Re: How can I generate this graphic?

• To: mathgroup at smc.vnet.net
• Subject: [mg113645] Re: How can I generate this graphic?
• From: Vicent <vginer at gmail.com>
• Date: Sat, 6 Nov 2010 04:58:26 -0500 (EST)

```On Thu, Nov 4, 2010 at 10:01, Vicent <vginer at gmail.com> wrote:

> On Wed, Nov 3, 2010 at 10:22, M.Roellig <markus.roellig at googlemail.com
> >wrote:
>
> > Hi, just an example to get you going:
> >
> > DataForPlotting = Table[Sin[x] y, {x, 1, 20, 1}, {y, 1, 20, 1}];
> > ListContourPlot[DataForPlotting, InterpolationOrder -> 0,
> >  ContourLabels -> All, PerformanceGoal -> "Quality", Mesh -> All,
> >  MeshStyle -> Directive[GrayLevel[0.2], Thickness[0.005]],
> >  ColorFunction -> (Blend[{Orange, Yellow, White}, #] &),
> >  FrameLabel -> {Style["x", Large, Italic, FontFamily -> "Arial"],
> >   Style[Rotate["y", -(\[Pi]/2)], Large, Italic,
> >    FontFamily -> "Arial"]}, FrameTicks -> None,
> >
> > Markus
> >
> >
>
> Markus,
>
> Thank you very much!!  There are so many options at Mathematica that
> sometimes it is difficult to know them all. That is a very good hint.
>
> Now I'll keep on tuning the graphic by myself.
>
> Thanks again.
>
> --
> Vicent Giner-Bosch
>
>
Just in case anyone is interested, I finally used  MatrixPlot  function in
order to represent my data --it fits my needs better than  ListContourPlot.

I needed a point or square to be printed for each (x,y) combination (with x
and y being integer), and I got that with MatrixPlot.

This is my code:

DataForPlotting = Table[Sin[x] y, {x, 1, 20, 1}, {y, 1, 20, 1}];

MatrixPlot[DataForPlotting, DataReversed -> {True, False},
Mesh -> All, PlotRangePadding -> 0,
MeshStyle -> Directive[GrayLevel[0.5], Thickness[0.0005]],
FrameTicksStyle ->
Directive[Medium, FontFamily -> "Calibri", Thickness[0.0005]],
FrameTicks -> {{All, None}, {All, None}},
FrameLabel -> {Style["x", Large, Italic, FontFamily -> "Calibri"],
Style["y", Large, Italic, FontFamily -> "Calibri"]}]

Thank you to all.

--
Vicent

```

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