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', '
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 subclass 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
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
On Mon, Dec 6, 2010 at 20:18, Mario Moura
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 _______________________________________________ NumPyDiscussion mailing list NumPyDiscussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpydiscussion
Just wondering, would using ogrid[] in numpy help the OP? Ben
On Mon, Dec 6, 2010 at 21:51, Benjamin Root
On Monday, December 6, 2010, Robert Kern
wrote: On Mon, Dec 6, 2010 at 20:18, Mario Moura
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)

Benjamin Root

Mario Moura

Robert Kern