Thanks for the answer.

Basically I was confused about the function of the random(). I knew it
was choosing a something from each playlist, but in turn I thought it
was also choosing a random element from each one.


On Fri, May 15, 2009 at 4:04 AM, David Baelde <[email protected]> wrote:
> On Thu, May 14, 2009 at 8:33 PM, Brandon Casci <[email protected]> 
> wrote:
>> all = random(weights=[1,8],strict=true,[ playlist_1, playlist_2 ])
>
> Let me develop a bit what this does.
>
> It does not make the playlists random. The random() operator doesn't
> even know if its children are playlists or external streams. What
> random() does is pick successive tracks from a random children. For
> example if playlist_1 plays tracks p1, p2, p3... and playlist_2 plays
> q1, q2, q3... then random([playlist_1,playlist_2]) might play tracks
> p1,q1,p2,p3,q2,... The relative order of p2 and p3, for example, will
> never be changed. If you want the playlist itself to be random, use
> the "mode" parameter of playlist().
>
> Now, in your example you pass weights=[1,8]: this means that
> playlist_2 has height times more chances to be selected than
> playlist_1.
>
> When you pass strict=true, the random() is not random anymore, but
> acts like a very deterministic quota operator. In you example it
> means: play 1 file of playlist_1, then 8 of playlist_2, then start
> again.
>
>> If you don't want the playlists to be random, would you use prepend
>> along with some custom selection logic to occasionally merge tracks
>> from play tracks from both playlists?
>
> I'm not sure to understand the question. Maybe you want to force two
> successive tracks of one playlist to be played consecutively in the
> random()? If so, the solution is indeed to merge those tracks before
> the random() node so that they become one. Prepending is one way of
> doing this. It should also be possible to play with transitions (using
> cross()) depending on what you want exactly.
>
> HTH
> --
> David
>

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