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.uint32)
I want to get the index of the elements of 'myData' where the following occurs:
Bit No. 0–1
Bit Combination: 00
How can I do it?
You can use np.binary_repr() to obtain the string representing each element and from this strings obtain the elements matching your criterion:
end = `00`
s = [np.binary_repr(ai, width=len(end)) for ai in myData]
indices = [i for (i, si) in enumerate(s) if si.endswith(end)]
After studying a bit more I found you can use np.unpackbits() after working with a uint8 view of your array:
myData = myData.view(np.uint32) # not needed if it is already np.uint32
tmp = np.unpackbits(myData.view(np.uint8)[::4][None, :], axis=0)
indices = np.where((tmp[-2, :] == 0) & (tmp[-1, :] == 0))[0]
Note that the slices are taken according to the bits you are comparing:
[::4] for the first 8 bits[1::4] for the 9th to te 16th bits[2::4] for the 17th to the 24th bits[3::4] for the 25th to the 32th bitsthis sequence could go on and on if your original data type was np.uint64, for example, but in this case using [n::8] instead.
I'd say the canonical way is this:
np.flatnonzero((a & 0b11) == 0)
I.e. mask out all bits except the last two and get the indices where it's all-zero.
This is short, fast, and doesn't make any assumption about endianness.
01 you'd do np.flatnonzero((a & 0b110000) == 0b010000), see also this question.