I wrote a simple function
someFunc list elem = do
list <- elem:elem:elem:list
return elem
Now, when i'm using it, i'm getting output like this
*Main> someFunc [] 'a'
"aaa"
Despite the fact, that this function has no practical use, why does it happen? Why does editing list have any effect in elem? And how to assign new value to list avoiding this situation?
Note that your function will be de-sugared into this:
someFunc :: [b] -> b -> [b]
someFunc list elem = (elem:elem:elem:list) >>= \list -> return elem
Now note that the list in \list -> return elem is different from the input list you pass to the function.
Now see how the Monad instance for list is defined:
instance Monad [] where
return x = [x]
xs >>= f = concat (map f xs)
fail _ = []
So, your code is translated to this form finally:
someFunc list elem = concat $ map (\list -> return elem) (elem:elem:elem:list)
Now can you understand why you are getting that output ?
someFunc [] 'a' will get be applied like this:
concat $ map (\list -> return 'a') ('a':'a':'a':[])
concat $ [['a'],['a'],['a']]
'a':'a':'a':[]
"aaa"
You can't assign a new value to list, what's happening is the list on the left of the <- is different than the list on the right of the <-. If you turn on warnings with -Wall you'll see
<interactive>:13:19: Warning:
This binding for `elem' shadows the existing binding
imported from `Prelude' (and originally defined in `GHC.List')
<interactive>:14:7: Warning:
This binding for `list' shadows the existing binding
bound at <interactive>:13:14
<interactive>:14:7: Warning: Defined but not used: `list'
You aren't using the name list defined by list <- ..., just defining it so that it happens to shadow the existing binding.
The reason someFunc [] 'a' returns "aaa" is due to how the list monad works. This do notation would be equivalent to
someFunc list e = (e:e:e:list) >>= \l -> return e
And for lists, >>= is essentially concatMap, so you have
someFunc list e = concatMap (\l -> return e) (e:e:e:list)
So substituting in [] for list and 'a' for e we get
someFunc [] 'a' = concatMap (\l -> return 'a') "aaa"
= concat $ map (\l -> return 'a') "aaa"
= concat [['a'], ['a'], ['a']]
= ['a', 'a', 'a']
= "aaa"
Your confusion probably comes from the use of return. In most languages return is a keyword, but in Haskell it's just a function. It does not exit a function call early, all it does is wrap a value in the context of the monad you're in. For lists return x = [x], that's the entire definition. Additionally, in Haskell you can't reassign values, but you can shadow them with a new definition. If you always compile with -Wall and -Werror though, you'll keep yourself from having this problem.
You are (apparently accidentally) using the list monad. In Python, your snippet would be expressed as:
for list in [elem, elem, elem] + list:
yield elem
That is, you create a list containing one elem for each item in the list, plus three more.
Here's what you probably intended:
someFunc list elem = do
list <- return (elem:elem:elem:list)
return elem
This just creates a new list variable shadowing your old one, and completely ignores it to return elem.
Related
The function valueOf gets the Int value of the corresponding String out of a tuple list.
Can someone explain how the third line works?
type State = [(String,Int)]
valueOf :: State -> String -> Int
valueOf xs var = head [b | (a,b) <- xs , a ==var ]
I've never seen a Haskell expression like that, I'm more used to expression like this:
(\xs -> length xs > 15)
There is no lambda expression. What you are seeing is a list comprehension, which is a way of creating a list of values satisfying a certain condition.
In this case the comprehension [b | (a,b) <- xs , a ==var ] means something like: create a list of all bs such that (a, b) is an element of list xs, where a is equal to var.
In imperative pseudo-code you could write it as
result = EMPTY_LIST
for (a,b) in xs:
if a == var:
result.add(b)
return result
So, the whole valueOf function works by generating a list of values that have the right String key, and then uses head function to get the first one. Note that if there is no match, the whole computation will crash.
(a,b) <- xs is not a lambda, it is a generator in a list comprehension
In this case it successively binds the pairs in the source list xs and matches them with (a, b). The boolean guard a == var retains all elements for which a == var is true.
The line
valueOf xs var = head [b | (a,b) <- xs , a ==var ]
is equivalent to the lambda expression
valueOf = \xs var -> head [b | (a,b) <- xs , a ==var ]
However, I guess that what really confuses you is the list comprehension [b | (a,b) <- xs , a ==var ]. This expression builds a list as follows: loop over all the pairs in xs (this is the part (a,b)<-xs), keep only those pairs where a==var, then take all the remaining b and build a list with those.
With head in front, this will simply look in xs for the first pair whose first component is var, and return its second component. If no such pair is found, a runtime error will be generated (unless xs is an infinite list, which will cause the program to run forever).
Say I have a List of integers l = [1,2]
Which I want to print to stdout.
Doing print l produces [1,2]
Say I want to print the list without the braces
map print l produces
No instance for (Show (IO ())) arising from a use of `print'
Possible fix: add an instance declaration for (Show (IO ()))
In a stmt of an interactive GHCi command: print it
`:t print
print :: Show a => a -> IO ()
So while I thought this would work I went ahead and tried:
map putStr $ map show l
Since I suspected a type mismatch from Integer to String was to blame. This produced the same error message as above.
I realize that I could do something like concatenating the list into a string, but I would like to avoid that if possible.
What's going on? How can I do this without constructing a string from the elements of the List?
The problem is that
map :: (a -> b) -> [a] -> [b]
So we end up with [IO ()]. This is a pure value, a list of IO actions. It won't actually print anything. Instead we want
mapM_ :: (a -> IO ()) -> [a] -> IO ()
The naming convention *M means that it operates over monads and *_ means we throw away the value. This is like map except it sequences each action with >> to return an IO action.
As an example mapM_ print [1..10] will print each element on a new line.
Suppose you're given a list xs :: [a] and function f :: Monad m => a -> m b. You want to apply the function f to each element of xs, yielding a list of actions, then sequence these actions. Here is how I would go about constructing a function, call it mapM, that does this. In the base case, xs = [] is the empty list, and we simply return []. In the recursive case, xs has the form x : xs. First, we want to apply f to x, giving the action f x :: m b. Next, we want recursively call mapM on xs. The result of performing the first step is a value, say y; the result of performing the second step is a list of values, say ys. So we collect y and ys into a list, then return them in the monad:
mapM :: Monad m => (a -> m b) -> [a] -> m [b]
mapM f [] = return []
mapM f (x : xs) = f x >>= \y -> mapM f ys >>= \ys -> return (y : ys)
Now we can map a function like print, which returns an action in the IO monad, over a list of values to print: mapM print [1..10] does precisely this for the list of integers from one through ten. There is a problem, however: we aren't particularly concerned about collecting the results of printing operations; we're primarily concerned about their side effects. Instead of returning y : ys, we simply return ().
mapM_ :: Monad m => (a -> m b) ->[a] -> m ()
mapM_ f [] = return ()
mapM_ f (x : xs) = f x >> mapM_ f xs
Note that mapM and mapM_ can be defined without explicit recursion using the sequence and sequence_ functions from the standard library, which do precisely what their names imply. If you look at the source code for mapM and mapM_ in Control.Monad, you will see them implemented that way.
Everything in Haskell is very strongly typed, including code to perform IO!
When you write print [1, 2], this is just a convenience wrapper for putStrLn (show [1, 2]), where show is a function that turns a (Show'able) object into a string. print itself doesn't do anything (in the side effect sense of do), but it outputs an IO() action, which is sort of like a mini unrun "program" (if you excuse the sloppy language), which isn't "run" at its creation time, but which can be passed around for later execution. You can verify the type in ghci
> :t print [1, 2]
print [1, 2]::IO()
This is just an object of type IO ().... You could throw this away right now and nothing would ever happen. More likely, if you use this object in main, the IO code will run, side effects and all.
When you map multiple putStrLn (or print) functions onto a list, you still get an object whose type you can view in ghci
> :t map print [1, 2]
map print [1, 2]::[IO()]
Like before, this is just an object that you can pass around, and by itself it will not do anything. But unlike before, the type is incorrect for usage in main, which expects an IO() object. In order to use it, you need to convert it to this type.
There are many ways to do this conversion.... One way that I like is the sequence function.
sequence $ map print [1, 2]
which takes a list of IO actions (ie- mini "programs" with side effects, if you will forgive the sloppy language), and sequences them together as on IO action. This code alone will now do what you want.
As jozefg pointed out, although sequence works, sequence_ is a better choice here....
Sequence not only concatinates the stuff in the IO action, but also puts the return values in a list.... Since print's return value is IO(), the new return value becomes a useless list of ()'s (in IO). :)
Using the lens library:
[1,2,3] ^! each . act print
You might write your own function, too:
Prelude> let l = [1,2]
Prelude> let f [] = return (); f (x:xs) = do print x; f xs
Prelude> f l
1
2
I am teaching myself Haskell and have run into a problem and need help.
Background:
type AInfo = (Char, Int)
type AList = [AInfo] (let’s say [(‘a’, 2), (‘b’,5), (‘a’, 1), (‘w’, 21)]
type BInfo = Char
type BList = [BInfo] (let’s say [‘a’, ‘a’, ‘c’, ‘g’, ‘a’, ‘w’, ‘b’]
One quick edit: The above information is for illustrative purposes only. The actual elements of the lists are a bit more complex. Also, the lists are not static; they are dynamic (hence the uses of the IO monad) and I need to keep/pass/"return"/have access to and change the lists during the running of the program.
I am looking to do the following:
For all elements of AList check against all elements of BList and where the character of the AList element (pair) is equal to the character in the Blist add one to the Int value of the AList element (pair) and remove the character from BList.
So what this means is after the first element of AList is checked against all elements of BList the values of the lists should be:
AList [(‘a’, 5), (‘b’,5), (‘a’, 1), (‘w’, 21)]
BList [‘c’, ‘g’, ‘w’, ‘b’]
And in the end, the lists values should be:
AList [(‘a’, 5), (‘b’,6), (‘a’, 1), (‘w’, 22)]
BList [‘c’, ‘g’]
Of course, all of this is happening in an IO monad.
Things I have tried:
Using mapM and a recursive helper function. I have looked at both:
Every element of AList checked against every element of bList -- mapM (myHelpF1 alist) blist and
Every element of BList checked against every element of AList – mapM (myHelpF2 alist) blist
Passing both lists to a function and using a complicated
if/then/else & helper function calls (feels like I am forcing
Haskell to be iterative; Messy convoluted code, Does not feel
right.)
I have thought about using filter, the character value of AList
element and Blist to create a third list of Bool and the count the
number of True values. Update the Int value. Then use filter on
BList to remove the BList elements that …… (again Does not feel
right, not very Haskell-like.)
Things I think I know about the problem:
The solution may be exceeding trivial. So much so, the more experienced Haskellers will be muttering under their breath “what a noob” as they type their response.
Any pointers would be greatly appreciated. (mutter away….)
A few pointers:
Don't use [(Char, Int)] for "AList". The data structure you are looking for is a finite map: Map Char Int. Particularly look at member and insertWith. toList and fromList convert from the representation you currently have for AList, so even if you are stuck with that representation, you can convert to a Map for this algorithm and convert back at the end. (This will be more efficient than staying in a list because you are doing so many lookups, and the finite map API is easier to work with than lists)
I'd approach the problem as two phases: (1) partition out the elements of blist by whether they are in the map, (2) insertWith the elements which are already in the map. Then you can return the resulting map and the other partition.
I would also get rid of the meaningless assumptions such as that keys are Char -- you can just say they are any type k (for "key") that satisfies the necessary constraints (that you can put it in a Map, which requires that it is Orderable). You do this with lowercase type variables:
import qualified Data.Map as Map
sieveList :: (Ord k) => Map.Map k Int -> [k] -> (Map.Map k Int, [k])
Writing algorithms in greater generality helps catch bugs, because it makes sure that you don't use any assumptions you don't need.
Oh, also this program has no business being in the IO monad. This is pure code.
import Data.List
type AInfo = (Char, Int)
type AList = [AInfo]
type BInfo = Char
type BList = [BInfo]
process :: AList -> BList -> AList
process [] _ = []
process (a:as) b = if is_in a b then (fst a,snd a + 1):(process as (delete (fst a) b)) else a:process as b where
is_in f [] = False
is_in f (s:ss) = if fst f == s then True else is_in f ss
*Main> process [('a',5),('b',5),('a',1),('b',21)] ['c','b','g','w','b']
[('a',5),('b',6),('a',1),('b',22)]
*Main> process [('a',5),('b',5),('a',1),('w',21)] ['c','g','w','b']
[('a',5),('b',6),('a',1),('w',22)]
Probably an important disclaimer: I'm rusty at Haskell to the point of ineptness, but as a relaxing midnight exercise I wrote this thing. It should do what you want, although it doesn't return a BList. With a bit of modification, you can get it to return an (AList,BList) tuple, but methinks you'd be better off using an imperative language if that kind of manipulation is required.
Alternately, there's an elegant solution and I'm too ignorant of Haskell to know it.
While I am by no means a Haskell expert, I have a partial attempt that returns that result of an operation once. Maybe you can find out how to map it over the rest to get your solution. The addwhile is clever, since you only want to update the first occurrence of an element in lista, if it exists twice, it will just add 0 to it. Code critiques are more than welcome.
import Data.List
type AInfo = (Char, Int)
type AList = [AInfo]
type BInfo = Char
type BList = [BInfo]
lista = ([('a', 2), ('b',5), ('a', 1), ('w', 21)] :: AList)
listb = ['a','a','c','g','a','w','b']
--step one, get the head, and its occurrences
items list = (eleA, eleB) where
eleA = length $ filter (\x -> x == (head list)) list
eleB = head list
getRidOfIt list ele = (dropWhile (\x -> x == ele) list) --drop like its hot
--add to lista
addWhile :: [(Char, Int)] -> Char -> Int -> [(Char,Int)]
addWhile [] _ _ = []
addWhile ((x,y):xs) letter times = if x == letter then (x,y+times) : addWhile xs letter times
else (x,y) : addWhile xs letter 0
--first answer
firstAnswer = addWhile lista (snd $ items listb) (fst $ items listb)
--[('a',5),('b',5),('a',1),('w',21)]
The operation you describe is pure, as #luqui points out, so we just define it as a pure Haskell function. It can be used inside a monad (including IO) by means of fmap (or do).
import Data.List
combine alist blist = (reverse a, b4) where
First we sort and count the B list:
b = map (\g->(head g,length g)) . group . sort $ blist
We need the import for group and sort to be available. Next, we roll along the alist and do our thing:
(a,b2) = foldl g ([],b) alist
g (acc,b) e#(x,c) = case pick x b of
Nothing -> (e:acc,b)
Just (n,b2) -> ((x,c+n):acc,b2)
b3 = map fst b2
b4 = [ c | c <- blist, elem c b3 ]
Now pick, as used, must be
pick x [] = Nothing
pick x ((y,n):t)
| x==y = Just (n,t)
| otherwise = case pick x t of Nothing -> Nothing
Just (k,r) -> Just (k, (y,n):r)
Of course pick performs a linear search, so if performance (speed) becomes a problem, b should be changed to allow for binary search (tree etc, like Map). The calculation of b4 which is filter (`elem` b3) blist is another potential performance problem with its repeated checks for presence in b3. Again, checking for presence in trees is faster than in lists, in general.
Test run:
> combine [('a', 2), ('b',5), ('a', 1), ('w', 21)] "aacgawb"
([('a',5),('b',6),('a',1),('w',22)],"cg")
edit: you probably want it the other way around, rolling along the blist while updating the alist and producing (or not) the elements of blist in the result (b4 in my code). That way the algorithm will operate in a more local manner on long input streams (that assuming your blist is long, though you didn't say that). As written above, it will have a space problem, consuming the input stream blist several times over. I'll keep it as is as an illustration, a food for thought.
So if you decide to go the 2nd route, first convert your alist into a Map (beware the duplicates!). Then, scanning (with scanl) over blist, make use of updateLookupWithKey to update the counts map and at the same time decide for each member of blist, one by one, whether to output it or not. The type of the accumulator will thus have to be (Map a Int, Maybe a), with a your element type (blist :: [a]):
scanl :: (acc -> a -> acc) -> acc -> [a] -> [acc]
scanning = tail $ scanl g (Nothing, fromList $ reverse alist) blist
g (_,cmap) a = case updateLookupWithKey (\_ c->Just(c+1)) a cmap of
(Just _, m2) -> (Nothing, m2) -- seen before
_ -> (Just a, cmap) -- not present in counts
new_b_list = [ a | (Just a,_) <- scanning ]
last_counts = snd $ last scanning
You will have to combine the toList last_counts with the original alist if you have to preserve the old duplicates there (why would you?).
I have a list of list of characters ::[[Char]].
I need to iterate both over the list of strings and also over each character in each string.
Say, my list is present in this variable.
let xs
Please suggest an easy way to iterate.
If you want to apply a function f to every element of a list like this:
[a, b, c, d] → [f a, f b, f c, f d]
then map f xs does the trick. map turns a function on elements to a function on lists. So, we can nest it to operate on lists of lists: if f transforms as into bs, map (map f) transforms [[a]]s into [[b]]s.
If you instead want to perform some IO action for every element of a list (which is more like traditional iteration), then you're probably looking for forM_:1
forM_ :: [a] -> (a -> IO b) -> IO ()
You give it a function, and it calls it with each element of the list in order. For instance, forM_ xs putStrLn is an IO action that will print out every string in xs on its own line. Here's an example of a more involved use of forM_:
main = do
...
forM_ xs $ \s -> do
putStrLn "Here's a string:"
forM_ s print
putStrLn "Now it's done."
If xs contains ["hello", "world"], then this will print out:
Here's a string:
'h'
'e'
'l'
'l'
'o'
Now it's done.
Here's a string:
'w'
'o'
'r'
'l'
'd'
Now it's done.
1 forM_ actually has a more general type, but the simpler version I've shown is more relevant here.
Just that:
[c | x <- xs, c <- x]
The "correct" way to iterate is actually fold. Anything you might ever want to do with a list can be done with a fold. Let's consider what you want to do. You're probably thinking of something like this:
for (row in xs):
for (c in row):
doSomething
The problem is, you're probably making use of mutable variables in doSomething. That's ok, we can deal with that. So suppose you have this.
def iter2d(xs):
outerVar = outerInit
for (row in xs):
innerVar = innerInit(row)
outerVar.adjust1(row)
for (c in row):
innerVar.adjust2(c)
outerVar.adjust3(c, innerVar)
return outerVar
Let's translate that to folds. And immutability.
iter2d :: [[Char]] -> Something
iter2d xs = foldl' outerStep outerInit xs
where outerInit = ... -- same as outerInit above
outerStep acc row = fst $ foldl' innerStep innerInit' row)
where innerInit' = ((adjust1 acc row), innerInit row)
innerInit row = ... -- same as innerInit above
innerStep (outAcc, inAcc) c = (outAcc', inAcc')
where inAcc' = adjust2 inAcc c
outAcc' = adjust3 outAcc c inAcc'
Notice with immutability, we are forced to indicate that outAc' depends on inAcc', rather than inAcc, meaning, the "state" of innerVar after it is updated.
Now you might say "wow that Haskell looks way ugly, why would I ever want to use Haskell". Yes, it does look ugly, but only because I tailored it to be a direct translation of imperative code. Once you get used to using folds instead of "iterating through a list", then you will find that folding is a very powerful technique that lets you do a lot of things in a more elegant way than for loops allow.
map (map f) l
where f :: Char -> Foo is a function to apply to each Char and l :: [[Char]]
returns l' :: [[Foo]]
I'm trying to understand how Haskell list comprehensions work "under the hood" in regards to pattern matching. The following ghci output illustrates my point:
Prelude> let myList = [Just 1, Just 2, Nothing, Just 3]
Prelude> let xs = [x | Just x <- myList]
Prelude> xs
[1,2,3]
Prelude>
As you can see, it is able to skip the "Nothing" and select only the "Just" values. I understand that List is a monad, defined as (source from Real World Haskell, ch. 14):
instance Monad [] where
return x = [x]
xs >>= f = concat (map f xs)
xs >> f = concat (map (\_ -> f) xs)
fail _ = []
Therefore, a list comprehension basically builds a singleton list for every element selected in the list comprehension and concatenates them. If a pattern match fails at some step, the result of the "fail" function is used instead. In other words, the "Just x" pattern doesn't match so [] is used as a placeholder until 'concat' is called. That explains why the "Nothing" appears to be skipped.
What I don't understand is, how does Haskell know to call the "fail" function? Is it "compiler magic", or functionality that you can write yourself in Haskell? Is it possible to write the following "select" function to work the same way as a list comprehension?
select :: (a -> b) -> [a] -> [b]
select (Just x -> x) myList -- how to prevent the lambda from raising an error?
[1,2,3]
While implemenatations of Haskell might not do it directly like this internally, it is helpful to think about it this way :)
[x | Just x <- myList]
... becomes:
do
Just x <- myList
return x
... which is:
myList >>= \(Just x) -> return x
As to your question:
What I don't understand is, how does Haskell know to call the "fail" function?
In do-notation, if a pattern binding fails (i.e. the Just x), then the fail method is called. For the above example, it would look something like this:
myList >>= \temp -> case temp of
(Just x) -> return x
_ -> fail "..."
So, every time you have a pattern-match in a monadic context that may fail, Haskell inserts a call to fail. Try it out with IO:
main = do
(1,x) <- return (0,2)
print x -- x would be 2, but the pattern match fails
The rule for desugaring a list comprehension requires an expression of the form [ e | p <- l ] (where e is an expression, p a pattern, and l a list expression) behave like
let ok p = [e]
ok _ = []
in concatMap ok l
Previous versions of Haskell had monad comprehensions, which were removed from the language because they were hard to read and redundant with the do-notation. (List comprehensions are redundant, too, but they aren't so hard to read.) I think desugaring [ e | p <- l ] as a monad (or, to be precise, as a monad with zero) would yield something like
let ok p = return e
ok _ = mzero
in l >>= ok
where mzero is from the MonadPlus class. This is very close to
do { p <- l; return e }
which desugars to
let ok p = return e
ok _ = fail "..."
in l >>= ok
When we take the List Monad, we have
return e = [e]
mzero = fail _ = []
(>>=) = flip concatMap
I.e., the 3 approaches (list comprehensions, monad comprehensions, do expressions) are equivalent for lists.
I don't think the list comprehension syntax has much to do with the fact that List ([]), or Maybe for that matter, happens to be an instance of the Monad type class.
List comprehensions are indeed compiler magic or syntax sugar, but that's possible because the compiler knows the structure of the [] data type.
Here's what the list comprehension is compiled to: (Well, I think, I didn't actually check it against the GHC)
xs = let f = \xs -> case xs of
Just x -> [x]
_ -> []
in concatMap f myList
As you can see, the compiler doesn't have to call the fail function, it can simply inline a empty list, because it knows what a list is.
Interestingly, this fact that the list comprehensions syntax 'skips' pattern match failures is used in some libraries to do generic programming. See the example in the Uniplate library.
Edit: Oh, and to answer your question, you can't call your select function with the lambda you gave it. It will indeed fail on a pattern match failure if you call it with an Nothing value.
You could pass it the f function from the code above, but than select would have the type:
select :: (a -> [b]) -> [a] -> [b]
which is perfectly fine, you can use the concatMap function internally :-)
Also, that new select now has the type of the monadic bind operator for lists (with its arguments flipped):
(>>=) :: [a] -> (a -> [b]) -> [b]
xs >>= f = concatMap f xs -- 'or as you said: concat (map f xs)