On Fri, Dec 3, 2010 at 6:31 AM, Mario Moura <moura.mario@gmail.com> wrote:

Hi Folks

I have this situation

>>> from timeit import Timer
>>> reps = 5
>>>
>>> t = Timer('itertools.combinations(range(1,10),3)', 'import itertools')
>>> print sum(t.repeat(repeat=reps, number=1)) / reps
1.59740447998e-05
>>> t = Timer('itertools.combinations(range(1,100),3)', 'import itertools')
>>> print sum(t.repeat(repeat=reps, number=1)) / reps
1.74999237061e-05
>>>
>>> t = Timer('list(itertools.combinations(range(1,10),3))', 'import itertools')
>>> print sum(t.repeat(repeat=reps, number=1)) / reps
5.31673431396e-05
>>> t = Timer('list(itertools.combinations(range(1,100),3))', 'import itertools')
>>> print sum(t.repeat(repeat=reps, number=1)) / reps
0.0556231498718
>>>

You can see list(itertools.combinations(range(1,100),3)) is terrible!!

If you change to range(1,100000) your computer will lock.

So I would like to know a good way to convert <itertools.combinations
object> to ndarray? fast! without use list
Is it possible?

>>> x = itertools.combinations(range(1,10),3)
>>> x
<itertools.combinations object at 0x25f1520>
>>>

I tried this from
http://docs.python.org/library/itertools.html?highlight=itertools#itertools.combinations

>>> numpy.fromiter(itertools.combinations(range(1,10),3), int, count=-1)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: setting an array element with a sequence.
>>>

and this from
http://docs.python.org/library/itertools.html?highlight=itertools#itertools.combinations

import numpy
from itertools import *
from numpy import *

def combinations(iterable, r):
pool = tuple(iterable)
n = len(pool)
for indices in permutations(range(n), r):
if sorted(indices) == list(indices):
yield tuple(pool[i] for i in indices)

numpy.fromiter(combinations(range(1,10),3), int, count=-1)

>>> numpy.fromiter(combinations(range(1,10),3), int, count=-1)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: setting an array element with a sequence.
>>>

I like itertools.combinations performance but I need convert it to numpy.

The docstring for numpy.fromiter() says it creates a 1D array. You can use it with itertools.combinations if you specify a dtype for a 1D structured array. Here's an example (I'm using ipython with the -pylab option, so the numpy functions have all been imported):

In [1]: from itertools import combinations

In [2]: dt = dtype('i,i,i')

In [3]: a = fromiter(combinations(range(100),3), dtype=dt, count=-1)

In [4]: b = array(list(combinations(range(100),3)))

In [5]: all(a.view(int).reshape(-1,3) == b)
Out[5]: True

In [6]: timeit a = fromiter(combinations(range(100),3), dtype=dt, count=-1)
10 loops, best of 3: 92.7 ms per loop

In [7]: timeit b = array(list(combinations(range(100),3)))
1 loops, best of 3: 627 ms per loop

In [8]: a[:3]
Out[8]:
array([(0, 1, 2), (0, 1, 3), (0, 1, 4)],
      dtype=[('f0', '<i4'), ('f1', '<i4'), ('f2', '<i4')])

In [9]: b[:3]
Out[9]:
array([[0, 1, 2],
       [0, 1, 3],
       [0, 1, 4]])

In the above example, 'a' is a 1D structured array; each element of 'a' holds one of the combinations. If you need it, you can create a 2D view with a.view(int).reshape(-1,3).

Warren