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I am trying to produce a correlation map for some NMR experiment with this data. I tried to plot the data with Mathematica but the result was not close to my expectation. Hence, I tried plotting the data with SigmaPlot (version 14) and I got something close to what I expect. What should I do to my code to get a plot close or better than the plot obtained with SigmaPlot? The plots are as follows:

Plot from SigmaPlot

and

Plot from Mathematica

My code is:

data = Import["correlationdata.csv"];
ListContourPlot[data, ColorFunction -> Hue, PlotRange -> All, FrameLabel -> {"x(m)", "y(m)"}, 
PlotLegends ->  Placed[BarLegend[Automatic, LegendMargins -> {{0, 0}, {10, 5}}, LegendLabel -> 
"Intensity"], Right]]
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    $\begingroup$ I'm not sure you are plotting the same data in the two cases. For instance, in the csv there are no points above {0.162, 0.644} (in agreement with the Mathematica plot), but your sigmaplot has data above that (for instance, near {0.162, 0.65}). $\endgroup$
    – bill s
    Commented Jul 13, 2020 at 18:53
  • $\begingroup$ @bills, you are correct. According to SigmaPlot documentation "When SigmaPlot graphs a mesh or contour graph, if the X and Y data forms a regular matrix, the graphs are usually as expected. When the X and Y values are not so spaced, SigmaPlot interpolates values, and can give unexpected results." $\endgroup$
    – Dean
    Commented Jul 13, 2020 at 18:58
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    $\begingroup$ But bill s seems to be right. For example, the lowest intensity value of your given csv is -0.129135. But from the SigmaPlot-Graph, we see values <-0.6 $\endgroup$
    – Julien Kluge
    Commented Jul 13, 2020 at 19:38
  • $\begingroup$ @Julien Kluge, yes! SigmaPlot usually make interpolations. My X and Y values were determined experimentally and each has the form: A+-B (where B is the standard deviation). An app was used to randomly distribute the measurements to generate X and Y values in my data. A model equation is used to generate Z values. Hence, I suppose the interpolations are not out of order. If I can do this (or even better) in Mathematica, I would be grateful. $\endgroup$
    – Dean
    Commented Jul 18, 2020 at 20:56

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