On 10/20/06, Travis Oliphant <[EMAIL PROTECTED]> wrote:
>
> How about this. To get the i,j,k,l element
>
> a[i:i+1,j:j+1,k:k+1,l:l+1].squeeze()
>
> -Travis
I think all this can be condensed in a method call or similar
mehcanism, natively provided by ndarray type. Or should this be seen
as a speci
Sebastien Bardeau wrote:
>>One possible solution (there can be more) is using ndarray:
>>
>>In [47]: a=numpy.array([1,2,3], dtype="i4")
>>In [48]: n=1# the position that you want to share
>>In [49]: b=numpy.ndarray(buffer=a[n:n+1], shape=(), dtype="i4")
>>
>>
>>
>Ok thanks. Actually that
Sebastien Bardeau wrote:
> Ooops sorry there was two mistakes with the 'hasslice' flag. This seems
> now to work for me.
>
>
[SNIP code]
That looks overly complicated. I believe that this (minimally tested in
a slightly different setting) or some variation should work:
return self[...,newaxi
Ooops sorry there was two mistakes with the 'hasslice' flag. This seems
now to work for me.
def __getitem__(self,index): # Index may be either an int or a tuple
# Index length:
if type(index) == int: # A single element through first dimension
ilen = 1
index = (ind
Francesc Altet wrote:
> A Divendres 20 Octubre 2006 11:42, Sebastien Bardeau va escriure:
> [snip]
>
>> I can understand that numpy.scalars do not provide inplace operations
>> (like Python standard scalars, they are immutable), so I'd like to use
>>
>> 0-d Numpy.ndarrays. But:
>> >>> d = numpy
> One possible solution (there can be more) is using ndarray:
>
> In [47]: a=numpy.array([1,2,3], dtype="i4")
> In [48]: n=1# the position that you want to share
> In [49]: b=numpy.ndarray(buffer=a[n:n+1], shape=(), dtype="i4")
>
Ok thanks. Actually that was also the solution I found. But t
On Fri, Oct 20, 2006 at 11:42:26AM +0200, Sebastien Bardeau wrote:
> >>> a = numpy.array((1,2,3))
> >>> b = a[:2]
Here you index by a slice.
> >>> c = a[2]
Whereas here you index by a scalar.
So you want to do
b = a[[2]]
b += 1
or in the general case
b = a[slice(2,3)]
b += 1
Regards
Stéf
A Divendres 20 Octubre 2006 11:42, Sebastien Bardeau va escriure:
[snip]
> I can understand that numpy.scalars do not provide inplace operations
> (like Python standard scalars, they are immutable), so I'd like to use
>
> 0-d Numpy.ndarrays. But:
> >>> d = numpy.array(a[2],copy=False)
> >>> d +=
Hi!
I am confused with Numpy behavior with its scalar or 0-d arrays objects:
>>> numpy.__version__
'1.0rc2'
>>> a = numpy.array((1,2,3))
>>> b = a[:2]
>>> b += 1
>>> b
array([2, 3])
>>> a
array([2, 3, 3])
>>> type(b)
To this point all is ok for me: subarrays share (by default) memory wit
