numarray :: multiplying all the elements in 1d array

[Suresh Jeevanandam]

Hi all,
Lets say I have an array:
from numarray import *
a = array([ 6, 7, 8, 9, 10, 11, 12])

I want to multiply out all the elements and get the result.

r = 1.0
for i in a:
r = r*i

Is there any faster, efficient way of doing this.

Thanks,

regards,
Suresh

[Mandus]

Tue, 20 Dec 2005 19:32:13 +0530 skrev Suresh Jeevanandam:

Hi all,
Lets say I have an array:
from numarray import *
a = array([ 6, 7, 8, 9, 10, 11, 12])

I want to multiply out all the elements and get the result.

r = 1.0
for i in a:
r = r*i

Is there any faster, efficient way of doing this.

You can use multiply.reduce (a) (multiply is a function imported from
numarray).

With regular python you can also do:

from operator import mul
reduce(mul,a)

This work even when 'a' is a plain python list.
--
Mandus - the only mandus around.

[Robert Kern]

Suresh Jeevanandam wrote:

Hi all,
Lets say I have an array:
from numarray import *
a = array([ 6, 7, 8, 9, 10, 11, 12])

I want to multiply out all the elements and get the result.

r = 1.0
for i in a:
r = r*i

Is there any faster, efficient way of doing this.

r = multiply.reduce (a)

--
Robert Kern
ro*********@gma il.com

"In the fields of hell where the grass grows high
Are the graves of dreams allowed to die."
-- Richard Harter

[Tim Hochberg]

Mandus wrote:

Tue, 20 Dec 2005 19:32:13 +0530 skrev Suresh Jeevanandam:

Hi all,
Lets say I have an array:
from numarray import *
a = array([ 6, 7, 8, 9, 10, 11, 12])

I want to multiply out all the elements and get the result.

r = 1.0
for i in a:
r = r*i

Is there any faster, efficient way of doing this.

You can use multiply.reduce (a) (multiply is a function imported from
numarray).

With regular python you can also do:

from operator import mul
reduce(mul,a)

This work even when 'a' is a plain python list.

Actually, they'll both work with a plain python list. Multiply.reduce is
much faster (see below) when operating on arrays, but somewhat slower
when operating on lists. I'd guess from the overhead of converting the
list to an array.

-tim

a = na.arange(10000 0)
l = range(100000)
t0 = time.clock(); na.multiply.red uce(a); print time.clock() - t0 0
0.0038896004939 2 t0 = time.clock(); na.multiply.red uce(l); print time.clock() - t0 0
0.042208716471 t0 = time.clock(); reduce(mul, a); print time.clock() - t0 0
0.0943455103931 t0 = time.clock(); reduce(mul, l); print time.clock() - t0

0
0.0146099574108