This is related to the issue brought up here, if you want to mark it as duplicate feel free to do so, but it is slightly separate.
Take two sets of data, identical except for the form,
dta1 = Table[Exp[-x^2] Sin[y], {y, -π, π, π/3}, {x, -8, 8, 1}];
dta2 = Table[{x, y, Exp[-x^2] Sin[y]}, {x, -8, 8, 1}, {y, -π, π, π/3}]~Flatten~1;
The first is a two-dimensional array and the second is a list of tuples. Now note the different effects of InterpolationOrder
on these data,
ListContourPlot[dta1, InterpolationOrder -> #, ImageSize -> 200,
DataRange -> {{-8, 8}, {-π, π}},PlotRange-> All] & /@ {0, 1, 2, 3}
ListContourPlot[dta2, InterpolationOrder -> #,
ImageSize -> 200,PlotRange -> All] & /@ {0, 1, 2, 3}
Going past an order of 1 has no effect. At order zero they are identical, at order 1 the 2D array is already better.
Notice that this is not a problem for ListLinePlot
with one-dimensional data
func = Exp[-#^2] Cos[4 #] &;
list1 = Table[func[x], {x, -5, 5, .5}];
list2 = Table[{x, func[x]}, {x, -5, 5, .5}];
ListLinePlot[list1, DataRange -> {-5, 5}, InterpolationOrder -> #,
PlotRange -> {-1, 1}, ImageSize -> 200] & /@ {0, 1, 2, 3}
ListLinePlot[list2, InterpolationOrder -> #, PlotRange -> {-1, 1},
ImageSize -> 200] & /@ {0, 1, 2, 3}
PlotRange -> All
. Also, withListContourPlot
replaced byListPlot3D
, the addition ofMesh -> All
produces what to me are surprising results. $\endgroup$ – bbgodfrey Dec 3 '15 at 5:02DataRange
option in the first use ofListContourPlot
is only telling Mathematica how you want the axes scaled. $\endgroup$ – djphd Dec 3 '15 at 5:08PlotRange->All
set in myinit.m
file. In my opinion that should be the default, since the contour plots look atrocious without it. That is also interesting howMesh->All
interacts with theInterpolationOrder
option. $\endgroup$ – Jason B. Dec 3 '15 at 7:45