On Friday 18 August 2006 16:51, Travis Oliphant wrote:
> Sebastian Haase wrote:
> > On Friday 18 August 2006 15:25, Travis Oliphant wrote:
> >> Sebastian Haase wrote:
> >>> On Friday 18 August 2006 11:38, Travis Oliphant wrote:
> >>>> Sebastian Haase wrote:
> >>>>> Hi,
> >>>>> array dtype descriptors have an attribute itemsize  that gives the
> >>>>> total number of bytes required for an item of that dtype.
> >>>>>
> >>>>> Scalar types, like  numy.int32, also have that attribute,
> >>>>> but it returns "something else" - don't know what:
> >>>>>
> >>>>>
> >>>>> Furthermore there are *lot's*  of more attributes to a scalar dtype,
> >>>>> e.g.
> >>>>
> >>>> The scalar types are actual Python types (classes) whereas the dtype
> >>>> objects are instances.
> >>>>
> >>>> The attributes you are seeing of the typeobject are very useful when
> >>>> you have an instance of that type.
> >>>>
> >>>> With numpy.int32.itemsize you are doing the equivalent of
> >>>> numpy.dtype.itemsize
> >>>
> >>> but why then do I not get the result 4 ?
> >>
> >> Because it's not a "class" attribute, it's an instance attribute.
> >>
> >> What does numpy.dtype.itemsize give you?
> >
> > I'm really sorry for being so dumb - but HOW can I get then the number of
> > bytes needed by a given scalar type ?
>
> Ah, the real question.  Sorry for not catching it earlier.  I've been in
> "make sure this isn't a bug mode" for a long time.
>
> If you have a scalar type you could create one and then check the itemsize:
>
> int32(0).itemsize
>
> Or you could look at the name and parse out how big it is.
>
> There is also a stored dictionary-like object that returns the number of
> bytes for any data-type recognized:
>
> numpy.nbytes[int32]

Thanks, that seems to be a handy "dictionary-like object"

Just for the record - in the meantime I found this:
>>> N.dtype(N.int32).itemsize
4

Cheers,
Sebastian

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