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	<title>Comments on: Realm of the Practical</title>
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		<title>By: Andres Cabrera</title>
		<link>http://codehop.com/realm-of-the-practical/#comment-38</link>
		<dc:creator><![CDATA[Andres Cabrera]]></dc:creator>
		<pubDate>Sun, 30 May 2010 08:30:52 +0000</pubDate>
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		<description><![CDATA[Do you mean that python iterators will be wrong for the graph because they are too slow?
 
Cheers,
Andrés]]></description>
		<content:encoded><![CDATA[<p>Do you mean that python iterators will be wrong for the graph because they are too slow?<br />
 <br />
Cheers,<br />
Andrés</p>
]]></content:encoded>
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		<title>By: Jacob Joaquin</title>
		<link>http://codehop.com/realm-of-the-practical/#comment-37</link>
		<dc:creator><![CDATA[Jacob Joaquin]]></dc:creator>
		<pubDate>Thu, 27 May 2010 18:53:00 +0000</pubDate>
		<guid isPermaLink="false">http://slipmat.noisepages.com/?p=178#comment-37</guid>
		<description><![CDATA[&lt;p&gt;In Python 3, map() returns an iterator, as opposed to python 2 where map() returns a list. So let&#039;s say you have this expression:&lt;/p&gt;
&lt;p&gt;
&lt;p&gt;foo = map(lambda x, y: x + y, [1, 2, 3], [4, 5, 6])&lt;/p&gt;
&lt;p&gt;Printing foo in Python 2 returns [5, 7, 9]. Printing foo in Python 3 returns &lt;map object at 0xd5f190&gt;.&lt;/p&gt;
&lt;p&gt;The * operator in the statement &quot;print(*f00)&quot; prints each value that is yielded by the map iterator object, outputting: 5 7 9&lt;/p&gt;
&lt;p&gt;That said, iterators/generators are completely inadequate for DSP graphs, though they did come in handy for developing a quick prototype.&lt;/p&gt;
&lt;p&gt;I think what might happen is that a graph on the Python side of Slipmat will merely be a description of how unit generators are connected. Perhaps done with XML? Once a graph description is passed to the C/C++ engine, the graph description is parsed, and an actual graph is created. This really isn&#039;t too different to how the Python Csound API works, or how Supercollider creates byte code that is passed to the server.&lt;/p&gt;
&lt;p&gt;And hopefully I answered your question.&lt;/p&gt;
&lt;/p&gt;]]></description>
		<content:encoded><![CDATA[<p>In Python 3, map() returns an iterator, as opposed to python 2 where map() returns a list. So let&#8217;s say you have this expression:</p>
</p>
<p>foo = map(lambda x, y: x + y, [1, 2, 3], [4, 5, 6])</p>
<p>Printing foo in Python 2 returns [5, 7, 9]. Printing foo in Python 3 returns &lt;map object at 0xd5f190&gt;.</p>
<p>The * operator in the statement &#8220;print(*f00)&#8221; prints each value that is yielded by the map iterator object, outputting: 5 7 9</p>
<p>That said, iterators/generators are completely inadequate for DSP graphs, though they did come in handy for developing a quick prototype.</p>
<p>I think what might happen is that a graph on the Python side of Slipmat will merely be a description of how unit generators are connected. Perhaps done with XML? Once a graph description is passed to the C/C++ engine, the graph description is parsed, and an actual graph is created. This really isn&#8217;t too different to how the Python Csound API works, or how Supercollider creates byte code that is passed to the server.</p>
<p>And hopefully I answered your question.</p>
]]></content:encoded>
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	<item>
		<title>By: Andrés</title>
		<link>http://codehop.com/realm-of-the-practical/#comment-36</link>
		<dc:creator><![CDATA[Andrés]]></dc:creator>
		<pubDate>Thu, 27 May 2010 17:15:10 +0000</pubDate>
		<guid isPermaLink="false">http://slipmat.noisepages.com/?p=178#comment-36</guid>
		<description><![CDATA[Hi,
I&#039;ve just checked this, and even though they are called generators, they seem to be evaluated when they are created. I&#039;m using python 2.6, so they * operator doesn&#039;t work... What is it for? (Is this the trick?)
 
Cheers,
Andrés
 ]]></description>
		<content:encoded><![CDATA[<p>Hi,<br />
I&#8217;ve just checked this, and even though they are called generators, they seem to be evaluated when they are created. I&#8217;m using python 2.6, so they * operator doesn&#8217;t work&#8230; What is it for? (Is this the trick?)<br />
 <br />
Cheers,<br />
Andrés<br />
 </p>
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