Numeric/Numarray equivalent to zip ?

G

George Sakkis

What's the fastest and most elegant equivalent of zip() in
Numeric/Numarray between two equally sized 1D arrays ? That is, how to
combine two (N,)-shaped arrays to one (N,2) (or (2,N)) shaped ? I
expect the fastest and the most elegant idiom to be identical, as it is
usually the case in this excellent library, but if not, both would be
useful to know. Thanks,

George
 
R

Robert Kern

George said:
What's the fastest and most elegant equivalent of zip() in
Numeric/Numarray between two equally sized 1D arrays ? That is, how to
combine two (N,)-shaped arrays to one (N,2) (or (2,N)) shaped ? I
expect the fastest and the most elegant idiom to be identical, as it is
usually the case in this excellent library, but if not, both would be
useful to know. Thanks,

Look at combining concatenate(), reshape(), and transpose(). In Scipy, I
would use hstack() and vstack().

--
Robert Kern
(e-mail address removed)

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

Peter Otten

George said:
What's the fastest and most elegant equivalent of zip() in
Numeric/Numarray between two equally sized 1D arrays ? That is, how to
combine two (N,)-shaped arrays to one (N,2) (or (2,N)) shaped ? I
expect the fastest and the most elegant idiom to be identical, as it is
usually the case in this excellent library, but if not, both would be
useful to know. Thanks,
import Numeric as nu
a = nu.array(range(3))
nu.array([a, a])
array([[0, 1, 2],
[0, 1, 2]])
array([[0, 0],
[1, 1],
[2, 2]])

Or am I missing something? As to speed, it seems to be the fastest to
write...

Peter
 
G

George Sakkis

Peter Otten said:
George said:
What's the fastest and most elegant equivalent of zip() in
Numeric/Numarray between two equally sized 1D arrays ? That is, how to
combine two (N,)-shaped arrays to one (N,2) (or (2,N)) shaped ? I
expect the fastest and the most elegant idiom to be identical, as it is
usually the case in this excellent library, but if not, both would be
useful to know. Thanks,
import Numeric as nu
a = nu.array(range(3))
nu.array([a, a])
array([[0, 1, 2],
[0, 1, 2]])
nu.transpose(nu.array([a, a]))
array([[0, 0],
[1, 1],
[2, 2]])

Or am I missing something? As to speed, it seems to be the fastest to
write...

Though not the fastest to execute; using concatenate instead of
initializing an array from a list [a,a] is more than 2,5 time faster in
my system (~4.6 vs 11.8 usec per loop according to timeit.py), and it's
not harder either. One difference is that the equivalent expression for
concatenate expects arrays of shape (1,len(a)) instead of 1D arrays os
shape (len(a),):
a = reshape(range(5), (1,5))
a array([ [0, 1, 2, 3, 4]])
concatenate((a,a))
array([[0, 1, 2, 3, 4],
[0, 1, 2, 3, 4]])


George
 
P

Peter Otten

George said:
Though not the fastest to execute; using concatenate instead of
initializing an array from a list [a,a] is more than 2,5 time faster in
my system (~4.6 vs 11.8 usec per loop according to timeit.py), and it's
not harder either.

That surprises me. I would expect essentially the same amount of
data-shuffling.
One difference is that the equivalent expression for
concatenate expects arrays of shape (1,len(a)) instead of 1D arrays os
shape (len(a),):

If you want to start out with 1D arrays, just reorder the operations:
array([[0, 1, 2, 3, 4],
[0, 1, 2, 3, 4]])
Peter
 
G

George Sakkis

Peter Otten said:
George said:
Though not the fastest to execute; using concatenate instead of
initializing an array from a list [a,a] is more than 2,5 time faster in
my system (~4.6 vs 11.8 usec per loop according to timeit.py), and it's
not harder either.

That surprises me. I would expect essentially the same amount of
data-shuffling.

Here are some timing comparisons of four versions I tried. The first
three work on 1D arrays directly and the fourth on 2D row arrays (i.e.
shape (1,len(a))):

from Numeric import *

# 11.5 usec/loop
def ziparrays_1(*arrays):
return array(arrays)

# 8.1 usec/loop
def ziparrays_2(*arrays):
a = zeros((len(arrays),len(arrays[0])))
for i in xrange(len(arrays)):
a = arrays
return a

# 13.6 usec/loop
def ziparrays_3(*arrays):
return reshape(concatenate(arrays), (len(arrays),len(arrays[0])))

# 4.6 usec/loop
def ziparrays_4(*arrays):
return concatenate(arrays)


So if one has the choice, it's better to start with 2D arrays instead
of 1D. Comparing versions 3 and 4, it's surprising that reshape takes
twice as much as concatenate.

George
 

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