[Numpy-discussion] trivial question: how to compare dtype - but ignoring byteorder ?

Sebastian Haase haase at msg.ucsf.edu
Mon Aug 14 13:02:53 CDT 2006


On Monday 24 July 2006 12:36, Travis Oliphant wrote:
> Sebastian Haase wrote:
> > Hi,
> > if I have a numpy array 'a'
> > and say:
> > a.dtype == numpy.float32
> >
> > Is the result independent of a's byteorder ?
> > (That's what I would expect ! Just checking !)
>
> I think I misread the question and saw "==" as "="
>
> But, the answer I gave should still help:  the byteorder is a property
> of the data-type.  There is no such thing as "a's" byteorder.   Thus,
> numpy.float32 (which is actually an array-scalar and not a true
> data-type) is interepreted  as a machine-byte-order IEEE floating-point
> data-type with 32 bits.   Thus, the result will depend on whether or not
> a.dtype is machine-order or not.
>
> -Travis
Hi,
I just realized that this question did actually not get sorted out.
Now I'm just about to convert my code to compare
arr.dtype.type to the (default scalar!) dtype numpy.uint8
like this:
if self.img.dtype.type == N.uint8:
  self.hist_min, self.hist_max = 0, 1<<8
elif self.img.dtype.type ==  N.uint16:
  self.hist_min, self.hist_max = 0, 1<<16
...

This seems to work independent of byteorder - (but looks ugly(er)) ...

Is this the best way of doing this ?

- Sebastian Haase




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