Hello. There is a random 1D array m_0 with size 3000, for example:
I need to generate two 1D arrays:
Is there faster way to do it than this one:
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What other ways have you tried?
On Sat, Jan 9, 2021 at 2:15 PM <[hidden email]> wrote: Hello. There is a random 1D array m_0 with size 3000, for example: _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion |
np.meshgrid, indexing, reshape 09.01.2021, 22:30, "Joseph Fox-Rabinovitz" <[hidden email]>:
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What about arange and then an integer divide or mod? Kevin On Sat, Jan 9, 2021, 20:54 <[hidden email]> wrote:
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Actually I would try a broadcast multiply followed by ravel first. Kevin On Sat, Jan 9, 2021, 21:12 Kevin Sheppard <[hidden email]> wrote:
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In reply to this post by klark--kent
On Sat, 2021-01-09 at 23:45 +0300, [hidden email] wrote:
> np.meshgrid, indexing, reshape I would expect e.g. reshape helps if you make use of the fact that this is very clear in 2-D. The other thing is ensuring to use the methods `arr.reshape`, `arr.repeat`, which avoids overheads (some of which may be optimized away in the future though). That said, with this type of code and the small arrays in the example, there are fairly significant overheads. So if you want to get serious input or a good solution, I would suggest to give more context and include how those lines of code are used in your program/function. Chances are that there is a much faster and possibly more elegant way to write it, e.g. by solving many of these at the same time, but nobody will be able to tell unless they know how you need it. Also with more context, the other possibility is that somone may hint at cython or numba/transonic. Cheers, Sebastian > > 09.01.2021, 22:30, "Joseph Fox-Rabinovitz" > <[hidden email]>: > What other ways have you tried? > > On Sat, Jan 9, 2021 at 2:15 PM <[hidden email]> wrote: > > Hello. There is a random 1D array m_0 with size 3000, for example: > > m_0 = np.array([0, 1, 2]) > > I need to generate two 1D arrays: > > m_1 = np.array([0, 1, 2, 0, 1, 2, 0, 1, 2]) > > m_2 = np.array([0, 0, 0, 1, 1, 1, 2, 2, 2]) > > Is there faster way to do it than this one: > > import numpy as np > > import time > > N = 3 > > m_0 = np.arange(N) > > > > t = time.time() > > m_1 = np.tile(m_0, N) > > m_2 = np.repeat(m_0, N) > > t = time.time() - t > > I tried other ways but they are slower or have the same time. Other > > NumPy operations in my code 10-100 times faster. Why the repeating > > an array is so slow? I need 10 times speed up. Thank you for your > > attantion to my problem. > > _______________________________________________ > > NumPy-Discussion mailing list > > [hidden email] > > https://mail.python.org/mailman/listinfo/numpy-discussion > , > _______________________________________________ > NumPy-Discussion mailing list > [hidden email] > https://mail.python.org/mailman/listinfo/numpy-discussion > _______________________________________________ > NumPy-Discussion mailing list > [hidden email] > https://mail.python.org/mailman/listinfo/numpy-discussion _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion signature.asc (849 bytes) Download Attachment |
In reply to this post by klark--kent
I guess the speed up, if any, will be machine dependent, but you can give a try at: import numpy as np On 2021-01-09 20:07, [hidden email] wrote:
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Thank you but the same time: import numpy as np import time N = 3000 m_0 = np.arange(N) a = np.ones(N) #===== first ========= t1 = time.time() m_1 = np.outer(m_0, a).ravel() m_2 = np.outer(a, m_0).ravel() t1 = time.time() - t1 #===== second ======= t2 = time.time() m_3 = np.tile(m_0, N) m_4 = np.repeat(m_0, N) t2 = time.time() - t2 #================== np.sum(m_1 - m_3) np.sum(m_2 - m_4) t1 t2 10.01.2021, 08:37, "V. Armando Sole" <[hidden email]>:
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As Sebastian said, it's going to take more information about context. It's very unlikely that you actually need these particular arrays-- probably, you're interested in some more important result which you're using these arrays as an intermediate step to obtain. Their structure makes that even more likely than usual. Without knowing how they're being used, no one can know whether they should be replaced by a nested numba or cython loop, or a clever use of einsum, or who knows what else. The minimum time code can take is no time. :-) Doug On Sun, 10 Jan 2021 at 02:00, <[hidden email]> wrote:
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