Adding two lists together with map - list

So I'm trying to write some minimal code to put two lists of strings together, and to do this I thought it was best to use the haskell map function.
Essentially I want to be able to do adders ["1","2"] ["3","4"] = ["1","2","3","4"]
So I have a function called adder, which takes a list, then adds a string to that list and returns the new list. Then I have a function called adders which replicates the adder function, but adds a list of strings instead of just one string, however at the moment it produces multiple lists instead of one list.
I thought
adder :: [String] -> String -> [String]
adder y x = y ++ [x]
adders y x = map (adder y) x
would work, but this just gives a list of two lists
[["1","2","3"],[["1","2","4"]]
How is the best way to go about this?

I thought it was best to use the haskell map function
No. map f applies f to every element of your list. But you don't want to change the elements at all, you want to change the list itself. That, however, is out of scope of the things that are possible with map. map cannot add more elements, neither can it remove some.
If you want to concatenate two lists, simply use ++:
adders :: [a] -> [a] -> [a]
adders x y = x ++ y

Related

Pattern matching on rest of list

I'm trying to pattern match on a list of pairs, where I'm trying to return a list from the list of pair, however I'm having trouble figuring out where to make the recursive call. Without the recursive call I have this:
let countriesInChart (cht: chart) =
match cht with
| [] -> []
| (x,y)::tt -> [x;y]
;;
But naturally this only applies to the first pair in the list and simply returns ["countryA"; "countryB"] without the rest of the list.
With the recursive call this simply only returns an empty list:
let rec countriesInChart (cht: chart) =
match cht with
| [] -> []
| (x,y)::tt -> [x;y]::countriesInChart tt
;;
How would I make the recursive call such that all the pairs in the list would return as a list?
You have this:
[x;y] :: countriesInChart tt
This says to add a new list of two elements onto the front of a list of lists.
I don't think you want a list of lists, you just want a list. So you shouldn't be making a list of two elements.
If x and y are the same type (as I suspect they are), what you probably want is this:
x :: y :: countriesInChart tt
This says to add both x and y individually to the front of the recursively generated list.
You can also write this, which is completely equivalent:
[x; y] # countriesInChart tt
However, this constructs a list of two elements only to throw it away. So it's a tiny bit of extra work for no benefit.

How to use if-then-else in a recursive function

I am writing a function that will take a list of list and merge it into sorted pairs of list. For example [[1],[9],[8],[7],[4],[5],[6]] would return [[1,9],[7,8],[4,5],[6]]. This is my first attempt at SML. I keep getting this error: operator and operand don't agree [overload conflict].
fun mergePass[] = []
| mergePass(x::[]) = x::[]
| mergePass(x::y::Z) =
if x<y
then (x # y)::mergePass(Z)
else (y # x)::mergePass(Z);
Edit: If mergePass is called on [[1,9],[7,8],[4,5],[6]] I will need it to return [[1,7,8,9],[4,5,6]].
This merge function takes two sorted lists
fun merge([],y) = y
| merge(x,[]) = x
| merge(a::x,b::y) =
if a < b then a::merge(x,b::y)
else b::merge(a::x,y);
You seem reasonably close. A few hints/remarks:
1) Aesthetically, using nil in one line and [] in others seems odd. Either use all nil or use all []
2) Since the input are lists of lists, in x::y::z, the identifiers x and y would be lists of integers, rather than individual integers. Thus, x<y wouldn't make sense. You can't compare lists of integers using <.
3) Your problem description strongly suggests that the inner-lists are all 1-element lists. Thus you could use the pattern [x]::[y]::z to allow you to compare x and y. In this case, x#y could be replaced by [x,y]
4) If the inner lists are allowed to be of arbitrary size, then your code needs major revision and would probably require a full-fledged sort function to sort the result of concatenating pairs of inner lists. Also, in this case, the single list in the one inner list case should probably be sorted.
5) You have a typo: mergeP isn't mergePass.
On Edit:
If the sublists are each sorted (and the name of the overall function perhaps suggests this) then you need a function called e.g. merge which will take two sorted lists and combine them into a single sorted list. If this is for a class and you have already seen a merge function as an example (perhaps in a discussion of merge-sort) -- just use that. Otherwise you will have to write your own before you write this function. Once you have the merge function, skip the part of comparing x and y and instead have something as simple as:
| mergePass (xs::ys::zss) = (merge xs ys) :: mergePass zss
If the sublists are not merged, then you will need a full-fledged sort in which case you would use something like:
| mergePass (xs::ys::zss) = sort(xs # ys) :: mergePass zss

"Non-exhaustive patterns" error when summing a list of lists

All I'm trying to do is sum a list of lists. Example of what I want to do:
Input: [[1,2,3],[2,5],[6,7]]
Output: [6,7,13]
The amount of lists inside the outer list can vary, and the amount of integers in each inner list can vary. I've tried a multitude of things, and this is the last one I tried but it doesn't work:
sumSubsets [[x]] = map sum [[x]]
Also, I wanted a base case of sumSubsets [] = [[]] but that causes errors as well. Any help would be appreciated.
You could use
sumSubsets x = map sum x
or even
sumSubsets = map sum
Your previous code,
sumSubsets [[x]] = map sum [[x]]
First performs a pattern match using [[x]] which matches a list containing a single element, which is itself a list containing a single element. Therefore it would work correctly on [[3]]
>> sumSubsets [[3]]
[3]
but not on [[1,2,3]] or [[1],[2]].
I think your problem stems primarily from mixing up types and values, which can happen easily to the beginner, in particular on lists. The whole confusion probably comes from the fact that in Haskell, [] is used as a data constructor as well as a type constructor.
For example, [Int] means "a list of Ints" (a type), but [1] means "the list that contains one element, namely the number 1" (a value -- meaning, the whole list is the value). Both things together:
xs :: [Int]
xs = [1]
When you write polymorphic functions, you abstract from something like the Int. For example, if you want to get the first element of a list, you can define a function that does that for any kind of list -- may they be lists of integers or lists of characters or even lists of lists:
firstElement :: [a] -> a
firstElement (x:xs) = x
[a] means "a list with elements of type a" (a type), and the a alone means "something of type a". firstElement is a function from a list with elements of type a to something of type a. a is a type variable. Since you're not saying what a should be, the function works for all kinds of lists:
*Main> firstElement [1,2,3]
1
*Main> firstElement ['a','b']
'a'
*Main> firstElement [[1,2],[3,4]]
[1,2]
When you wrote [[x]] you were perhaps thinking of the type of the first argument of the function, which would be a list of lists of elements of some type x (x is a type variable). You can still use that, but you have to put it into the type signature of your function (the line that contains the double colon):
sumSubsets :: Num a => [[a]] -> [a]
sumSubsets xs = map sum xs
I've used a here instead of x, since it's more commonly done, but you could use x, too. Unfortunately, the whole thing gets a bit complicated with the Num a which describes additional requirements on the type a (that it belongs to the numbers, since for other things, sum is not defined). To simplify matters, you could write:
sumSubsetsInts :: [[Int]] -> [Int]
sumSubsetsInts xs = map sum xs

Replacing element in a list of lists in Haskell

I have a list of lists like so:
[["BBBBBBBB",
"BWFFFPFGB",
"BWFFFPFGB",
"BWFFMPFGB",
"BWFFFPF_B",
"BWFFFPF6B",
"BBBBBBB"]]
I've done a little research and have found out how to access individual elements using the !! operator. But when it comes to searching for a certain element 'M' I'm not sure how to go about that. My friend said I need to use something like (x:xs):xss on a list, but when I try this in the WinGHCi haskell program I get this.
Prelude> let list = [["BBBBBBBB",
"BWFFFPFGB",
"BWFFFPFGB",
"BWFFMPFGB",
"BWFFFPF_B",
"BWFFFPF6B",
"BBBBBBB"]]
Prelude> head(x:xs):xss
<interactive>:192:2: Not in scope: `x'
<interactive>:192:4: Not in scope: `xs'
<interactive>:192:8: Not in scope: `xss'
I understand that I declare the name as list and not x:xs but even when I declare it as x:xs I still get the errors. I'm probably still a little new to haskell to really understand what to do so I may be going about this way wrong.
I've looked here Replace individual list elements in Haskell? because eventually I want to replace the M with something different but I'm not completely sure how I would implement that.
Any help/guidance is appreciated, thanks!
First let's see how to replace a W with M
charWM :: Char -> Char
charWM 'W' = 'M' -- If you see W, put M.
charWM x = x -- If you see anything else, put it back as is.
You can rewrite that function how you like by adding other letter transformations.
Now let's make that work over a list. There's a great function map :: (a ->b) -> [a] -> [b] that lets you apply a function on every element on a list.
stringWM :: String -> String
stringWM xs = map charWM xs -- do charWM to everything in xs.
For example stringWM "QWERTY WILL WIN" = "QMERTY MILL MIN"
Next we can do that to a list of lists:
lolWM :: [String] -> [String]
lolWM xss = map stringWM xss
(String is a type synonym for [Char].)
Let's test that out in ghci:
*Main> list'
["BBBBBBBB","BWFFFPFGB","BWFFFPFGB","BWFFMPFGB","BWFFFPF_B","BWFFFPF6B","BBBBBBB"]
*Main> lolWM list'
["BBBBBBBB","BMFFFPFGB","BMFFFPFGB","BMFFMPFGB","BMFFFPF_B","BMFFFPF6B","BBBBBBB"]
All good.
Your example wasn't exactly list', it was [list'] which has 1 element, so to work on that we'd need to map lolWM. Often we wouldn't bother writing stringWM or lolWM and go directly to lists of lists of lists, if that's what we needed:
lololWM = (map.map.map) charWM
map.map.map means map the map of the map. You can allow that to blow your mind a little, or you can just say list of list of list of Char, so map map map - one map per list level.
In the future, maybe you'll want to replace W with Strings instead of characters.
rewriteChar :: Char -> String
rewriteChar 'W' = "--^--"
rewriteChar x = [x] -- put x in a list to make it a string
This time, map isn't enough: map rewriteChar "QWERTY WILL WIN" gives
["Q","--^--","E","R","T","Y"," ","--^--","I","L","L"," ","--^--","I","N"]
We could use concat on that to flatten it into a single list, but it's more fun to do
rewriteString = concatMap rewriteChar
So now rewriteString "QWERTY WILL WIN" give us "Q--^--ERTY --^--ILL --^--IN".
For more mindblowing things to try, there's "QWERTY WILL WIN" >>= rewriteChar and "Hello Mum" >>= \x -> [x,x,x]
First of all, virtually all "variables" in Haskell are immutable, so there's no "changing a list", there are modified copies.
Second, you need to find an element by some criteria. To do that, you need to traverse a list. - This can be done using recursion. Filtering can be done using a function passed as an argument of your traversing function (this function must take an element and return a boolean value).
Try to put the above together and make your own function. Start with a type signature, it shows what you want to do: to take a list of Char (it's better to generalize to a generic type) and a function which possibly changes an element and return a modified list:
replaceFunc :: (Char -> Char) -> String -> String
Also, read http://www.haskell.org/haskellwiki/How_to_work_on_lists , there's a hint there how to apply some function to specific elements only.

Recursive function that returns all values in list (In OCaml)

I need a function that recursively returns (not prints) all values in a list with each iteration. However, every time I try programming this my function returns a list instead.
let rec elements list = match list with
| [] -> []
| h::t -> h; elements t;;
I need to use each element each time it is returned in another function that I wrote, so I need these elements one at a time, but I can't figure this part out. Any help would be appreciated.
Your function is equivalent to :
let rec elements list =
match list with
| [] -> []
| h :: t -> elements t
This happens because a ; b evaluates a (and discards the result) and then evaluates and returns b. Obviously, this is in turn equivalent to:
let elements (list : 'a list) = []
This is not a very useful function.
Before you try solving this, however, please understand that Objective Caml functions can only return one value. Returning more than one value is impossible.
There are ways to work around this limitation. One solution is to pack all the values you wish to return into a single value: a tuple or a list, usually. So, if you need to return an arbitrary number of elements, you would pack them together into a list and have the calling code process that list:
let my_function () = [ 1 ; 2; 3; 4 ] in (* Return four values *)
List.iter print_int (my_function ()) (* Print four values *)
Another less frequent solution is to provide a function and call it on every result:
let my_function action =
action 1 ;
action 2 ;
action 3 ;
action 4
in
my_function print_int
This is less flexible, but arguably faster, than returning a list : lists can be filtered, sorted, stored...
Your question is kind of confusing - you want a function that returns all the values in a list. Well the easiest way of returning a variable number of values is using a list! Are you perhaps trying to emulate Python generators? OCaml doesn't have anything similar to yield, but instead usually accomplishes the same by "passing" a function to the value (using iter, fold or map).
What you have currently written is equivalent to this in Python:
def elements(list):
if(len(list) == 0):
return []
else:
list[0]
return elements(list[1:])
If you are trying to do this:
def elements(list):
if(len(list) > 0):
yield list[0]
# this part is pretty silly but elements returns a generator
for e in elements(list[1:]):
yield e
for x in elements([1,2,3,4,5]):
dosomething(x)
The equivalent in OCaml would be like this:
List.iter dosomething [1;2;3;4;5]
If you are trying to determine if list a is a subset of list b (as I've gathered from your comments), then you can take advantage of List.mem and List.for_all:
List.for_all (fun x -> List.mem x b) a
fun x -> List.mem x b defines a function that returns true if the value x is equal to any element in (is a member of) b. List.for_all takes a function that returns a bool (in our case, the membership function we just defined) and a list. It applies that function to each element in the list. If that function returns true for every value in the list, then for_all returns true.
So what we have done is: for all elements in a, check if they are a member of b. If you are interested in how to write these functions yourself, then I suggest reading the source of list.ml, which (assuming *nix) is probably located in /usr/local/lib/ocaml or /usr/lib/ocaml.