[Numpy-discussion] rank-0 arrays ideas

[Darren Dale]

On Friday 18 February 2005 06:04 pm, Travis Oliphant wrote:
>  From the current  PEP:
>
>
>     Proposed Solution:
>
>         The solution proposed by this PEP is to fix the places in Python
>         that could use rank-0 arrayobjects to allow for them before raising
>         an exception (this will be in the core after-all).
>
>         A Python scalar will never be returned unless explicitly
>         requested.  I think this is the cleanest, easiest to understand
>         and code for solution.  It may require some explicity conversion
>         to int here and there, but it seems like a case of "explicit is
>         better than implicit"
>
>         Possible compromises:
>             -  a compromise could be made for OBJECT arrays and
>                perhaps LONG arrays if needed.
>
>             - a special flag could be defined which is the default
>               when an array of integers is constructed and which when
>               set rank-0 array returns behave differently.
>
>         It is also proposed that slicing, indexing and len() do
>         not work (i.e. they raise an error) for rank-0 arrays.
>
>
> Comments:

I have one:

The following script measures the time it takes for in-place array operations, 
comparing 0-D arrays with scalars. The results are 

0.81 0D double
0.85 0D int
0.11 Scalar float
0.12 Scalar int

Would Numeric3 address this issue?

---

import Numeric as N
import time


a=N.zeros(100000,'d')

one=N.array(1,'d')
d=time.clock()
for i in N.arange(100000):
    a[i]+=one
d = time.clock()-d
print d,'0D double'

one=N.array(1)
d=time.clock()
for i in N.arange(100000):
    a[i]+=one
d = time.clock()-d
print d,'0D int'

d=time.clock()
for i in N.arange(100000):
    a[i]+=1.
d = time.clock()-d
print d,'Scalar float'

d=time.clock()
for i in N.arange(100000):
    a[i]+=1
d = time.clock()-d
print d,'Scalar int'

-- 

Darren