2009/7/29 Zachary Pincus <[email protected]>:
>> Does numpy have functions to convert between e.g. an array of uint32
>> and
>> uint8, where the uint32 array is a packed version of the uint8 array
>> (selecting little/big endian)?
>
>
> You could use the ndarray constructor to look at the memory differently:
>
> In : a = numpy.arange(240, 260, dtype=numpy.uint32)
> In : a
> Out:
> array([240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252,
> 253, 254, 255, 256, 257, 258, 259], dtype=uint32)
>
> In : b = numpy.ndarray(shape=(len(a)*4,), dtype=numpy.uint8, buffer=a)
>
> In : b
> Out:
> array([240, 0, 0, 0, 241, 0, 0, 0, 242, 0, 0, 0, 243,
> 0, 0, 0, 244, 0, 0, 0, 245, 0, 0, 0, 246, 0,
> 0, 0, 247, 0, 0, 0, 248, 0, 0, 0, 249, 0, 0,
> 0, 250, 0, 0, 0, 251, 0, 0, 0, 252, 0, 0, 0,
> 253, 0, 0, 0, 254, 0, 0, 0, 255, 0, 0, 0, 0,
> 1, 0, 0, 1, 1, 0, 0, 2, 1, 0, 0, 3, 1,
> 0, 0], dtype=uint8)
>
>
> I assume for selecting little/big-endian going the other way, you
> could use the other methods of specifying dtypes that allow for byte-
> order descriptors. (Like dtype objects or the format strings.)
Something like:
In [17]: np.dtype(np.int32).newbyteorder('>')
Out[17]: dtype('>i4')
In [18]: dt = np.dtype(np.int32).newbyteorder('>')
In [19]: x = np.array([123, 456, 789], dtype=dt)
In [20]: x.view(np.uint8)
Out[20]: array([ 0, 0, 0, 123, 0, 0, 1, 200, 0, 0, 3,
21], dtype=uint8)
Regards
Stéfan
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