CODEWITHSUNDEEP
ocaml
user1280282
user1280282

Reputation: 303

How to read from a file in ocaml?

I have an input file in following format : 

q0;q1
a;b
(q0,x);(q1;x)

I want to have three lists as follows:

a = ["q0";"q1"];
b = ["a";"b"];
c = [("q0","x");("q1","y")];

Here is my code :

let buf = Queue.create ();;

let catfile filename =
let rec print_all_lines in_chan =
    Queue.add (input_line in_chan) buf;
    print_all_lines in_chan
in
let in_file = open_in filename in
try
    print_all_lines in_file
with End_of_file -> close_in in_file;;

catfile "test.txt";;

let rec pvt_rec_split skipf rv str c i limit =
if (List.length rv < (limit - 1)) && (i < String.length str) then (
    if String.contains_from str i c then
        let o = String.index_from str i c in
        pvt_rec_split skipf
            (rv @ [ String.sub str i (o - i)])
            str c
            (skipf str c o)
            limit;
    else
        rv @ [ String.sub str i ((String.length str) - i) ]
) else (
    if i < String.length str then
        rv @ [ String.sub str i ((String.length str) - i) ]
    else
        rv
);;

let split s c limit =
let rec pvt_skip_char s c i =
    if (i >= String.length s ) then (
        String.length s
    ) else (
        if ((String.get s i) == c) then (
            pvt_skip_char s c (i +1)
        ) else (
            i
        )
    )
in
pvt_rec_split pvt_skip_char [] s c 0 limit ;;

let a = split (Queue.take buf) ';' 100;;

let b = split (Queue.take buf) ';' 100;;

let c = split (Queue.take buf) ';' 100;;

Basically split function splits the string with delimiter and returns a list.

So, I am able to generate list a and b properly.

But in case of List c, I get a type string list. Actually I want a list with type ('string*'string).

How do I do that?

Upvotes: 2

Views: 3344

Answers (1)

Ribtoks
Ribtoks

Reputation: 6922

File contents (notice comma-separations inside brackets)

q0;q1
a;b
(q0,x);(q1,x)

Ocaml code 

let split_str separator s =
  let list = ref [] in
  let start = ref 0 in
  let () = try
    while true do
      let index = String.index_from s !start separator in
      list := (String.sub s !start (index - !start)) :: !list;
      start := index + 1
    done
  with Not_found -> list := (String.sub s !start ((String.length s) - !start)) :: !list
  in List.rev !list;;

let maybe_input_line stdin =
  try Some (input_line stdin) with
    End_of_file -> None;;

let input_lines stdin =
  let rec input lines =
    match maybe_input_line stdin with
      Some line -> input (line :: lines)
    | None -> List.rev lines
  in
  input [];;

let rec parse_list_line delim line =
  if (String.length line) > 0 then
    let parts = split_str delim line in
    parse_tokens parts
  else
    []

and parse_tokens tokens =
  let rec inner_parse_tokens buffer = function
      [] -> List.rev buffer
    | h :: t ->
        let parsed = parse_line h in
        inner_parse_tokens (parsed :: buffer) t
  in
  inner_parse_tokens [] tokens

and parse_line str =
  if (String.length str) > 1 then
    if str.[0] = '(' && str.[(String.length str) - 1] = ')' then
      let substr = String.sub str 1 ((String.length str) - 2) in
      split_str ',' substr
    else
      str :: []
  else
    str :: []

and parse_lines lines =
  let rec inner_parse chunks = function
      [] -> List.rev chunks
    | head :: rest ->
        let parsed = parse_list_line ';' head in
        inner_parse (parsed :: chunks) rest
  in
  inner_parse [] lines;;

let file_channel = open_in("read_file2");;
let all_lines = input_lines file_channel;;
let chunks = parse_lines all_lines;;
let () = close_in file_channel;;

And the output

# val file_channel : in_channel = <abstr>
# val all_lines : string list = ["q0;q1"; "a;b"; "(q0,x);(q1,x)"]
# val chunks : string list list list =
  [[["q0"]; ["q1"]]; [["a"]; ["b"]]; [["q0"; "x"]; ["q1"; "x"]]]

Please, notice, that in last output we have to use lists for single string items in order to return same type from ocaml functions. If you want, you can write something similar to matlab squeeze function to change [["q0"]; ["q1"]] to ["q0"; "q1"]

Upvotes: 2

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