There is a case mapping two vectors into a single vector. I expected that the result of both ML should be same. Unfortunately, the result of ReasonML is different. Please help and comment how to fix it.
OCaml
List.map2 (fun x y -> x+y) [1;2;3] [10;20;30];;
[11;;22;;33]
ReasonML
Js.log(List.map2 ( (fun (x,y) => x+y), [1,2,3], [10,20,30]))
[11,[22,[33,0]]]
This is the same result. If you run:
Js.log([11,22,33]);
You'll get:
[11,[22,[33,0]]]
The result is the same, but you're using different methods of printing them. If instead of Js.log you use rtop or sketch.sh, you'll get the output you expect:
- : list(int) = [11, 22, 33]
Js.log prints it differently because it is a BuckleScript binding to console.log, which will print the JavaScript-representation of the value you give to it. And lists don't exist in JavaScript, only arrays do.
The way BuckleScript represents lists is pretty much the same way it is done natively. A list in OCaml and Reason is a "cons-cell", which is essentially a tuple or a 2-element array, where the first item is the value of that cell and the last item is a pointer to the next cell. The list type is essentially defined like this:
type list('a) =
| Node('a, list('a))
| Empty;
And with this definition could have been constructed with:
Node(11, Node(22, Node(33, Empty)))
which is represented in JavaScript like this:
[11,[22,[33,0]]]
^ ^ ^ ^
| | | The Empty list
| | Third value
| Second value
First value
Lists are defined this way because immutability makes this representation very efficient. Because we can add or remove values without copying all the items of the old list into a new one. To add an item we only need to create one new "cons-cell". Using the JavaScript representation with imagined immutability:
const old = [11,[22,[33,0]]];
const new = [99, old];
And to remove an item from the front we don't have to create anything. We can just get a reference to and re-use a sub-list, because we know it won't change.
const old = [11,[22,[33,0]]];
const new = old[1];
The downside of lists is that adding and removing items to the end is relatively expensive. But in practice, if you structure your code in a functional way, using recursion, the list will be very natural to work with. And very efficient.
#Igor Kapkov, thank you for your help. Base on your comment, I found a pipeline statement in the link, there is a summary.
let a = List.map2 ( (fun (x,y) => x+y), [1,2,3], [10,20,30] )
let logl = l => l |> Array.of_list |> Js.log;
a |> logl
[11,22,33]
Related
So, I found out that Ocaml supports the creation of circular lists using let rec.
utop # let rec ones = 1::ones;;
val ones : int list = [1; <cycle>]
That is pretty neat, and it even prints out in utop without blowing up.
But when I try to use List.map on this kind of data it does blow up:
utop # let twos = List.map ((+) 1) ones;;
Stack overflow during evaluation (looping recursion?).
Raised by primitive operation at Stdlib__List.map in file "list.ml", line 92, characters 32-39
Called from Stdlib__List.map in file "list.ml", line 92, characters 32-39
...
That is somewhat disapointing, though not totally unexpected.
Now the question, would it be possible to implement a 'better' map function that can handle this properly. I.e. you would do something like:
let twos = betterMap ((+) 1) ones;;
And instead of blowing up it would be able to detect the cycle properly and produce:
val twos : int list = [2; <cycle>]
Since the list of ones, though looping back on itself, is effectively a finite structure, it feels like this should be possible. But how?
It is only possible to create cyclic lists when the cycle is statistically known. It is thus impossible to create a map function that works on any cyclic lists without knowing in advance the topology of cycles in the list. For instance, this function works for lists that are 1-cycle:
let map_1_cycle f = function
| [] -> []
| a :: l ->
let rec answer = f a :: answer in
answer
The generic solution is to use sequences since as a form of lazy list, they have a much better support for infinite sequences of elements:
let ones = Seq.repeat 1
let twos = Seq.map ((+) 1) ones
I'm trying to implement a function to add like terms of a sorted list of tuples (first number represents polynomial's constant, the second represents the power). I'm an ocaml noob and don't really know what I'm doing wrong or how to do this correctly.
I tried to write it, but it doesn't work
https://gyazo.com/d37bb66d0e6813537c34225b6d4048d0
let rec simp list =
match list with
| (a,b)::(c,d)::remainder where b == d -> (a+c,b)::simp(remainder)
| (a,b)::(c,d)::remainder where b != d -> (a,b)::(c,d)::simp(remainder)
| _ -> list;;
This should combine all the terms with the same second value and just return one tuple with their first values added to the new list. ie: [(3,2);(4,2)] -> [(7,2)].
I am not familiar with the where keyword - there is ocaml-where which provides it, but it seems to be doing something different than what you are expecting. As such, the syntax is just wrong, and where is unexpected.
You probably meant when instead of where.
I have a list of coordinates that I need to put on map. Is it possible in julius to iterate the list ? Right now I am creating an hidden table in hamlet and accessing that table in julius which does not seem to be an ideal solution.
Could some one point to a better solution ? Thanks.
edit: Passing a JSON string for the list (which can be read by julius) seems to solve my problem.
As far as I know, you can't directly iterate over a list in julius. However, you can use the Monoid instance for the Javascript type to accomplish a similar effect. For example:
import Text.Julius
import Data.Monoid
rows :: [Int] -> t -> Javascript
rows xs = mconcat $ map row xs
where
row x = [julius|v[#{show x}] = #{show x};
|]
Then you can use rows xs wherever you'd normally put a julius block. For example, in ghci:
> renderJavascript $ rows [1..5] ()
"v[1] = 1;\nv[2] = 2;\nv[3] = 3;\nv[4] = 4;\nv[5] = 5;\n"
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.
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.