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Hi, I'm using the setFunction method of ColorControl to set a custom color map. This mostly works, but I have a couple of questions. One is, is there a simple way to "unset" the function (that is, to have it go back to the default colormap)? I tried naively setting the function to null but that did not work. I want to allow the user to choose between several colormaps, including the default. I know I can build a custom map that looks like the default, but wanted to avoid that. Second, I use a FlatField to represent the colormap, which gets passed to the setFunction method, with it constructed like this: RealType cp = RealType.getRealType("cp"); RealType cr = RealType.getRealType("cr"); RealType cg = RealType.getRealType("cg"); RealType cb = RealType.getRealType("cb"); try { RealTupleType crgb = new RealTupleType(cr, cg, cb); FunctionType colormapfunction = new FunctionType(cp, crgb); float[][] cpos = new float[1][pos.size()]; for (int i=0; i<pos.size(); i++) { cpos[0][i] = pos.get(i); } Irregular1DSet colorbarpos = new Irregular1DSet(cp, cpos); FlatField mycolormap = new FlatField(colormapfunction, colorbarpos); double[][] samples = new double[3][colors.size()]; ColorRGBValue col; for (int i=0; i<colors.size(); i++) { col = colors.get(i); samples[0][i] = col.r; samples[1][i] = col.g; samples[2][i] = col.b; } mycolormap.setSamples(samples); "pos" and "colors" are simply Java Vectors holding the 1D positons of the colormap or the 3D rgb colors of the map. It seems that this creates a segmented colormap, in that there are abrupt boundaries between the colors, instead of smooth. Is this how it works? i.e. do I have to have lots of points in my colormap to get a smooth appearance? Thanks for any help, Donna Donna L. Gresh, Ph.D. IBM T.J. Watson Research Center (914) 945-2472 http://www.research.ibm.com/people/g/donnagresh gresh@xxxxxxxxxx
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