From: Sebastien B. <Seb...@ob...> - 2006-10-20 13:25:59
|
> 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 |