[Numpy-discussion] Structured array dtype

Nicolas Rougier Nicolas.Rougier@inria...
Sat Aug 31 02:05:11 CDT 2013



Thanks for the explanation.
In fact, since the 4 components of 'b' are contiguous in memory, I wanted to find a way to express that fact in the dtype.

Z = np.zeros(10, [('a', np.float32, 3), ('b', np.float32, 4)])
Z['b'].strides
(28,4)

Z = np.zeros((10,4), np.float32)
Z.strides
(16,4)

Z = np.zeros(10, (np.float32,4))
Z.strides
(16,4)




Nicolas



On Aug 31, 2013, at 7:51 AM, Stéfan van der Walt <stefan@sun.ac.za> wrote:

> Hi Nicolas
> 
> On Fri, 30 Aug 2013 17:26:51 +0200, Nicolas Rougier wrote:
>>>>> Z = np.zeros(10, [('a', np.float32, 3), ('b', np.float32, 4)])
>> 
>>>>> Z['a'].dtype
>> dtype('float32')
>> 
>>>>> Z.dtype['a']
>> dtype(('<f4', (3,)))
>> 
>> 
>> Does that mean that dtype['a'] is the dtype of field 'a' when in Z, while Z['a'].dtype is the dtype of field 'a' when "extracted" or my way of thinking is totally wrong ?
> 
> Apologies if this is a duplicate response; I'm sending it offline.
> 
> In case 1, you are indexing into the array, and querying its dtype.  In case
> two, you are indexing into a dtype.
> 
> I.e., in case two, you are doing this:
> 
> In [18]: dtype = np.dtype([('a', float, 3), ('b', int)])
> 
> In [19]: dtype['a']
> Out[19]: dtype(('<f8', (3,)))
> 
>> What bothers me the most is that I cannot do:
>> 
>>>>> Z['a'].view(Z.dtype['a'])
>> ValueError: new type not compatible with array.
> 
> That's quite a tricky operation to perform, since it has to take into account
> the underlying strides of the old array as well as calculate a shape for the
> new array.  It should be possible to make it work using something similar to
> `np.lib.stride_tricks.as_strided`, but my quick attempt failed because of the
> following:
> 
> In [13]: class Foo:
>    __array_interface__ = Z.__array_interface__
>   ....:
> 
> In [14]: f = Foo()
> 
> In [15]: np.asarray(f)
> Out[15]:
> array([, ,
>       , ,
>       , ,
>       , ,
>       , ],
>      dtype='|V28')
> 
> This does not seem right.
> 
> Stéfan
> 
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