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Today's Topics:
1. Re: converting a json encoded radix tree to a haskell data
type (David McBride)
2. applicative style (Williams, Wes(AWF))
3. Re: applicative style (Brandon Allbery)
4. applicative style (Williams, Wes(AWF))
----------------------------------------------------------------------
Message: 1
Date: Fri, 28 Aug 2015 12:45:35 -0400
From: David McBride <toa...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] converting a json encoded radix tree
to a haskell data type
Message-ID:
<CAN+Tr41G=O5O3Ah_3EuJi1PGg+=dyqua_jqq9cdt7ubg4n6...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Well I went ahead and completed that function, but I didn't use your data
types exactly, but it should be a one to one mapping, just modify this
function with your constructors.
radix2things :: RadixTree -> [(Text, (Maybe Int, Maybe Int))]
radix2things r = conv' mempty r
where
conv' :: Text -> RadixTree -> [(Text, (Maybe Int, Maybe Int))]
conv' acc (Leaf (Times a b)) = [(acc, (a, b))]
conv' acc r@(Node ns) = P.concatMap (\(t,r) -> conv' (acc <> t) r) ns
And you'll get a result like:
*Main> case decode teststr of Nothing -> undefined; Just a -> conv a
[("a2b2",(Just 4,Just 5)),("ab",(Just 1,Just 2)),("acd",(Just 3,Nothing))]
Good luck.
On Thu, Aug 27, 2015 at 4:31 PM, Adam Flott <a...@adamflott.com> wrote:
> I am attached to the data structure as it's what our Thrift message
> spits out and has to be mapped that way for the down stream consumers.
>
>
> On 08/27/2015 01:45 PM, David McBride wrote:
> > I was trying this but ran into a bit of trouble. Are you super
> > attached to that data structure? I would expect a radix tree as
> > you've described it to look more like this:
> >
> > data RadixTree = Node [(Text, RadixTree)] | Leaf Times
> > data Times = Times (Maybe Int) (Maybe Int)
> >
> > In which case it is much easier to write the json instances. From
> > there you shouldn't have too much of a problem writing a recursive
> > function to do the rest, without dealing with all the aeson stuff at
> > the same time. Here's what I ended up with (I think it could be
> > cleaned up a bit).
> >
> > import Control.Monad
> > import Data.Text as T
> > import Data.Aeson
> > import Data.HashMap.Strict as HM
> > import Data.Vector as V hiding (mapM)
> >
> > data RadixTree = Node [(Text, RadixTree)] | Leaf Times deriving Show
> > data Times = Times (Maybe Int) (Maybe Int) deriving Show
> >
> > instance FromJSON RadixTree where
> > parseJSON (Object o) = do
> > let els = HM.toList o
> > contents <- mapM (\(t,v) -> do v' <- parseJSON v; return (t, v'))
> > (HM.toList o)
> > return $ Node contents
> > parseJSON a@(Array _) = Leaf <$> parseJSON a
> > parseJSON _ = mzero
> >
> > instance FromJSON Times where
> > parseJSON (Array v) | (V.length v) >= 2 =
> > let v0 = v V.! 0
> > v1 = v V.! 1
> > in Times <$> parseJSON v0 <*> parseJSON v1
> > parseJSON _ = mzero
> >
> > {-
> > tree2things :: RadixTree -> [(Text, (Maybe Int, Maybe Int))]
> > tree2things (Node xs) = _
> > tree2things (Leaf t) = _
> > -}
> >
> > On Thu, Aug 27, 2015 at 11:30 AM, Adam Flott <a...@adamflott.com
> > <mailto:a...@adamflott.com>> wrote:
> >
> > On 08/27/2015 11:18 AM, Karl Voelker wrote:
> > > On Thu, Aug 27, 2015, at 08:04 AM, Adam Flott wrote:
> > >> data Things = MkThings {
> > >> thing :: TL.Text,
> > >> times :: ThingTimes
> > >> } deriving (Show, Eq, Typeable)
> > >>
> > >> data ThingTimes = MkThingtimes {
> > >> ml :: V.Vector Times
> > >> } deriving (Show, Eq, Typeable)
> > >>
> > >> data Times = MkTimes {
> > >> t1 :: Maybe Int32,
> > >> t2 :: Maybe Int32
> > >> } deriving (Show, Eq, Typeable)
> > >>
> > >> -- radix.json --
> > >> {
> > >> "a" : {
> > >> "b" : [ 1, 2 ],
> > >> "c" : {
> > >> "d" : [ 3, null ]
> > >> }
> > >> },
> > >> "a2" : { "b2" : [ 4, 5 ] }
> > >> }
> > >> -- radix.json --
> > > It looks like your input file has Things nested inside Things,
> > but your
> > > data types don't allow for that. Is that intentional? What value
> > is that
> > > example input supposed to parse to?
> >
> > Vector [
> > MkThings "ab" (MkThingTimes (Vector [ Just 1, Just 2 ])),
> > MkThings "abcd" (MkThingsTimes (Vector [ Just 3, Nothing))
> > MkThings "a2b2" (MkThingTimes (Vector [ Just 4, Just 5 ])) ]
> > _______________________________________________
> > Beginners mailing list
> > Beginners@haskell.org <mailto:Beginners@haskell.org>
> > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
> >
> >
> >
> >
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Message: 2
Date: Fri, 28 Aug 2015 17:12:29 +0000
From: "Williams, Wes(AWF)" <wewilli...@paypal.com>
To: "beginners@haskell.org" <beginners@haskell.org>
Subject: [Haskell-beginners] applicative style
Message-ID: <d205e38b.dcca%wewilli...@paypalcorp.com>
Content-Type: text/plain; charset="iso-8859-1"
Hi haskellers,
I am trying to understand why I get the following error in learning applicative
style.
Prelude> let estimates = [5,5,8,8,2,1,5,2]
Prelude> (/) <$> Just $ foldl (+) 0 estimates <*> Just . fromIntegral $ length
estimates
<interactive>:54:1:
Non type-variable argument in the constraint: Fractional (Maybe r)
(Use FlexibleContexts to permit this)
When checking that 'it' has the inferred type
it :: forall a r.
(Fractional (Maybe r), Num a, Num (Int -> Maybe a -> r)) =>
Maybe r -> Maybe r
All the parts work individually. If use let and assign the parts to x and y it
also works.
E.g. This works
let x = Just $ foldl (+) estimates
Let y = Just . fromIntegral $ length estimates
(/) <$> x <*> y
I clearly do not understand exactly how these work. :-)
Thanks for any help,
-wes
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Message: 3
Date: Fri, 28 Aug 2015 13:23:10 -0400
From: Brandon Allbery <allber...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] applicative style
Message-ID:
<CAKFCL4Xc+Vw22+s=tojr7swcjca8gqrkfqgdq95emstmuq6...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
On Fri, Aug 28, 2015 at 1:12 PM, Williams, Wes(AWF) <wewilli...@paypal.com>
wrote:
> Num (Int -> Maybe a -> r))
That looks highly suspect. If it infers a function Num instance, you
probably got your parentheses wrong. Or your $-s...
...in fact, that is the problem. That final $ does not do what you think;
it produces
(foldl (+) 0 estimates <*> Just . fromIntegral) (length estimates)
when you presumably intended
foldl (+) 0 estimates <+> (Just . fromIntegral) (length estimates)
--
brandon s allbery kf8nh sine nomine associates
allber...@gmail.com ballb...@sinenomine.net
unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net
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Message: 4
Date: Fri, 28 Aug 2015 18:31:39 +0000
From: "Williams, Wes(AWF)" <wewilli...@paypal.com>
To: "beginners@haskell.org" <beginners@haskell.org>
Subject: [Haskell-beginners] applicative style
Message-ID: <d205f718.dcdc%wewilli...@paypalcorp.com>
Content-Type: text/plain; charset="iso-8859-1"
Awesome, I clearly am off on my understanding.
How could I do this: "foldl (+) 0 estimates <+> (Just . fromIntegral) (length
estimates)" without the parenthesis?
Thanks,
Wes
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