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The main issue which needs to be solved is:

Let's say I have an array with 8 numbers, e.g. [2,4,8,3,5,4,9,2] and I use them as values for my x axis in an coordinate system to draw a line. But I can only display 3 of this points. What I need to do now is do reduce the number of points (8) to 3, without manipulating the line too much - so using an average or something like that should be an option. I am NOT looking for the average of the array in a whole - I still need 3 points of the amount of 8 in total.

For an array like [2,4,2,4,2,4,2,4] and 4 numbers out of that array, I could simply use the average "3" of each pair I guess - but that's not possible if the number is uneven.

But how would I do that? Do you know any algorithms to do this? How is that process even called?

To give you some more realistic details about this issue: I have an x axis, which is 720px long and let's say I get 1000 points. Now I have to reduce this 1000 points (2 arrays, one for x and one for y values) to a maximum of 720 points.

Thought about interpolation and stuff like that, but I'm still not quite sure if this is what I am looking for.

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    $\begingroup$ I'm not sure I follow. Something like this? en.wikipedia.org/wiki/… $\endgroup$ – Szabolcs May 25 '16 at 11:45
  • $\begingroup$ Take a look at MovingMap + Mean $\endgroup$ – Kuba May 25 '16 at 11:49
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ArrayResample may be what you are after, as it keeps the same general shape of the original data

list1 = Sin[#] + RandomReal[0.1] & /@ Subdivide[2 π, 999];
Length@list
list2 = ArrayResample[list, 720];
Length@list2
ListLinePlot[{list1, list2}, DataRange -> {1, 1000}]
(* 1000 *)
(* 720 *)

Mathematica graphics

But if you want to do some smoothing, then MovingAverage will do the job.

MovingAverage gives a list of length Length[list]-r+1.

So if you have a list of 1000 elements you want reduced to 720 elements, then you need to give an r value of 281. But note that the new data does not cover the same range as the old data

list = Sin[#] + RandomReal[0.1] & /@ Subdivide[2 π, 999];
Length@list
list2 = MovingAverage[list, 281];
Length@list2
Show[ListPlot@list,
 ListLinePlot[list2, DataRange -> {140, 860}, PlotStyle -> Red]
 ]
(* 1000 *)
(* 720 *)

Mathematica graphics

| improve this answer | |
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  • $\begingroup$ Thanks for your helpful hints so that I can go further. $\endgroup$ – Christopher Wagner Jun 1 '16 at 8:44

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