Make ListDensityPlot cover a full rectangle area

Questions
1

I’ve taken this piece of code from the documentation:

data = Table[Sin[i + j], {j, 0, 2 Pi, 0.5}, {i, -1, 1}];
ListDensityPlot[data, Mesh -> All]

I am working in something similar, but I want to make the ListDensityPlot to cover the full rectangle showed in the plot (no white space). Can anyone help me?

Thank you in advance for any help!

2

data = Table[Sin[i + j], {j, 0, 2 Pi, 0.5}, {i, -1, 1}];
ListDensityPlot[data, Mesh → All, Prolog → {Blue, Rectangle[{-1, -1}, {1, 1}]}]

ListDensityPlotBackground.png
ListDensityPlotBackground.png370×360 34.5 KB

3

Thank you for your answer!

I was thinking about an extension of the interpolation done to cover the full rectangle.

4

OK. Try by adding corner points to the list:

data = Table[Sin[i + j], {j, 0, 2 Pi, 0.5}, {i, -1, 1}];
corners = {{-1, -1, 0}, {1, -1, 0}, {1, 1, 0}, {-1, 1, 0}};
ListDensityPlot[Join[data, corners], Mesh -> All]

ListDensityPlotExtended.png
ListDensityPlotExtended.png375×361 49.1 KB

5

Ok, I don’t know why but using InterpolationOrder->0 it works.

6

Yes, InterpolationOrder->0 does make the plot cover the whole square. But then it looks a bit strange:

ListDensityPlotIO0.png
ListDensityPlotIO0.png331×311 18.7 KB

7

Yes, it’s true. It looks a bit weird. However, in my case study it seems to fit well.

It’d be nice if there were an option to full the rectangle with a higher interpolation order though.