If I have a function, for example:

p1 = Plot[Evaluate[S[t]*10^-12 /. sol], {t, 20 10^-9, 49 10^-9}]

and this function is setting options->Automatic so the function arguments are implicitly:

p1 = Plot[Evaluate[S[t]*10^-12 /. sol], {t, 20 10^-9, 49 10^-9}, 
          PlotPoints->Automatic, MaxRecursion->Automatic]

how do I get the value to which Automatic has been set?

  • $\begingroup$ You can look at Options[p1] although this can sometime also show Automatic for some of internal options. $\endgroup$
    – Nasser
    Commented Nov 30, 2014 at 0:37
  • 1
    $\begingroup$ @Nasser That will only tell me that an option has been set to Automatic. I want to know what that value is. $\endgroup$ Commented Nov 30, 2014 at 0:39
  • 4
    $\begingroup$ It should be AbsoluteOptions[p1, PlotPoints], but AbsoluteOptions[] has a lot of bugs and doesn't work for PlotPoints $\endgroup$ Commented Nov 30, 2014 at 0:45
  • $\begingroup$ @belisarius AbsoluteOptions[p1, PlotPoints] returns empty for p1 = Plot[Sin[x], {x, 0, Pi}, PlotPoints -> Automatic] on windows, V 10.01. So you are right, it has a bug. $\endgroup$
    – Nasser
    Commented Nov 30, 2014 at 0:49

3 Answers 3


Another option is to use Trace to find occurrences of the relevant rules which appear during evaluation:

 Plot[Sin[t], {t, 0, 2 Pi}],
 HoldPattern[PlotPoints | MaxRecursion -> _],
 TraceInternal -> True] // Flatten // Union

{MaxRecursion -> 6, MaxRecursion -> Automatic, PlotPoints -> 50, PlotPoints -> Automatic}

Or you can set "Verbose" -> True in the system visualisation options to get lots of internal information as the plot is created:

SetSystemOptions["VisualizationOptions" -> {"Verbose" -> True}]
Plot[Sin[t], {t, 0, 2 Pi}]

(0.): >>>>>>>>>>>>>>> Plot <<<<<<<<<<<<<<<

(0.): Expr to Evaluate:

{Sin[t],{{t,0,2π}, Mesh->None, Exclusions->Automatic, PlotPoints->50, MaxRecursion->6, Filling->None, ColorFunction->...


"10.0 for Mac OS X x86 (64-bit) (September 10, 2014)"

From the documentation at http://reference.wolfram.com/language/tutorial/Options.html the default for PlotPoints is 50.

To determine the default for MaxRecursion, let

f[x_] = Product[x + n (-1)^n, {n, -4, 5}] E^(-x^2/2);

The number of points for the Plot of f[x] with the defaults for PlotPoints and MaxRecursion is

auto = Cases[Plot[f[x], {x, -5, 4}, PlotRange -> All], 
   Line[pts_] :> Length[pts], Infinity][[1]]


With the default PlotPoints and varying MaxRecursion check for the same number of points:

     Plot[f[x], {x, -5, 4},
      PlotRange -> All,
      MaxRecursion -> #],
     Line[pts_] :> {#, Length[pts]},
     Infinity][[1]] & /@ Range[15],
 #[[2]] == auto &]

{{6, 1154}}

Hence, the default MaxRecursion is 6

  • $\begingroup$ I remembered now the Cases[...Line[pts_] method. I forgot all about it before. Much better than using index. $\endgroup$
    – Nasser
    Commented Nov 30, 2014 at 5:01

When all else fail, i.e. Options[p1] or AbsoluteOptions[p1, PlotPoints], you can always grab the Line itself and find how many points it has:

 p1 = Plot[Sin[x], {x, 0, Pi}, PlotPoints -> Automatic]

Mathematica graphics

 line = p1[[1, 1, 3, 2]];

Mathematica graphics


Mathematica graphics

So, 259 points are used in this case.

 ListPlot[line[[1]], Mesh -> All, PlotStyle -> Red]

Mathematica graphics

You can see that M oversamples where the function changes most, as expected. ps. this is not the recommended way to find number of points used, since internal data structure can change, and index used above can become invalid in future versions.

  • $\begingroup$ @belisarius I am not following you, sorry. Let me think. I need coffee. But what I see is more points are used where the function changes most. (I mean near the top) $\endgroup$
    – Nasser
    Commented Nov 30, 2014 at 1:02
  • $\begingroup$ Get your coffee. The function is almost flat at the top. $\endgroup$ Commented Nov 30, 2014 at 1:06
  • $\begingroup$ @Nasser This helps, but I still would like to know the value of MaxRecursion->Automatic. Please let me know if you have any ideas. $\endgroup$ Commented Nov 30, 2014 at 1:10
  • $\begingroup$ @Nasser Actually, I think this method doesn't return the value of PlotPoints, rather PlotPoints*(MaxRecursion+1). $\endgroup$ Commented Nov 30, 2014 at 1:36
  • $\begingroup$ Sorry, that gives the maximum number of points used in the plot. The code: Length[line[[1]]], will give a value between PlotPoints and PlotPoints*(MaxRecursion+1)...I think. $\endgroup$ Commented Nov 30, 2014 at 1:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.