On Wed, Dec 10, 2014 at 4:20 PM, Shiyao Ma <i...@introo.me> wrote: > from itertools import product > space_len = 580 > space = product(xrange(space_len), xrange(space_len), xrange(space_len)) > > sparse_cloud = product(xrange(1000), xrange(1000), xrange(1000)) > for i, j, k in sparse_cloud: > ts = timeit.default_timer() > if (i, j, k) in space: pass > te = timeit.default_timer()
On Wed, Dec 10, 2014 at 4:23 PM, Shiyao Ma <i...@introo.me> wrote: > One thing to note, the logic of using "in" is not of concern here. > This is a *contrived* example, the problem is the slowness of the first > iteration. Are you sure it isn't? Your 'space' is an iterable cubic cross-product. Your first loop checks (0,0,0) which is the first element returned, and is thus fast... but it also *consumes* that first element. The next time you test it, the entire space gets consumed, looking for another (0,0,0), which won't exist. That means iterating over 580**3 == 195112000 (two hundred million) tuples, and that *is* going to be slow. Is product() really the right way to represent your space? ChrisA -- https://mail.python.org/mailman/listinfo/python-list
