scale bottle-lag?

Jun 23, 2009 at 2:08pm

scale bottle-lag?

I always used iter/scale/group to scale my lists, now with very large and fast input-lists (up to 4096, see patch) could it be a bottle-lag? Is vexpr much more faster? What’s the vexpr formula for simple linear scaling?
I know the Lobjects, but I try not to use 3rdParty objects if possible…

big thx,

– Pasted Max Patch, click to expand. –
Jun 23, 2009 at 3:31pm

[vexpr ($f1-$f2)*($f5-$f4)/($f3-$f2) + $f4 @scalarmode 1]

Jun 23, 2009 at 3:42pm

I don’t think [vexpr] can handle lists that long but I might be mistaken.


Jun 23, 2009 at 4:35pm

It sure can (>= 5.0.6). You can also have a look to my ej.lscale.

Jun 23, 2009 at 4:37pm

big thx!
works fine even with very big ones!

Jun 23, 2009 at 4:41pm

Ah, that would be the problem, I’m still using 5.0.4 where it appears the maximum is 999. Thanks for pointing that out Emmanuel.


Jun 23, 2009 at 4:50pm

That’s right. It has been fixed for 5.0.6. You can have access to the what’s new information by using the “Latest Support Information” item from the Help menu.

Jun 24, 2009 at 12:49am

the scale formula for [vexpr] is the same as for expr,
but i always find it a bit brrrr to send [vexpr] a
list of 100 identical elements, which is why i do
the [zl len] [zl iter 1] [zl group] thing.


Jul 11, 2009 at 5:32am

You can measure the difference in time, and if your input has a fixed range, you can simplify the vexpr. Even if you get rid of the [zl len] you’re still more than twice as fast with vexpr…


– Pasted Max Patch, click to expand. –

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