On 10/7/2014 1:01 PM, Ned Batchelder wrote:
On 10/7/14 2:10 AM, Gelonida N wrote:
Disadvantage of itertools.product() is, that it makes a copy in memory.
Reason ist, that itertools also makes products of generators (meaning of
objects, that one can't iterate several times through)
There are
On 8/6/2014 1:39 PM, Tim Chase wrote:
On 2014-08-06 11:04, Gayathri J wrote:
Below is the code I tried to check if itertools.product() was
faster than normal nested loops...
they arent! arent they supposed to be...or am i making a mistake?
I believe something like this was discussed a while
On 10/7/14 2:10 AM, Gelonida N wrote:
Disadvantage of itertools.product() is, that it makes a copy in memory.
Reason ist, that itertools also makes products of generators (meaning of
objects, that one can't iterate several times through)
There are two use cases, that I occasionaly stumble
Dear Peter
Below is the code I tried to check if itertools.product() was faster than
normal nested loops...
they arent! arent they supposed to be...or am i making a mistake? any idea?
**
*# -*- coding: utf-8 -*-*
*import numpy as np*
On Wed, Aug 6, 2014 at 3:34 PM, Gayathri J usethisid2...@gmail.com wrote:
Below is the code I tried to check if itertools.product() was faster than
normal nested loops...
they arent! arent they supposed to be...or am i making a mistake? any idea?
Don't worry about what's faster. That almost
On 06/08/2014 06:34, Gayathri J wrote:
Dear Peter
Below is the code I tried to check if itertools.product() was faster
than normal nested loops...
they arent! arent they supposed to be...or am i making a mistake? any idea?
*
*
*
*
*#
Gayathri J wrote:
Dear Peter
Below is the code I tried to check if itertools.product() was faster than
normal nested loops...
they arent! arent they supposed to be...
I wouldn't have expected product() to be significantly faster, but neither
did I expect it to be slower.
or am i
You might check numpy it is really powerful tool for working with multi
dimensional arrays:
ex.
a = arange(81).reshape(3,3,3,3)
a
array( 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8]],
[[ 9, 10, 11],
[12, 13, 14],
[15, 16, 17]],
[[18, 19,
Dear Peter
Yes the f[t] or f[:,:,:] might give a marginal increase, but then i need
to do further operations using the indices, in which case this wouldnt help
Dear Wojciech
np.flat() works if u dont care about the indices and only the matrix/array
values matter.
but if the i,j,k matters,
Gayathri J wrote:
Dear Peter
Yes the f[t] or f[:,:,:] might give a marginal increase,
The speedup compared itertools.product() is significant:
$ python -m timeit -s 'from itertools import product; from numpy.random
import rand; N = 100; a = rand(N, N, N); r = range(N)' 'for x in
On 2014-08-06 11:04, Gayathri J wrote:
Below is the code I tried to check if itertools.product() was
faster than normal nested loops...
they arent! arent they supposed to be...or am i making a mistake?
I believe something like this was discussed a while ago and there was
a faster-but-uglier
Dear Peter
thanks . But thats what I was trying to say
just taking them to zero by f[:,:,:] = 0.0 or using np.zeros is surely
going to give me a time gain...
but my example of using the itertools.product() and doing f[x] =0.0 is just
to compare the looping timing with the traditional nested
I need to evaluate a complicated function over a multidimensional space
as part of an optimization problem. This is a somewhat general problem
in which the number of dimensions and the function being evaluated can
vary from problem to problem.
I've got a working version (with loads of
Frank Miles wrote:
I need to evaluate a complicated function over a multidimensional space
as part of an optimization problem. This is a somewhat general problem
in which the number of dimensions and the function being evaluated can
vary from problem to problem.
I've got a working version
On Tue, 05 Aug 2014 20:06:05 +, Frank Miles wrote:
I need to evaluate a complicated function over a multidimensional space
as part of an optimization problem. This is a somewhat general problem
in which the number of dimensions and the function being evaluated can
vary from problem to
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