Benjamin Root, on 2011-01-27 13:04, wrote: > I believe he would rather call the core function that contour uses to > do the heavy lifting. This was something that one can do in matlab, > btw. I don't have access to the source right now. What does contour > call to perform this calculation?
matplotlib.contour.QuadContourSet - which in turn uses ContourSet, and both take ax as a required argument. They use matplotlib.contour._cntr which is Daniel Fulger, on 2011-01-27 20:21, wrote: > no, I would like to suppress plotting entirely, avoid changing of > active figure and avoid handling figures or axis completely. > I m only interested in the contourset. I wonder if my post was > somehow sloppy. > > Yes, there are work-arounds like creating a dummy figure, similar to > your suggestion, and return focus to > the previously active figure. But plotting takes time and memory, is > not needed and requires several code lines. Once might be ok but > speed and memory is important. > Plotting with alpha=0 still requires figure and axis handling. > > So how can I switch off all figure and axis related actions and > savely call contourset = contour(x,y,...) that does nothing else than > return the contours? I understand better now, but as far as I could tell from poking inside the QuadContourSet code, there isn't a simple way to call the underlying machinery which generates the contours. You'll have to look at what QuadContourSet._contour_args does internally to see what what x, y, z should be, and then create a contour using C = matplotlib.contour._cntr.Cntr(x,y,z) and then for each level, do something like what QuadContourSet._get_allsegs_and_allkinds does C.trace(..) best, -- Paul Ivanov 314 address only used for lists, off-list direct email at: http://pirsquared.org | GPG/PGP key id: 0x0F3E28F7
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