On Mar 1, 8:53 am, "Bart Ogryczak" <[EMAIL PROTECTED]> wrote:
> On Mar 1, 7:52 am, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]>
> wrote:
>
> > It seems like this would be easy but I'm drawing a blank.
>
> > What I want to do is be able to open any file in binary mode, and read
> > in one byte (8 bits) at a time and then count the number of 1 bits in
> > that byte.
>
> > I got as far as this but it is giving me strings and I'm not sure how
> > to accurately get to the byte/bit level.
>
> > f1=file('somefile','rb')
> > while 1:
> > abyte=f1.read(1)
>
> import struct
> buf = open('somefile','rb').read()
> count1 = lambda x: (x&1)+(x&2>0)+(x&4>0)+(x&8>0)+(x&16>0)+(x&32>0)+
> (x&64>0)+(x&128>0)
> byteOnes = map(count1,struct.unpack('B'*len(buf),buf))
>
> byteOnes[n] is number is number of ones in byte n.
This solution looks nice, but how does it work? I'm guessing
struct.unpack will provide me with 8 bit bytes (will this work on any
system?)
How does count1 work exactly?
Thanks for the help.
-Greg
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