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I use ListLinePlot to plot data:

data = {1.76703857907629*10^-05, 4.81140563002228*10^-05, 
0.000289014453115104, 0.00197771091043209, 0.0110423194539717, 
0.0498810246783513, 0.190154244971562, 0.627874813716429, 
1.80989253153018, 4.54933071011039, 9.93984896528626, 
18.8366788617287, 30.9406437229311, 44.0665972030142, 
54.4589381973665, 58.4368628024863, 54.4589381973665, 
44.0665972030142, 30.9406437229311, 18.8366788617287, 
9.93984896528626, 4.54933071011039, 1.80989253153018, 
0.627874813716429, 0.190154244971562, 0.0498810246783513, 
0.0110423194539717, 0.00197771091043209, 0.000289014453115104, 
4.81140563002228*10^-05, 1.76703857907629*10^-05}
x1=ListLinePlot[data]
x2=ListLinePlot[data, Filling -> Axis]

But now I only want to fill the data from x=10 to x=20. I didn't find anything liek partial filling depending on the x-axis in the documentary or here at mathematica.SE.

Is there a simple solution or do I have to play with partial plots, which I combine with Show?

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3
  • $\begingroup$ You can overlay two plots. Make one unfilled plot with all the data, one filled plot with a subset of your data, and combine with Show[]. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jul 3, 2015 at 10:49
  • $\begingroup$ @Guesswhoitis. ok thanks. that was one option which i thought might work. good to have a confirmation that there is no simpler way. thanks! $\endgroup$
    – Mario Krenn
    Commented Jul 3, 2015 at 10:57
  • $\begingroup$ There might be a simpler way I'm missing; I just threw out one of the things that could be tried in this situation. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jul 3, 2015 at 11:01

1 Answer 1

Reset to default
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data2 = data; data2[[;; 9]] = Null; data2[[21 ;;]] = Null;
ListLinePlot[{data, data2}, Filling -> {2 -> Axis}]

Mathematica graphics

Or

data3 = MapIndexed[{First@#2, #} &, data];
data3b = Select[data3, 10 <= #[[1]] <= 20 &];
ListLinePlot[{data3, data3b}, Filling -> {2 -> Axis}]
(* same picture *)
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2
  • $\begingroup$ Ah the first one is a nice solution. I expected that one could do it somehow like that. And I'm surprised that partial filling is not implemented more directly; expected that it might be used quite often. Thanks a lot! $\endgroup$
    – Mario Krenn
    Commented Jul 3, 2015 at 11:29
  • 1
    $\begingroup$ Oh, and now I learned that Null!=0. That is very important! Thanks :) $\endgroup$
    – Mario Krenn
    Commented Jul 4, 2015 at 14:27

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