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Consider this list plot

ls = Table[Sin[t], {t, 0, 100, 0.1}];
p = ListPlot[ls, PlotRange -> {{0., 0.1}, All}, DataRange -> {0, 1}, Joined -> True]

enter image description here

If we look at the data in the plot, we see that all the data in ls are included in the plot.

ls2 = p[[1, 2, 1, 3, 3, 1]];
Dimensions[ls2]
(* {1001, 2} *)

ListPlot[ls2, Joined -> True]

enter image description here

Sometimes this is really helpful because we can just retrieve the data at a later time from the plots and don't need to regenerate the data again. But sometimes I'm only intreated in the range set in the ListPlot, and saving all the data is kindly of wasting of space, especially when I have many plots in a notebook, the notebook size becomes very large and not very responsive.

So are there simple ways to tell ListPlot to not include the data outside the plot range, other than manually select the data before plot?

P.S. Sometimes, ListPlot does drop the data outside the range, see example here.

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You could assign p[[1, 2, 1, 3, 3, 1]] to whatever you want ... –  belisarius Apr 7 at 21:49
    
@belisarius What do you mean? I think maybe I didn't say clearly, I meant to make p[[1, 2, 1, 3, 3, 1]] only contains the data points that in the plot range I set. –  xslittlegrass Apr 7 at 23:08
    
Whether save the graphics as raster image fulfill your need? Anyway you can always choose to plot only part of your data (ls). –  Silvia Apr 7 at 23:58
2  
No, there's no documented (or undocumented, AFAIK) feature in plot to do this. If you don't need a full tank, don't fill it (truncate the data list at the source). Fiddling with it ex post facto means possibly breaking things later in the chain (e.g., carrying the "baggage" allows one to use say things like Show with differing options from the original plot). –  rasher Apr 8 at 0:06
1  
So why not just plot the data you want? –  Silvia Apr 8 at 3:15

1 Answer 1

I think @rasher's comment is correct and you can plot the points you need by truncating the list. You can make a function that does that. I named mine (ironically) economicListPlot but it took a lot of copy-pasting to account for different values of the options so it could do with a tidy up

economicListPlot[data_ /; VectorQ[data], opts : OptionsPattern[ListPlot]] :=

 Module[{newopts, plotRange1, dataRange1, ind1, ind2},
  With[{
    plotRange = PlotRange /. FilterRules[{opts}, PlotRange], 
    dataRange = DataRange /. FilterRules[{opts}, DataRange]
    },
   If[dataRange === DataRange,



    If[plotRange === PlotRange || plotRange === All || plotRange === Automatic,

     ListPlot[data, Evaluate[opts]],(*no plotrange or datarange defined*)

     newopts = Sequence @@ Cases[{opts}, Except[HoldPattern[PlotRange -> _]]];
     plotRange1 = If[VectorQ@plotRange, plotRange, First@plotRange];
     ind1 = IntegerPart@Max[First@plotRange1, 1];
     ind2 = IntegerPart@Max[Last@plotRange1, 1];
     ListPlot[data[[ind1 ;; ind2]], 
      Evaluate@newopts, 
      PlotRange -> {Automatic, 
        Evaluate@
         Complement[ 
          plotRange, {plotRange1}]}(*plotrange defined only*)]
     ],


    If[plotRange === All || plotRange === Automatic,
     ListPlot[data, Evaluate[opts]],
     newopts = Sequence @@ Cases[{opts}, Except[HoldPattern[PlotRange -> _]]];
     newopts = Sequence @@ Cases[{newopts}, Except[HoldPattern[DataRange -> _]]];
     plotRange1 = If[VectorQ@plotRange, plotRange, First@plotRange];
     dataRange1 = If[VectorQ@dataRange, dataRange, First@dataRange];
     ind1 = Max[Ceiling[(plotRange1[[1]] - dataRange1[[1]])/
         Subtract @@ Reverse@dataRange1 (Length[data])], 1];
     ind2 = Min[Ceiling[
        ind1 + (Subtract @@ Reverse@plotRange1* (Length[data]))/
         Subtract @@ Reverse@dataRange1 ], Length[data]
       ];
     ListPlot[data[[ind1 ;; ind2]], Evaluate@newopts, 
      PlotRange -> {All, 
        Evaluate@Complement[ plotRange, {plotRange1}]}, 
      DataRange -> 
       Evaluate@
        plotRange1](*both plotrange and datarange defined*)
     ]
    ]
   ]
  ]

Anyhow, it does what you want:

    data = Table[Sin[t], {t, 0, 100, 0.1}];
p = ListPlot[data, PlotRange -> {{0., 0.1}, All}, DataRange -> {0, 1},
    Joined -> True];
p2 = economicListPlot[data, PlotRange -> {{0., 0.1}, All}, 
   DataRange -> {0, 1}, Joined -> True];

p2[[1, 2, 1, 3, 3, 1]] // Dimensions
(*{102 , 2}*)

p[[1, 2, 1, 3, 3, 1]] // Dimensions
(*{1001 , 2}*)

and the plots are the "same":

GraphicsRow[{p, p2}]

enter image description here

but if you go to wider plot/data ranges there is some difference (my index gymnastics are a little off) so I'd be cautious on how I'd use this:

Show[ListPlot[data, Joined -> True, PlotRange -> {{1.1, 1.3}, All}, 
  DataRange -> {0, 20}, PlotStyle -> Thickness[.01]], 
 economicListPlot[data, Joined -> True, 
  PlotRange -> {{1.1, 1.3}, All}, DataRange -> {0, 20}, 
  PlotStyle -> Orange]]

discrepancy

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