Thanks for the reply. Rethinking the question ... wasn't there an attribute named something like: is_native() ??
The a.dtype.str[0] is certainly better than nothing ... just doesn't look very good ;-) -Sebastian On 7/3/07, Francesc Altet <[EMAIL PROTECTED]> wrote: > El dt 03 de 07 del 2007 a les 08:35 +0200, en/na Sebastian Haase va > escriure: > > any comments !? > > > > On 6/25/07, Sebastian Haase <[EMAIL PROTECTED]> wrote: > > > Hi, > > > Suppose I'm on a little-edian system. > > > Could I have a little-endian numpy array arr, where > > > arr.dtype.byteorder > > > would actually be "<" > > > instead of "=" !? > > You can always use arr.dtype.str[0], which I think it always returns a > '<', '>' or '|': > > In [2]:a=numpy.array([1]) > In [3]:a.dtype.byteorder > Out[3]:'=' > In [4]:a.dtype.str > Out[4]:'<i4' > In [5]:a.dtype.str[0] > Out[5]:'<' > > > > There are two kinds of systems: little edian and big endian. > > > But there are three possible byteorder values: "<", ">" and "=" > > > > > > I assume that if arr.dtype.byteorder is "=" > > > then, even on a little endian system > > > the comparison arr.dtype.byteorder == "<" still fails !? > > > Or are the == and != operators overloaded !? > > No, this will fail. The == and != are not overloaded because > dtype.byteorder is a pure python string: > > In [11]:type(a.dtype.byteorder) > Out[11]:<type 'str'> _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
