On Sep 26, 2011 1:49 AM, wrote:
> Is there an equivalent command in python that would immediately provide
the number of set bits in a large bit vector/string
Besides what others have said, if you expect the number of bits set to be
small, you might just use a set.
bits = set()
# set bit 1000
bit
On Sep 26, 12:56 am, Nizamov Shawkat
wrote:
> > Is there an equivalent command in python that would immediately provide the
> > number of set bits in a large bit vector/string
>
> You might be able to achieve this using numpy boolean array and, e.g,
> the arithmetic sum function or something simil
b = numpy.zeros(10**7, dtype=bool)
for x in 3, 4, 6: b[10**x] = True
> ...
b.sum()
> 3
Without numpy:
>>> counts = [bin(i).count('1') for i in range(256)]
>>> bytes = b"hello python"*10
>>> len(bytes)*8
960
>>> sum(map(counts.__getitem__, bytes))
480
Pretty fast as wel
bmacin...@comcast.net wrote:
> In Perl I can create a large bit vector as follows:
> vec($bitmap,1000,1) = 0;# this will create a bit string of all
> zeros
> To set bits I may using commands like:
> vec($bitmap,1000, 1) = 1 # turn on bit 1000
> vec($bitmap,1, 1) =
> Is there an equivalent command in python that would immediately provide the
> number of set bits in a large bit vector/string
>
You might be able to achieve this using numpy boolean array and, e.g,
the arithmetic sum function or something similar.
There is also another library http://pypi.pytho
In Perl I can create a large bit vector as follows:
vec($bitmap,1000,1) = 0;# this will create a bit string of all zeros
To set bits I may using commands like:
vec($bitmap,1000, 1) = 1 # turn on bit 1000
vec($bitmap,1, 1) = 1# turn on bit 1
