On 18.10.2014 14:14, Artur Bercik wrote:
>
>
> On Sat, Oct 18, 2014 at 9:00 PM, Artur Bercik <vbubbly21@gmail.com
> <mailto:vbubbly21@gmail.com>> wrote:
>
>
>
> On Sat, Oct 18, 2014 at 8:28 PM, Julian Taylor
> <jtaylor.debian@googlemail.com
> <mailto:jtaylor.debian@googlemail.com>> wrote:
>
> On 18.10.2014 07:58, Artur Bercik wrote:
> > Dear Python and Numpy Users:
> >
> > My data are in the form of '32-bit unsigned integer' as follows:
> >
> > myData = np.array([1073741824, 1073741877, 1073742657, 1073742709,
> > 1073742723, 1073755137, 1073755189,1073755969],dtype=np.int32)
> >
> > I want to get the index of my data where the following occurs:
> >
> > Bit No. 0–1
> > Bit Combination: 00
> >
> > How can I do it? I heard this type of problem first time, please help me.
> >
> > Artur
> >
>
> not sure I understand the problem, maybe this?
>
> np.where((myData & 0x3) == 0)
>
>
> yes, it works greatly for the following case:
>
> myData = np.array([1073741824, 1073741877, 1073742657, 1073742709,
> 1073742723, 1073755137, 1073755189,1073755969],dtype=np.uint32)
> Bit No. 0–1
> Bit Combination: 00
>
> Can you make such automation for the following case as well?
>
> Bit No. 2–5
> Bit Combination: 1101
>
sure, you can do any of these with the right masks:
np.where((myData & 0x3c) == 0x34)
you can use bin(number) to check if your numbers are correct.
>
> Also wondering why np.where((myData & 0x3) == 0) instead of
> just np.where((myData & 3) == 0)
>
its the same, 0x means the number is in hexadecimal representation, for
3 they happen to be equal (as 3 < 10)
It is often easier to work in the hexadecimal representation when
dealing with binary data as its base is a power of two. So two digits in
hexadecimal represent one byte.
In the case above: 0x3c
c is 12 -> 1100
3 is 3 -> 11
together you get 111100, mask for bits 2-5
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