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Today's Topics:
1. Re: Laziness and XML parsing (Michael Snoyman)
2. Problem with an if statement (Karol Samborski)
3. Re: Problem with an if statement (Karol Samborski)
4. Re: Problem with an if statement (Karol Samborski)
----------------------------------------------------------------------
Message: 1
Date: Tue, 8 Nov 2011 23:15:30 -0800
From: Michael Snoyman <[email protected]>
Subject: Re: [Haskell-beginners] Laziness and XML parsing
To: Sean Hess <[email protected]>
Cc: [email protected]
Message-ID:
<caka2jgjfd1e-vxn4nd3qinxj_hia3xwwsgybwemrabqwdw_...@mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
The following should work. The basic idea is:
* Try to parse a <person>
* If it's not a <person>, recursively try again.
{-# LANGUAGE OverloadedStrings #-}
import Text.XML.Stream.Parse
import Data.Text (Text, unpack)
import Control.Monad (join)
import Data.Enumerator (Iteratee)
import Data.XML.Types (Event)
data Person = Person { age :: Int, name :: Text }
deriving Show
parsePerson :: Monad m => Iteratee Event m (Maybe [Person])
parsePerson = tagName "person" (requireAttr "age") $ \age -> do
name <- content
return [Person (read $ unpack age) name]
parseWrapper :: Monad m => Iteratee Event m (Maybe [Person])
parseWrapper =
parsePerson `orE`
(fmap . fmap) concat (tagPredicate (const True) ignoreAttrs (const
$ many parseWrapper))
main = parseFile_ def "people.xml" $ force "people required" parseWrapper
Michael
On Tue, Nov 8, 2011 at 7:28 AM, Sean Hess <[email protected]> wrote:
> Thanks all for your help so far. Using xml-enumerator, is there any way to
> parse the following xml, and ignore the people tag? In other words, can I
> parse it by only providing an Iteratee for Person, no matter where a
> <person> tag appears nested within a document?
>
> <?xml version="1.0" encoding="utf-8"?>
> <people>
> <person age="25">Michael</person>
> <person age="2">Eliezer</person>
> </people>
>
>
>
>
> On Nov 8, 2011, at 7:33 AM, Michael Snoyman wrote:
>
> Thanks for the heads-up, it's just a few minor tweaks in the 0.3->0.4
> transition. I'll update later, and add a link to the blog post, and
> release a new version to Hackage.
>
> On Tue, Nov 8, 2011 at 6:03 AM, Sean Hess <[email protected]> wrote:
>
> Thanks so much to both of you that sent that link.
>
> Sorry, my email totally wasn't clear. I meant that the example in the
>
> package description doesn't
>
> run:?http://hackage.haskell.org/packages/archive/xml-enumerator/0.4.3.1/doc/html/Text-XML-Stream-Parse.html#t:ParseSettings
>
> I'll read through that article.
>
> On Nov 8, 2011, at 7:01 AM, Michael Snoyman wrote:
>
> Here's a blog post on the package:
>
> http://www.yesodweb.com/blog/2011/10/xml-enumerator . It doesn't cover
>
> the streaming interface, but it might give you a good overview of the
>
> package in general. I'm not sure what you mean by "it doesn't run,"
>
> but you'll need at least a basic understanding of enumerators to get
>
> off the ground.
>
> On Tue, Nov 8, 2011 at 5:38 AM, Sean Hess <[email protected]> wrote:
>
> I cannot seem to find a working example of xml-enumerator. It doesn't run:
>
> the names seem to have changed for some things, and I'm too much of a
>
> beginner to figure it out easily.
>
> http://hackage.haskell.org/packages/archive/xml-enumerator/0.4.3.1/doc/html/Text-XML-Stream-Parse.html#t:ParseSettings
>
> On Nov 7, 2011, at 7:59 PM, Felipe Almeida Lessa wrote:
>
> On Tue, Nov 8, 2011 at 12:45 AM, Sean Hess <[email protected]> wrote:
>
> I want to parse a large xml file (2GB), without putting the whole thing into
>
> memory. It's pretty simple with a sax parser in most languages, you just
>
> stream bytes to the sax parser, and wait for sax events.
>
> I recommend you taking a look at xml-enumerator [1] and
>
> libxml-enumerator [2]. ?They are the SAX parsers you know from the
>
> imperative world but much easier to write =). ?In particular, you
>
> don't need to rely on lazyness.
>
> Cheers,
>
> [1] http://hackage.haskell.org/package/xml-enumerator
>
> [2] http://hackage.haskell.org/package/libxml-enumerator
>
> --
>
> Felipe.
>
>
> _______________________________________________
>
> Beginners mailing list
>
> [email protected]
>
> http://www.haskell.org/mailman/listinfo/beginners
>
>
>
>
>
>
------------------------------
Message: 2
Date: Wed, 9 Nov 2011 10:04:03 +0100
From: Karol Samborski <[email protected]>
Subject: [Haskell-beginners] Problem with an if statement
To: Biginners Haskell Mailinglist <[email protected]>
Message-ID:
<CACe2dTvA1Su5sLELaozUdEvX0mnk=yfmKxwQ+q-ycPh3=jn...@mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
Hi,
With a function defined like this (colors is type of Array Int [Int]
and edges is [Int]):
saturation (_,edges,_) = foldr (\ed ac ->if null (colors!ed) then ac
else ac+1) 0 edges
I'm getting an infinite loop.
But when I remove the if statement like this:
saturation (_,edges,_) = foldr (\ed ac -> ac+1) 0 edges
everything's working.
How should I define my function with the if statement to not getting
an infinite loop?
Best Regards,
Karol Samborski
------------------------------
Message: 3
Date: Wed, 9 Nov 2011 10:17:46 +0100
From: Karol Samborski <[email protected]>
Subject: Re: [Haskell-beginners] Problem with an if statement
To: Biginners Haskell Mailinglist <[email protected]>
Message-ID:
<cace2dttzqzvtvnbluzfsppobfnhkuqt_-hd-yp2s_ole+3a...@mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
I forgot to mention that this definition:
saturation (_,edges,_) = length $ foldr (\ed ac -> colors!ed : ac) [] edges
is working too.
I don't understand why the version with an 'if' is not working.
Best Regards,
Karol Samborski
2011/11/9 Karol Samborski <[email protected]>:
> Hi,
>
> With a function defined like this (colors is type of Array Int [Int]
> and edges is [Int]):
> saturation (_,edges,_) = foldr (\ed ac ->if null (colors!ed) then ac
> else ac+1) 0 edges
> I'm getting an infinite loop.
>
> But when I remove the if statement like this:
> saturation (_,edges,_) = foldr (\ed ac -> ac+1) 0 edges
> everything's working.
>
> How should I define my function with the if statement to not getting
> an infinite loop?
>
> Best Regards,
> Karol Samborski
>
------------------------------
Message: 4
Date: Wed, 9 Nov 2011 11:32:38 +0100
From: Karol Samborski <[email protected]>
Subject: Re: [Haskell-beginners] Problem with an if statement
To: Biginners Haskell Mailinglist <[email protected]>
Message-ID:
<cace2dts9uo8hcqtn6njdmhbw3lbbqmbfdjvcnj+xstq_ya6...@mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
Oh, nevermind. The problem was in other place not in that function.
I'm sorry for this...
Best Regards,
Karol Samborski
2011/11/9 Karol Samborski <[email protected]>:
> I forgot to mention that this definition:
> saturation (_,edges,_) = length $ foldr (\ed ac -> colors!ed : ac) [] edges
> is working too.
>
> I don't understand why the version with an 'if' is not working.
>
> Best Regards,
> Karol Samborski
>
> 2011/11/9 Karol Samborski <[email protected]>:
>> Hi,
>>
>> With a function defined like this (colors is type of Array Int [Int]
>> and edges is [Int]):
>> saturation (_,edges,_) = foldr (\ed ac ->if null (colors!ed) then ac
>> else ac+1) 0 edges
>> I'm getting an infinite loop.
>>
>> But when I remove the if statement like this:
>> saturation (_,edges,_) = foldr (\ed ac -> ac+1) 0 edges
>> everything's working.
>>
>> How should I define my function with the if statement to not getting
>> an infinite loop?
>>
>> Best Regards,
>> Karol Samborski
>>
>
------------------------------
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