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

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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 !)

Thanks,
Sebastian Haase

On Monday 24 July 2006 06:47, 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 !)
>
> Thanks,
> Sebastian Haase

The condition will always be False, because you're comparing wrong things 
here, numpy.float32 is a scalar type, not a dtype.

>>> numpy.float32
<type 'float32scalar'>
>>> type(numpy.dtype('>f4'))
<type 'numpy.dtype'>

And I think byteorder matters when comparing dtypes:
>>> numpy.dtype('>f4') == numpy.dtype('<f4')
False

Cheers,
Karol

-- 
written by Karol Langner
pon lip 24 12:05:54 CEST 2006

On Monday 24 July 2006 03:18, Karol Langner wrote:
> On Monday 24 July 2006 06:47, 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 !)
> >
> > Thanks,
> > Sebastian Haase
>
> The condition will always be False, because you're comparing wrong things
> here, numpy.float32 is a scalar type, not a dtype.

Hi !
 Thanks for the reply.
Did you actually run this ? I get:
#>>> a=N.arange(10, dtype=N.float32)
#>>> a.dtype == N.float32
#True
#>>> N.__version__
#'0.9.9.2823'

>
> >>> numpy.float32
>
> <type 'float32scalar'>
>
> >>> type(numpy.dtype('>f4'))
>
> <type 'numpy.dtype'>
>
> And I think byteorder matters when comparing dtypes:
> >>> numpy.dtype('>f4') == numpy.dtype('<f4')
> False

OK - I did a test now:
#>>> b= a.copy()
#>>> b=b.newbyteorder('big')
#>>> a.dtype == b.dtype
#False
#>>> a.dtype
#'<f4'
#>>> b.dtype
#'>f4'

How can I do a comparison showing that both a and b are float32 ??

Thanks,
Sebastian Haase

On Monday 24 July 2006 20:10, Sebastian Haase wrote:
> Hi !
>  Thanks for the reply.
> Did you actually run this ? I get:
> #>>> a=N.arange(10, dtype=N.float32)
> #>>> a.dtype == N.float32
> #True
> #>>> N.__version__
> #'0.9.9.2823'

Hi,

Looks like I need to upgrade my working version or something...

>>> import numpy as N
>>> N.__version__
'0.9.8'
>>> a=N.arange(10, dtype=N.float32)
>>> a.dtype == N.float32
False

Cheers,
Karol

-- 
written by Karol Langner
pon lip 24 20:36:05 CEST 2006

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

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

> 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
> ...
>
>   
Yes, you can do this and it should work independent of byteorder.  The 
dtype comparison will take into account the byte-order but comparing the 
type objects directly  won't.  So, if that is your intent, then great.

-Travis

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 ?
>   
The byteorder is a property of the data-type (not of the array) --- this 
is different from numarray where byteorder is a property of the array.

a.dtype == numpy.float32 will always set the data-type to a 
machine-native float32 data-type.

-Travis

> > And I think byteorder matters when comparing dtypes:
> > >>> numpy.dtype('>f4') == numpy.dtype('<f4')
> > False

Ohhhhh -- that '<' part is indicating *byte order* ?!
I thought it was odd that numpy could only tell me the type was "less
than f4", which I assumed must be shorthand for "less than or equal to
f4".  Makes much more sense now!

--bb

On Monday 24 July 2006 16:42, Bill Baxter wrote:
> > > And I think byteorder matters when comparing dtypes:
> > > >>> numpy.dtype('>f4') == numpy.dtype('<f4')
> > >
> > > False
>
> Ohhhhh -- that '<' part is indicating *byte order* ?!
> I thought it was odd that numpy could only tell me the type was "less
> than f4", which I assumed must be shorthand for "less than or equal to
> f4".  Makes much more sense now!
>
> --bb
Which is why I was trying to change the str() representation of a type to 
something more intuitive. 
If nothing else one could even leave 
repr(a.dtype)  --> '<i4'
but 
str(a.dtype)    --> 'int32 (little endian)'

I do now understand that (as opposed to numarray and numeric) the byteorder is 
now part of the data-type  - but I would really encourage keeping the string 
for such an important (and often used !) thing more readable than
"<i4".   
Most people will thankfully never have to think about byteorder - it should be 
like an implementation detail that numpy can transparently handle ! 

What what it's worth ,
Sebastian Haase

On Tuesday 25 July 2006 01:42, Bill Baxter wrote:
> > > And I think byteorder matters when comparing dtypes:
> > > >>> numpy.dtype('>f4') == numpy.dtype('<f4')
> > >
> > > False
>
> Ohhhhh -- that '<' part is indicating *byte order* ?!
> I thought it was odd that numpy could only tell me the type was "less
> than f4", which I assumed must be shorthand for "less than or equal to
> f4".  Makes much more sense now!
>
> --bb

Yep! And there are then four possiblities.

 '>' - big-endian
'<' - little
'|' - not-applicable
'=' - native

Karol

-- 
written by Karol Langner
wto lip 25 08:54:51 CEST 2006

Sebastian Haase wrote:
> Which is why I was trying to change the str() representation of a type to 
> something more intuitive. 
> If nothing else one could even leave 
> repr(a.dtype)  --> '<i4'
> but 
> str(a.dtype)    --> 'int32 (little endian)'

+1

that's the whole point of __str__. It should be human readable!

-Chris

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

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