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Timeline for Estimating progress on plots

Current License: CC BY-SA 3.0

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May 24, 2012 at 12:44 comment added celtschk BTW, further experiments make me think that the automatic MaxRecursion for Plot is just 6 anyway.
May 24, 2012 at 10:53 comment added celtschk Of course it was not because of the function name. I did a few tests, and surprisingly often the progress bar gives a quite good estimate.
May 24, 2012 at 10:33 history edited Rojo CC BY-SA 3.0
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May 24, 2012 at 10:33 comment added Rojo Thanks for the accept @celtschk, I hope it was not because of the function name. I'll be editing the "which I you're probably not" out
May 24, 2012 at 10:31 comment added celtschk Thanks, that's a quite nice solution. Indeed I am willing to sacrifice automaticity on MaxRecursion because for those plots where it matters I have to do that anyway to get my plot in a reasonable time.
May 24, 2012 at 10:28 vote accept celtschk
May 24, 2012 at 10:24 comment added Rojo @Szabolcs, in any case, if it works for most cases, even when you don't know in which, it's probably good enough to be useful as an "estimate" I think
May 24, 2012 at 10:19 comment added Szabolcs I wanted to give a counterexample or the computation steps growing exponentially, but it turned out that even in my example they do exactly that ... ListPlot[Reverse@ Table[Part[ Reap@Plot[1/(1 + Exp[-10 x]), {x, -3, 3}, Mesh -> All, EvaluationMonitor :> Sow[x], MaxRecursion -> k], 2, 1], {k, 0, 3}], PlotRange -> All]
May 24, 2012 at 10:14 comment added Rojo @Heike, there we go, thanks
May 24, 2012 at 10:12 history edited Rojo CC BY-SA 3.0
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May 24, 2012 at 10:04 comment added Rojo @Heike, thank god I have an alternative. My code is disgusting
May 24, 2012 at 10:01 comment added Heike As an alternative you could use EvaluationMonitor :> (If[v<ant, i++]; xant = v).
May 24, 2012 at 9:59 history edited Rojo CC BY-SA 3.0
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May 24, 2012 at 9:44 history answered Rojo CC BY-SA 3.0