Actually, it's already working pretty well but it slows down when you're doing a lot of zoom in/out.
The trick is that rendering is done using shader (OpenGL) and this computation is used to give information to the shader to where to draw antialiased lines. In the end, this shader is able to draw any amiunt of grids/ticks (as in matplotlib). Some old example are available from here: https://github.com/rougier/gl-agg I tested your solution and it is faster by only a tiny amount but the way you wrote it might open the door for other improvements. Thanks. Nicolas On 22 Jun 2014, at 21:14, Eelco Hoogendoorn <[email protected]> wrote: > Protip: if you are writing your own rasterization code in python, be prepared > to forget about performance altogether. > > Something like numba or other c-like extension will be necessary unless you > are willing to leave big gobs of performance on the table; and even with pure > C you will get nowhere close to the performance of super-duper optimized > library code you are used to. > > But before you go down that rabbit hole, its probably worth thinking about > whether you can get an existing rendering framework to do what you want to do. > > > On Sun, Jun 22, 2014 at 8:30 PM, Nicolas P. Rougier > <[email protected]> wrote: > > Thanks, I'll try your solution. > > Data (L) is not so big actually, it represents pixels on screen and (I) > represents line position (for grids). I need to compute this quantity > everytime the user zoom in or out. > > > Nicolas > > > On 22 Jun 2014, at 19:05, Eelco Hoogendoorn <[email protected]> > wrote: > > > Well, if the spacing is truly uniform, then of course you don't really need > > the search, and you can do away with the extra log-n, and there is a purely > > linear solution: > > > > def find_closest_direct(start, end, count, A): > > Q = (A-start)/(end-start)*count > > mid = ((Q[1:]+Q[:-1]+1)/2).astype(np.int) > > boundary = np.zeros(count, np.int) > > boundary[mid] = 1 > > return np.add.accumulate(boundary) > > > > I expect this to be a bit faster, but nothing dramatic, unless your > > datasets are huge. It isn't really more or less elegant either, id say. > > Note that the output isn't 100% identical; youd need to do a little > > tinkering to figure out the correct/desired rounding behavior. > > > > > > On Sun, Jun 22, 2014 at 5:16 PM, Nicolas P. Rougier > > <[email protected]> wrote: > > > > Thanks for the answer. > > I was secretly hoping for some kind of hardly-known numpy function that > > would make things faster auto-magically... > > > > > > Nicolas > > > > > > On 22 Jun 2014, at 10:30, Eelco Hoogendoorn <[email protected]> > > wrote: > > > > > Perhaps you could simplify some statements, but at least the algorithmic > > > complexity is fine, and everything is vectorized, so I doubt you will get > > > huge gains. > > > > > > You could take a look at the functions in scipy.spatial, and see how they > > > perform for your problem parameters. > > > > > > > > > On Sun, Jun 22, 2014 at 10:22 AM, Nicolas P. Rougier > > > <[email protected]> wrote: > > > > > > > > > Hi, > > > > > > I have an array L with regular spaced values between 0 and width. > > > I have a (sorted) array I with irregular spaced values between 0 and > > > width. > > > > > > I would like to find the closest value in I for any value in L. > > > > > > Currently, I'm using the following script but I wonder if I missed an > > > obvious (and faster) solution: > > > > > > > > > import numpy as np > > > > > > def find_closest(A, target): > > > idx = A.searchsorted(target) > > > idx = np.clip(idx, 1, len(A) - 1) > > > left = A[idx - 1] > > > right = A[idx] > > > idx -= target - left < right - target > > > return idx > > > > > > n, width = 256, 100.0 > > > > > > # 10 random sorted values in [0,width] > > > I = np.sort(np.random.randint(0,width,10)) > > > > > > # n regular spaced values in [0,width] > > > L = np.linspace(0, width, n) > > > > > > print I[find_closest(I,L)] > > > > > > > > > > > > Nicolas > > > _______________________________________________ > > > NumPy-Discussion mailing list > > > [email protected] > > > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > > > > > _______________________________________________ > > > NumPy-Discussion mailing list > > > [email protected] > > > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > > > _______________________________________________ > > NumPy-Discussion mailing list > > [email protected] > > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > > > _______________________________________________ > > NumPy-Discussion mailing list > > [email protected] > > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > _______________________________________________ > NumPy-Discussion mailing list > [email protected] > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > _______________________________________________ > NumPy-Discussion mailing list > [email protected] > http://mail.scipy.org/mailman/listinfo/numpy-discussion _______________________________________________ NumPy-Discussion mailing list [email protected] http://mail.scipy.org/mailman/listinfo/numpy-discussion
