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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Using Tomcat but need to do more? Need to support web services, security?
Get stuff done quickly with pre-integrated technology to make your job easier
Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo
http://sel.as-us.falkag.net/sel?cmd=lnk&kid=120709&bid=263057&dat=121642
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