then once the bounds i and j have been entered one easily locates the relevant sublist and its maximum. The problem is that the initial pre-processing of the large stream is immensely time-consuming.
Is there an efficient way of locating the small sublist directly from the large stream i.e. processing the input along with the main procedure?
OCaml's standard library is rather small. It provides necessary and sufficient set of orthogonal features, as should do any good standard library. But, usually, this is not enough for a casual user. That's why there exist libraries, that do the stuff, that is rather common.
I would like to mention two the most prominent libraries: Jane Street's Core library and Batteries included (aka Core and Batteries).
Both libraries provides a bunch of high-level I/O functions, but there exists a little problem. It is not possible or even reasonable to try to address any use case in a library. Otherwise the library's interface wont be terse and comprehensible. And your case is non-standard. There is a convention, a tacit agreement between data engineers, to represent a set of things with a set of lines in a file. And to represent one "thing" (or a feature) with a line. So, if you have a dataset where each element is a scalar, you should represent it as a sequence of scalars separated by a newline. Several elements on a single line is only for multidimensional features.
So, with a proper representation, your problem can be solve as simple as (with Core):
open Core.Std
let () =
let filename = "data" in
let max_number =
let open In_channel in
with_file filename
~f:(fold_lines ~init:0
~f:(fun m s -> Int.(max m ## of_string s))) in
printf "Max number is %s is %d\n" filename max_number
You can compile and run this program with corebuild test.byte -- assuming that code is in a file name test.byte and core library is installed (with opam install core if you're using opam).
Also, there exists an excellent library Lwt, that provides a monadic high-level interface to the I/O. With this library, you can parse a set of scalars in a following way:
open Lwt
let program =
let filename = "data" in
let lines = Lwt_io.lines_of_file filename in
Lwt_stream.fold (fun s m -> max m ## int_of_string s) lines 0 >>=
Lwt_io.printf "Max number is %s is %d\n" filename
let () = Lwt_main.run program
This program can be compiled and run with ocamlbuild -package lwt.unix test.byte --, if lwt library is installed on your system (opam install lwt).
So, that is not to say, that your problem cannot be solved (or is hard to be solved) in OCaml, it is just to mention, that you should start with a proper representation. But, suppose, you do not own the representation, and cannot change it. Let's look, how this can be solved efficiently with OCaml. As previous examples represent, in general your problem can be described as a channel folding, i.e. an consequential application of a function f to each value in a file. So, we can define a function fold_channel, that will read an integer value from a channel and apply a function to it and the previously read value. Of course, this function can be further abstracted, by lifting the format argument, but for the demonstration purpose, I suppose, this will be enough.
let rec fold_channel f init ic =
try Scanf.fscanf ic "%u " (fun s -> fold_channel f (f s init) ic)
with End_of_file -> init
let () =
let max_value = open_in "atad" |> fold_channel max 0 in
Printf.printf "max value is %u\n" max_value
Although, I should note that this implementation is not for a heavy duty work. It is even not tail-recursive. If you need really efficient lexer, you can use ocaml's lexer generator, for example.
Update 1
Since there is a word "efficient" in the title, and everybody likes benchmarks, I've decided to compare this three implementations. Of course, since pure OCaml implementation is not tail-recursive it is not comparable to others. You may wonder, why it is not tail-recursive, as all calls to fold_channel is in a tail position. The problem is with exception handler - on each call to the fold channel, we need to remember the init value, since we're going to return it. This is a common issue with recursion and exceptions, you may google it for more examples and explanations.
So, at first we need to fix the third implementation. We will use a common trick with option value.
let id x = x
let read_int ic =
try Some (Scanf.fscanf ic "%u " id) with End_of_file -> None
let rec fold_channel f init ic =
match read_int ic with
| Some s -> fold_channel f (f s init) ic
| None -> init
let () =
let max_value = open_in "atad" |> fold_channel max 0 in
Printf.printf "max value is %u\n" max_value
So, with a new tail-recursive implementation, let's try them all on a big-data. 100_000_000 numbers is a big data for my 7 years old laptop. I've also added a C implementations as a baseline, and an OCaml clone of the C implementation:
let () =
let m = ref 0 in
try
let ic = open_in "atad" in
while true do
let n = Scanf.fscanf ic "%d " (fun x -> x) in
m := max n !m;
done
with End_of_file ->
Printf.printf "max value is %u\n" !m;
close_in ic
Update 2
Yet another implementation, that uses ocamllex. It consists of two files, a lexer specification lex_int.mll
{}
let digit = ['0'-'9']
let space = [' ' '\t' '\n']*
rule next = parse
| eof {None}
| space {next lexbuf}
| digit+ as n {Some (int_of_string n)}
{}
And the implementation:
let rec fold_channel f init buf =
match Lex_int.next buf with
| Some s -> fold_channel f (f s init) buf
| None -> init
let () =
let max_value = open_in "atad" |>
Lexing.from_channel |>
fold_channel max 0 in
Printf.printf "max value is %u\n" max_value
And here are the results:
implementation time ratio rate (MB/s)
plain C 22 s 1.0 12.5
ocamllex 33 s 1.5 8.4
Core 62 s 2.8 4.5
C-like OCaml 83 s 3.7 3.3
fold_channel 84 s 3.8 3.3
Lwt 143 s 6.5 1.9
P.S. You can see, that in this particular case Lwt is an outlier. This doesn't mean that Lwt is slow, it is just not its granularity. And I would like to assure you, that to my experience Lwt is a well suited tool for a HPC. For example, in one of my programs it processes a 30 MB/s network stream in a real-time.
Update 3
By the way, I've tried to address the problem in an abstract way, and I didn't provide a solution for your particular example (with j and k). Since, folding is a generalization of the iteration, it can be easily solved by extending the state (parameter init) to hold a counter and check whether it is contained in a range, that was specified by a user. But, this leads to an interesting consequence: what to do, when you have outran the range? Of course, you can continue to the end, just ignoring the output. Or you can non-locally exit from a function with an exception, something like raise (Done m). Core library provides such facility with a with_return function, that allows you to break out of your computation at any point.
open Core.Std
let () =
let filename = "data" in
let b1,b2 = Int.(of_string Sys.argv.(1), of_string Sys.argv.(2)) in
let range = Interval.Int.create b1 b2 in
let _,max_number =
let open In_channel in
with_return begin fun call ->
with_file filename
~f:(fold_lines ~init:(0,0)
~f:(fun (i,m) s ->
match Interval.Int.compare_value range i with
| `Below -> i+1,m
| `Within -> i+1, Int.(max m ## of_string s)
| `Above -> call.return (i,m)
| `Interval_is_empty -> failwith "empty interval"))
end in
printf "Max number is %s is %d\n" filename max_number
You may use the Scanf module family of functions. For instance, Scanf.fscanf let you read tokens from a channel according to a string format (which is a special type in OCaml).
Your program can be decomposed in two functions:
one which skip a number i of tokens from the input channel,
one which extract the maximum integer out of a number j from a channel
Let's write these:
let rec skip_tokens c i =
match i with
| i when i > 0 -> Scanf.fscanf c "%s " (fun _ -> skip_tokens c ## pred i)
| _ -> ()
let rec get_max c j m =
match j with
| j when j > 0 -> Scanf.fscanf c "%d " (fun x -> max m x |> get_max c (pred j))
| _ -> m
Note the space after the token format indicator in the string which tells the scanner to also swallow all the spaces and carriage returns in between tokens.
All you need to do now is to combine them. Here's a small program you can run from the CLI which takes the i and j parameters, expects a stream of tokens, and print out the maximum value as wanted:
let _ =
let i = int_of_string Sys.argv.(1)
and j = int_of_string Sys.argv.(2) in
skip_tokens stdin (pred i);
get_max stdin j min_int |> print_int;
print_newline ()
You could probably write more flexible combinators by extracting the recursive part out. I'll leave this as an exercise for the reader.
I am trying to do something fairly simple. I want to take a string such as "1,000" and return the string "1000".
Here was my attempt:
String.map (function x -> if x = ',' then '' else x) "1,000";;
however I get a compiler error saying there is a syntax error wrt ''
Thanks for the insight!
Unfortunately, there's no character like the one you're looking for. There is a string that's 0 characters long (""), but there's no character that's not there at all. All characters (so to speak) are 1 character.
To solve your problem you need a more general operation than String.map. The essence of a map is that its input and output have the same shape but different contents. For strings this means that the input and output are strings of the same length.
Unless you really want to avoid imperative coding (which is actually a great thing to avoid, especially when starting out with OCaml), you would probably do best using String.iter and a buffer (from the Buffer module).
Update
The string_map_partial function given by Andreas Rossberg is pretty nice. Here's another implementation that uses String.iter and a buffer:
let string_map_partial f s =
let b = Buffer.create (String.length s) in
let addperhaps c =
match f c with
| None -> ()
| Some c' -> Buffer.add_char b c'
in
String.iter addperhaps s;
Buffer.contents b
Just an alternate implementation with different stylistic tradeoffs. Not faster, probably not slower either. It's still written imperatively (for the same reason).
What you'd need here is a function like the following, which unfortunately is not in the standard library:
(* string_map_partial : (char -> char option) -> string -> string *)
let string_map_partial f s =
let buf = String.create (String.length s) in
let j = ref 0 in
for i = 0 to String.length s - 1 do
match f s.[i] with
| None -> ()
| Some c -> buf.[!j] <- c; incr j
done;
String.sub buf 0 !j
You can then write:
string_map_partial (fun c -> if c = ',' then None else Some c) "1,000"
(Note: I chose an imperative implementation for string_map_partial, because a purely functional one would require repeated string concatenation, which is fairly expensive in OCaml.)
A purely functional version could be this one:
let string_map_partial f s =
let n = String.length s in
let rec map_str i acc =
if i < n then
map_str (i + 1) (acc ^ (f (String.make 1 s.[i])))
else acc
in map_str 0 ""
Which is terminal recursive, but less performant than the imperative version.
I've been trying many different ways to do this in Haskell, and I can't for the life of me figure this out.
I want to get a list of names from the user, and if I know the length of the list (let's assume that is n), I want to prompt the user n times and ask for the i th item at the i'th time.
So far, I have this:
getinput a b
| a == b = []
| otherwise = input:getinput (a+1) b
where input = do
a <- getLine
return a
but I keep getting errors.
Strongly appreciate any help!
Problem with your code
The return type of input is IO String so you just can not append it to a list.
Similarly the return type of getinput (a+1) b is IO [String] and not just [String].
Here I have corrected your code
getinput a b | a == b = return []
| otherwise = do
i <- getLine
rest <- getinput (a+1) b
return (i:rest)
A better and more haskellish way
getinput2 n = sequence $ replicate n getLine
Satvik had a good answer, but your code is also 100% correct.
You can append an IO String to the beginning of a list, as long as all elements are of the same type -- so you end up with something that has the type [IO String]
All you need to do, with the code you've written, is apply it with sequence -- for example:
sequence $ getinput 0 4