# Numpy-scalars vs Numpy 0-d arrays: copy or not copy?

Sebastien Bardeau Sebastien.Bardeau at obs.u-bordeaux1.fr
Fri Oct 20 08:49:15 CDT 2006

```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 = (index,)    # A tuple
else:
ilen = len(index)
# Array rank:
arank = len(self.shape)
# Check if there is a slice:
hasslice = False
for i in index:
if type(i) == slice:
hasslice = True
# Array is already a 0-d array:
if arank == 0 and index == (0,):
return self
elif arank == 0:
raise IndexError, "0-d array has only one element at index 0."
# This will return a single element as a 0-d array:
elif arank == ilen and not hasslice:
# This ugly thing returns a numpy 0-D array AND NOT a numpy scalar!
# (Numpy scalars do not share their data with the parent array)
newindex = list(index)
newindex[0] = slice(index[0],index[0]+1,None)
newindex = tuple(newindex)
return self[newindex].reshape(())
# This will return a n-D subarray (n>=1):
else:
return self[index]

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 was also the solution I found. But this is much
> more complicated when arrays are N dimensional with N>1, and above all
> if user asks for a slice in one or more dimension. Here is how I
> redefine the __getitem__ method for my arrays. Remember that the goal is
> to return a 0-d array rather than a numpy.scalar when I extract a single
> element out of a N-dimensional (N>=1) array:
>
>    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 = (index,)    # A tuple
>       else:
>          ilen = len(index)
>       # Array rank:
>       arank = len(self.shape)
>       # Check if there is a slice:
>       for i in index:
>          if type(i) == slice:
>             hasslice = True
>          else:
>             hasslice = False
>       # Array is already a 0-d array:
>       if arank == 0 and index == (0,):
>          return self[()]
>       elif arank == 0:
>          raise IndexError, "0-d array has only one element at index 0."
>       # This will return a single element as a 0-d array:
>       elif arank == ilen and hasslice:
>          # This ugly thing returns a numpy 0-D array AND NOT a numpy scalar!
>          # (Numpy scalars do not share their data with the parent array)
>          newindex = list(index)
>          newindex[0] = slice(index[0],index[0]+1,None)
>          newindex = tuple(newindex)
>          return self[newindex].reshape(())
>       # This will return a n-D subarray (n>=1):
>       else:
>          return self[index]
>
>  Well... I do not think this is very nice. Someone has another idea? My
> question in my first post was: is there a way to get a single element of
> an array into
> a 0-d array which shares memory with its parent array?
>
>  Sebastien
>
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Sebastien Bardeau
L3AB - CNRS UMR 5804
2 rue de l'observatoire
BP 89
F - 33270 Floirac
Tel: (+33) 5 57 77 61 46
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