Hey, I just created a pull request: https://github.com/numpy/numpy/pull/5874 Best, Stefan On Tue, May 12, 2015 at 3:29 PM Stefan Otte <stefan.otte@gmail.com> wrote:
Hey,
here is an ipython notebook with benchmarks of all implementations (scroll to the bottom for plots):
https://github.com/sotte/ipynb_snippets/blob/master/2015-05%20gridspace%20-%...
Overall, Jaime's version is the fastest.
On Tue, May 12, 2015 at 2:01 PM Jaime Fernández del Río < jaime.frio@gmail.com> wrote:
On Tue, May 12, 2015 at 1:17 AM, Stefan Otte <stefan.otte@gmail.com> wrote:
Hello,
indeed I was looking for the cartesian product.
I timed the two stackoverflow answers and the winner is not quite as clear:
n_elements: 10 cartesian 0.00427 cartesian2 0.00172 n_elements: 100 cartesian 0.02758 cartesian2 0.01044 n_elements: 1000 cartesian 0.97628 cartesian2 1.12145 n_elements: 5000 cartesian 17.14133 cartesian2 31.12241
(This is for two arrays as parameters: np.linspace(0, 1, n_elements)) cartesian2 seems to be slower for bigger.
On my system, the following variation on Pauli's answer is 2-4x faster than his for your test cases:
def cartesian4(arrays, out=None): arrays = [np.asarray(x).ravel() for x in arrays] dtype = np.result_type(*arrays)
n = np.prod([arr.size for arr in arrays]) if out is None: out = np.empty((len(arrays), n), dtype=dtype) else: out = out.T
for j, arr in enumerate(arrays): n /= arr.size out.shape = (len(arrays), -1, arr.size, n) out[j] = arr[np.newaxis, :, np.newaxis] out.shape = (len(arrays), -1)
return out.T
I'd really appreciate if this was be part of numpy. Should I create a pull request?
There hasn't been any opposition, quite the contrary, so yes, I would go ahead an create that PR. I somehow feel this belongs with the set operations, rather than with the indexing ones. Other thoughts?
Also for consideration: should it work on flattened arrays? or should we give it an axis argument, and then "broadcast on the rest", a la generalized ufunc?
Jaime
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