On Mar 1, 4:58 pm, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]>
wrote:
> 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
unpack with 'B' format gives you int value equivalent to unsigned char
(1 byte).
> (will this work on any system?)
Any system with 8-bit bytes, which would mean any system made after
1965. I'm not aware of any Python implementation for UNIVAC, so I
wouldn't worry ;-)
> How does count1 work exactly?
1,2,4,8,16,32,64,128 in binary are
1,10,100,1000,10000,100000,1000000,10000000
x&1 == 1 if x has first bit set to 1
x&2 == 2, so (x&2>0) == True if x has second bit set to 1
... and so on.
In the context of int, True is interpreted as 1, False as 0.
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