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Bug introduced in 10.0 or earlier and fixed in 12.0

I know that one can retrieve PlotRange of a plot by using

AbsoluteOptions[plot,PlotRange]

but that won't work on Histogram. Here an example:

In[1099]:=

data = {-1.2056, -1.46192, -1.30053, -2.52879, -0.99636, -1.73904, -1.164,
-1.83398,-0.97505, -0.503256, -0.63802, -0.785963, -0.711821, -0.820439, -1.8699,
-3.9659, -1.4456, -1.67021, -1.42009, -2.5644, -1.45002, -1.27806, -1.66529,
-1.67073, -3.31102, -3.38638};
Histogram[%, PlotRange -> Automatic];
AbsoluteOptions[%, PlotRange]

When running the code I get the following message.

PlotRange::prng: Value of option PlotRange -> {{All,All},{-4.,0.}} is not All,
Full, Automatic, a positive machine number, or an appropriate list of range
specifications. >>

As I understood the documentation, PlotRange need to be of a certain format (e.g. two numbers) and {{All,All},{-4.,0.}} apparently does not fit to that format, for which reason Mathematica won't give me back the PlotRange of my histogram.

Does anybody know how I can get the PlotRange of a Histogram anyway? By the way: In the first place, it will only make sense to get hold of the second value of PlotRange (in my example: {-4.,0.}) since one can calculate the first one for instance through {0, Length[data]}.

Many thanks!

John

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3 Answers 3

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data = {-1.2056, -1.46192, -1.30053, -2.52879, -0.99636, -1.73904, \
-1.164, -1.83398, -0.97505, -0.503256, -0.63802, -0.785963, \
-0.711821, -0.820439, -1.8699, -3.9659, -1.4456, -1.67021, -1.42009, \
-2.5644, -1.45002, -1.27806, -1.66529, -1.67073, -3.31102, -3.38638};
hist = Histogram[data, PlotRange -> Automatic]
First[PlotRange /. Options[hist, PlotRange]]
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  • $\begingroup$ Works perfect. Thank you Rolf! One more issue appeared after setting the plot range as I wanted it to be. For the second histogram Mathematica draws at the level of the minimum value a line which I cannot get rid of (e.g. by GridLines->None). Do you have any suggestions on that issue? Code: HistogramFirst=Histogram[data, PlotRange -> Automatic] HistogramSecond=Histogram[data, PlotRange -> {-8, 0}] $\endgroup$ Commented Feb 10, 2012 at 14:47
  • $\begingroup$ I do not see such a line. Which Mathematica version and which operating system are you using? $\endgroup$ Commented Feb 10, 2012 at 23:22
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Update: getting PlotRange in both directions:

hist = Histogram[data, PlotRange -> Automatic, BarOrigin -> Left]

enter image description here

PlotRange /. Options[hist, PlotRange]

{{All, All}, {-4., 0.}}

To get the values for {All, All} is more involved than suggested in OP:

it will only make sense to get hold of the second value of PlotRange (in my example: {-4.,0.}) since one can calculate the first one for instance through {0, Length[data]}

{0, Length[data]}

{0, 26}

Obviously not the same as the horizontal plot range in the picture above.

To get plot ranges in both horizontal and vertical dimensions, we can use the functions prF1,prF2, or prF3 from this answer to a closely related Q/A:

ClearAll[prF1]
prF1 = Charting`CommonDump`getplotrange[#, AxesOrigin /. Options[#, AxesOrigin]] &;

prF1 @ hist

{{0, 14.},{-4., 0.}}

ClearAll[prF2]
prF2 = MinMax/@Transpose[Join@@Cases[ToBoxes@#, RectangleBox[x_, y_, ___] :> {x, y}, ∞]]&;

prF2 @ hist

{{0, 14.}, {-4., 0.}}

prF3 = Module[{boundingbox}, 
    Histogram[#, ChartElementFunction -> ((boundingbox = 
          Charting`ChartStyleInformation["BoundingBox"]; 
         ChartElementData["GlassRectangle"][##]) &)]; boundingbox] &;

prF3 @ data

{{0, 14.}, {-4., 0.}}

Original answer:

You can also extract the PlotRange from the second Part of hist:

PlotRange /. hist[[2]] // First
(* {-4.`, 0.`} *)

Note: hist[[2]] contains the options

hist[[2]]
(* {AspectRatio->1/GoldenRatio, Axes->{True,True}, AxesLabel->{None,None}, 
   AxesOrigin->{-4.,0}, FrameTicks->{{Automatic, Automatic},{Automatic, Automatic}}, 
   GridLines->{None,None}, PlotRange->{{-4.,0.},{All,All}},
   PlotRangePadding->{{Scaled[0.02],Scaled[0.02]}, 
       {Scaled[0.02],Scaled[0.1]}},Ticks->{Automatic,Automatic}} *)
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2
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You could use my function graphicsInformation to do this:

data = {-1.2056,-1.46192,-1.30053,-2.52879,-0.99636,-1.73904,-1.164,
-1.83398,-0.97505,-0.503256,-0.63802,-0.785963,-0.711821,-0.820439,
-1.8699,-3.9659,-1.4456,-1.67021,-1.42009,-2.5644,-1.45002,-1.27806,
-1.66529,-1.67073,-3.31102,-3.38638};

hist=Histogram[data,PlotRange->Automatic]

graphicsInformation[hist]

enter image description here

{"ImagePadding" -> {{13.7242, 1.5}, {12.7383, 0.5}}, "ImageSize" -> {360., 226.322}, "PlotRangeSize" -> {344.776, 213.083}, "ImagePaddingSize" -> {15.2242, 13.2383}, "PlotRange" -> {{-4.08333, 0.0833333}, {-0.301075, 14.7527}}}

Note that the PlotRange returned is different than that given by the other answers, e.g.:

First[PlotRange /. Options[hist]]

{-4., 0.}

Looking at the actual plot, it is clear that the value returned by graphicsInformation is more accurate.

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