Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]
Groups > comp.lang.python > #100244
| From | Oscar Benjamin <oscar.j.benjamin@gmail.com> |
|---|---|
| Newsgroups | comp.lang.python |
| Subject | Re: np.searchSorted over 2D array |
| Date | 2015-12-10 19:55 +0000 |
| Message-ID | <mailman.120.1449777339.12405.python-list@python.org> (permalink) |
| References | <2b660168-3322-4bf6-bd8b-2dc846a3a874@googlegroups.com> <mailman.95.1449679003.12405.python-list@python.org> <168a11ec-dd1e-481c-827b-8f5021a315cb@googlegroups.com> |
On 10 Dec 2015 14:46, "Heli" <hemla21@gmail.com> wrote: > > Thanks Peter, > > I will try to explain what I really need. > > I have a 3D numpy array of 100*100*100 (1M elements). Then I have another numpy array of for example 10*2*10 (200 elements). I want to know if in the bigger dataset of 100*100*100, there is anywhere, where the second numpy array of 200 elements with shape 10*2*10 appears. If it does, then I want the indices of the bigger dataset where this happens. > So you want to find N in M. First find all occurrences of N[0][0][0] in M[:90][:98][:90]. Then for each of those extract the same size subview from M and check if (Ms == N).all(). -- Oscar
Back to comp.lang.python | Previous | Next — Previous in thread | Find similar | Unroll thread
np.searchSorted over 2D array Heli <hemla21@gmail.com> - 2015-12-09 08:06 -0800
Re: np.searchSorted over 2D array Peter Otten <__peter__@web.de> - 2015-12-09 17:36 +0100
Re: np.searchSorted over 2D array Heli <hemla21@gmail.com> - 2015-12-10 06:40 -0800
Re: np.searchSorted over 2D array Oscar Benjamin <oscar.j.benjamin@gmail.com> - 2015-12-10 19:55 +0000
csiph-web