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Today's Topics:
1. Re: \x -> x < 0.5 && x > -0.5 (Daniel Fischer)
2. Re: Haskell Output Help (Jan Jakubuv)
3. Re: Caching evaluation of lazy lists (Daniel Fischer)
4. Re: \x -> x < 0.5 && x > -0.5 (Darrin Thompson)
5. Re: [Haskell-cafe] Re: [Haskell-beginners] using quickcheck
for blackbox testing for 3rd party apps. (Srikanth K)
6. random (John Moore)
7. Re: random (Brent Yorgey)
8. Re: random (Tom Davie)
9. Re: random (aditya siram)
----------------------------------------------------------------------
Message: 1
Date: Fri, 23 Oct 2009 17:32:29 +0200
From: Daniel Fischer <[email protected]>
Subject: Re: [Haskell-beginners] \x -> x < 0.5 && x > -0.5
To: [email protected]
Message-ID: <[email protected]>
Content-Type: text/plain; charset="iso-8859-1"
Am Freitag 23 Oktober 2009 17:25:57 schrieb Darrin Thompson:
> On Fri, Oct 23, 2009 at 10:25 AM, pl <[email protected]> wrote:
> > filter ((<=0.5) . abs) xs
>
> pure (&&) <*> (< 0.5) <*> (> -0.5)
>
> liftM2 (&&) (< 0.5) (> -0.5)
>
> Someone suggested that this was an example of the reader monad but I
> don't get that.
It's because ((->) r) *is* the reader monad.
Control.Monad.Reader's Reader r a is just that wrapped in a newtype:
newtype Reader r a = Reader { runReader :: r -> a }
>
> > :i (->)
>
> data (->) a b -- Defined in GHC.Prim
> instance Monad ((->) r) -- Defined in Control.Monad.Instances
> instance Functor ((->) r) -- Defined in Control.Monad.Instances
> instance Applicative ((->) a) -- Defined in Control.Applicative
>
> That's what I see working here.
>
> --
> Darrin
------------------------------
Message: 2
Date: Fri, 23 Oct 2009 16:42:48 +0100
From: Jan Jakubuv <[email protected]>
Subject: Re: [Haskell-beginners] Haskell Output Help
To: Chandni Navani <[email protected]>
Cc: [email protected]
Message-ID: <[email protected]>
Content-Type: text/plain; charset=iso-8859-1
Hi,
seems to me like a job for `Text.PrettyPrint`:
import Text.PrettyPrint
ppString :: String -> Doc
ppString = doubleQuotes . text
ppList :: [Doc] -> Doc
ppList = brackets . vcat . punctuate (text ",")
pretty = ppList . map (ppList . map ppString)
The code is hopefully almost self-explaining (`vcat` does the line
breaking). The result looks as follows:
*Main> pretty [["abc", "cde"], ["fgh", "ghi"]]
[["abc",
"cde"],
["fgh",
"ghi"]]
Sincerely,
jan.
On Thu, Oct 22, 2009 at 12:11:07PM -0700, Chandni Navani wrote:
> I have a list of lists which all contain strings. [[String]]. I need to
> figure out how to print them so that after each individual string, there is a
> new line.
>
> If this is the initial list [["abc", "cde"] ["fgh", "ghi"]]
> [["abc"
> "cde"]
> ["fgh",
> "ghi"]]
>
> Can anyone help me figure this out? Thanks.
--
Heriot-Watt University is a Scottish charity
registered under charity number SC000278.
------------------------------
Message: 3
Date: Fri, 23 Oct 2009 19:34:44 +0200
From: Daniel Fischer <[email protected]>
Subject: Re: [Haskell-beginners] Caching evaluation of lazy lists
To: Philip Scott <[email protected]>
Cc: [email protected]
Message-ID: <[email protected]>
Content-Type: text/plain; charset="iso-8859-1"
Am Freitag 23 Oktober 2009 18:30:53 schrieb Philip Scott:
> Hello again,
>
> > then, barring memory pressure forcing it out, it will be computed only
> > once (each list element will be computed only once, when it's first
> > needed).
>
> Thanks Daniel, that was what I was after. Is there any way of
> investigating these things without using the profiler? E.g. is there any
> way to stick a debug print statement inside a function without moving
> over to sideeffects and IO Monads etc.. I know printing is a side
> effect, but it would be nice to say 'I can has itsy sneeky side effect
> plz Haskell, just for little testing while'
>
> Cheers,
>
> Philip
import Debug.Trace
infixl 0 `debug`
debug = flip trace
dfib :: Int -> Integer
dfib =
let fib 0 = 0
fib 1 = 1
fib n = dfib (n-2) + dfib (n-1) `debug` "eval fib " ++ show n
in (map fib [0 .. ] !!)
Ok, modules loaded: MFib.
*MFib> dfib 4
eval fib 4
eval fib 2
eval fib 3
3
*MFib> dfib 7
eval fib 7
eval fib 5
eval fib 6
13
*MFib> dfib 15
eval fib 15
eval fib 13
eval fib 11
eval fib 9
eval fib 8
eval fib 10
eval fib 12
eval fib 14
610
*MFib>
The trick with debug = flip trace makes commenting out the debug-code easier:
fun x = trace ("fun " ++ show x) $ body x
~>
fun x = {- trace ("fun " ++ show x) $ -} body x
vs.
fun x = body x `debug` "fun " ++ show x
~>
fun x = body x -- `debug` "fun " ++ show x
But beware, including the argument in the trace message can lead to
recalculation of
values which would be cached without it, it's a hairy issue.
------------------------------
Message: 4
Date: Fri, 23 Oct 2009 16:24:38 -0400
From: Darrin Thompson <[email protected]>
Subject: Re: [Haskell-beginners] \x -> x < 0.5 && x > -0.5
To: Daniel Fischer <[email protected]>
Cc: [email protected]
Message-ID:
<[email protected]>
Content-Type: text/plain; charset=ISO-8859-1
On Fri, Oct 23, 2009 at 11:32 AM, Daniel Fischer
<[email protected]> wrote:
> It's because ((->) r) *is* the reader monad.
> Control.Monad.Reader's Reader r a is just that wrapped in a newtype:
>
> newtype Reader r a = Reader { runReader :: r -> a }
>
So I was thinking:
:t runReader $ liftM2 (&&) (Reader (< 0.5)) (Reader (> -0.5))
Thanks.
--
Darrin
------------------------------
Message: 5
Date: Sat, 24 Oct 2009 21:25:10 +0530
From: Srikanth K <[email protected]>
Subject: Re: [Haskell-cafe] Re: [Haskell-beginners] using quickcheck
for blackbox testing for 3rd party apps.
To: Daniel Fischer <[email protected]>
Cc: [email protected], [email protected]
Message-ID:
<[email protected]>
Content-Type: text/plain; charset="iso-8859-1"
Thanks. unsafePerformIO seems to suffice me for the moment...
However, I am ignorant about what would happen when multiple such
unsafePerformIO are done inside one function.
On Tue, Oct 13, 2009 at 11:04 PM, Daniel Fischer
<[email protected]>wrote:
> Am Dienstag 13 Oktober 2009 18:04:52 schrieb Brent Yorgey:
> > Brent
> >
> > * Some smart-alecks might pipe up with something about unsafePerformIO
> > here. But that's not a cure, it's more like performing an emergency
> > tracheotomy with a ballpoint pen.
>
> Quote of the month!
> _______________________________________________
> Haskell-Cafe mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/haskell-cafe
>
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Message: 6
Date: Sat, 24 Oct 2009 17:59:35 +0100
From: John Moore <[email protected]>
Subject: [Haskell-beginners] random
To: [email protected]
Message-ID:
<[email protected]>
Content-Type: text/plain; charset="iso-8859-1"
Hi All,
Can anyone help me I want to produce a list of three random
numbers for e.g. [7,8,1]
I tried using x <- getStdRandom $ randomR (1,10) but don't really understand
this and it only generates one number. Any help greatly appreciated.
Regards
John
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Message: 7
Date: Sat, 24 Oct 2009 20:05:40 -0400
From: Brent Yorgey <[email protected]>
Subject: Re: [Haskell-beginners] random
To: [email protected]
Message-ID: <[email protected]>
Content-Type: text/plain; charset=us-ascii
On Sat, Oct 24, 2009 at 05:59:35PM +0100, John Moore wrote:
> Hi All,
> Can anyone help me I want to produce a list of three random
> numbers for e.g. [7,8,1]
> I tried using x <- getStdRandom $ randomR (1,10) but don't really understand
> this and it only generates one number. Any help greatly appreciated.
replicateM is your friend:
replicateM :: (Monad m) => Int -> m a -> m [a]
so if 'foo' produces a single random number, then 'replicateM 3 foo'
produces a list of three.
-Brent
------------------------------
Message: 8
Date: Sun, 25 Oct 2009 02:25:58 +0200
From: Tom Davie <[email protected]>
Subject: Re: [Haskell-beginners] random
To: Brent Yorgey <[email protected]>
Cc: [email protected]
Message-ID:
<[email protected]>
Content-Type: text/plain; charset="iso-8859-1"
Or just randomRs :: (Random a, RandomGen g) => (a,a) -> g -> [a]
Bob
On Sun, Oct 25, 2009 at 2:05 AM, Brent Yorgey <[email protected]>wrote:
> On Sat, Oct 24, 2009 at 05:59:35PM +0100, John Moore wrote:
> > Hi All,
> > Can anyone help me I want to produce a list of three random
> > numbers for e.g. [7,8,1]
> > I tried using x <- getStdRandom $ randomR (1,10) but don't really
> understand
> > this and it only generates one number. Any help greatly appreciated.
>
> replicateM is your friend:
>
> replicateM :: (Monad m) => Int -> m a -> m [a]
>
> so if 'foo' produces a single random number, then 'replicateM 3 foo'
> produces a list of three.
>
> -Brent
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/beginners
>
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Message: 9
Date: Sat, 24 Oct 2009 20:04:08 -0500
From: aditya siram <[email protected]>
Subject: Re: [Haskell-beginners] random
Cc: [email protected]
Message-ID:
<[email protected]>
Content-Type: text/plain; charset="iso-8859-1"
Hi John,
When I was first encountered replicateM I found it really hard to
understand. So,of course, I am audacious enough to assume that it is hard
for you too!
The code suggested by Brent , 'replicateM 3 foo' is a nicer way of writing
the following:
foo = do
x <- getStdRandom $ randomR (1,10)
y <- getStdRandom $ randomR (1,10)
z <- getStdRandom $ randomR (1,10)
return [x,y,z]
Hope this helped.
-deech
On Sat, Oct 24, 2009 at 7:05 PM, Brent Yorgey <[email protected]>wrote:
> On Sat, Oct 24, 2009 at 05:59:35PM +0100, John Moore wrote:
> > Hi All,
> > Can anyone help me I want to produce a list of three random
> > numbers for e.g. [7,8,1]
> > I tried using x <- getStdRandom $ randomR (1,10) but don't really
> understand
> > this and it only generates one number. Any help greatly appreciated.
>
> replicateM is your friend:
>
> replicateM :: (Monad m) => Int -> m a -> m [a]
>
> so if 'foo' produces a single random number, then 'replicateM 3 foo'
> produces a list of three.
>
> -Brent
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/beginners
>
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