Hi Folks Is it possible some example how deal with strides with combinations, let see:
from numpy import * import itertools dt = dtype('i,i,i') a = fromiter(itertools.combinations(range(10),3), dtype=dt, count=-1) a array([(0, 1, 2), (0, 1, 3), (0, 1, 4), (0, 1, 5), (0, 1, 6), (0, 1, 7), (0, 1, 8), (0, 1, 9), (0, 2, 3), (0, 2, 4), (0, 2, 5), (0, 2, 6), (0, 2, 7), (0, 2, 8), (0, 2, 9), (0, 3, 4), (0, 3, 5), (0, 3, 6), (0, 3, 7), (0, 3, 8), (0, 3, 9), (0, 4, 5), (0, 4, 6), (0, 4, 7), (0, 4, 8), (0, 4, 9), (0, 5, 6), (0, 5, 7), (0, 5, 8), (0, 5, 9), (0, 6, 7), (0, 6, 8), (0, 6, 9), (0, 7, 8), (0, 7, 9), (0, 8, 9), (1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 2, 7), (1, 2, 8), (1, 2, 9), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 3, 7), (1, 3, 8), (1, 3, 9), (1, 4, 5), (1, 4, 6), (1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 5, 6), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 7, 8), (1, 7, 9), (1, 8, 9), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 3, 7), (2, 3, 8), (2, 3, 9), (2, 4, 5), (2, 4, 6), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 5, 6), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 6, 7), (2, 6, 8), (2, 6, 9), (2, 7, 8), (2, 7, 9), (2, 8, 9), (3, 4, 5), (3, 4, 6), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 5, 6), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 6, 7), (3, 6, 8), (3, 6, 9), (3, 7, 8), (3, 7, 9), (3, 8, 9), (4, 5, 6), (4, 5, 7), (4, 5, 8), (4, 5, 9), (4, 6, 7), (4, 6, 8), (4, 6, 9), (4, 7, 8), (4, 7, 9), (4, 8, 9), (5, 6, 7), (5, 6, 8), (5, 6, 9), (5, 7, 8), (5, 7, 9), (5, 8, 9), (6, 7, 8), (6, 7, 9), (6, 8, 9), (7, 8, 9)], dtype=[('f0', '<i4'), ('f1', '<i4'), ('f2', '<i4')])
Many thanks Mr. Warren about this ((itertools.combinations(range(10),3), dtype=dt, count=-1)) But as you can see itertools.combinations are emitted in lexicographic sort order but NOT with "power of" strides. So what I see is every element in this array into one memory spot but I would like to know if is possible, use "the power of strides"!
x = a.reshape(120,1) x = stride_tricks.as_strided(a,shape=(120,),strides=(4,4))
Should I use some sub-class like record array, scalar array? So what I want is repetitive elements on same memory spot. I want save memory in big arrays (main reason) and want go fast. How can I deal with this in random arrays but with repetitive elements? Is it possible have custom strides in subclass(that change in dimension) ? How do this? Best Regards Mario
On Mon, Dec 6, 2010 at 20:18, Mario Moura <moura.mario@gmail.com> wrote:
Hi Folks
Is it possible some example how deal with strides with combinations, let see:
No, sorry. It is not possible to generate combinations just using strides. -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco
On Monday, December 6, 2010, Robert Kern <robert.kern@gmail.com> wrote:
On Mon, Dec 6, 2010 at 20:18, Mario Moura <moura.mario@gmail.com> wrote:
Hi Folks
Is it possible some example how deal with strides with combinations, let see:
No, sorry. It is not possible to generate combinations just using strides.
-- Robert Kern
"I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Just wondering, would using ogrid[] in numpy help the OP? Ben
On Mon, Dec 6, 2010 at 21:51, Benjamin Root <ben.root@ou.edu> wrote:
On Monday, December 6, 2010, Robert Kern <robert.kern@gmail.com> wrote:
On Mon, Dec 6, 2010 at 20:18, Mario Moura <moura.mario@gmail.com> wrote:
Hi Folks
Is it possible some example how deal with strides with combinations, let see:
No, sorry. It is not possible to generate combinations just using strides.
Just wondering, would using ogrid[] in numpy help the OP?
No. The limitations of the stride mechanisms apply with ogrid just the same. Generating combinations is not regular enough. -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco
participants (3)
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Benjamin Root
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Mario Moura
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Robert Kern