I need to calculate the lrc of a string using an exclusive or on each
byte in the string. How would I do this in python?
Chris
byte in the string. How would I do this in python?
Chris
-
-
Re: XOR on string
snacktime wrote:[color=blue]
> I need to calculate the lrc of a string using an exclusive or on each
> byte in the string. How would I do this in python?[/color]
lrc == Linear Redundancy Check? or Longitudinal? Note that
such terms are not precisely defined... generally just acronyms
people make up and stick in their user manuals for stuff. :-)
import operator
lrc = reduce(operator .xor, [ord(c) for c in string])
Note that this returns an integer, so if you plan
to send this as a byte or compare it to a character
received, use chr(lrc) on it first.
-Peter
-
Re: XOR on string
> lrc == Linear Redundancy Check? or Longitudinal? Note that[color=blue]
> such terms are not precisely defined... generally just acronyms
> people make up and stick in their user manuals for stuff. :-)
>[/color]
Longitudinal
[color=blue]
> import operator
> lrc = reduce(operator .xor, [ord(c) for c in string])[/color]
That's better than what I had, which as it turned out was working I
was just calculating the lrc on one extra digit that I should have
been.
Chris
Comment
-
Re: XOR on string
Peter Hansen <peter@engcorp. com> wrote:[color=blue]
> snacktime wrote:[color=green]
> > I need to calculate the lrc of a string using an exclusive or on each
> > byte in the string. How would I do this in python?[/color]
>
> lrc == Linear Redundancy Check? or Longitudinal? Note that
> such terms are not precisely defined... generally just acronyms
> people make up and stick in their user manuals for stuff. :-)
>
> import operator
> lrc = reduce(operator .xor, [ord(c) for c in string])[/color]
Or for the full functional programming effect...
lrc = reduce(operator .xor, map(ord, string))
which is slightly faster and shorter...
$ python2.4 -m timeit -s'import operator; string = "abcdefghij1312 3kj12l3k1j23lk1 2j3l12kj3"' \
'reduce(operato r.xor, [ord(c) for c in string])'
10000 loops, best of 3: 20.3 usec per loop
$ python2.4 -m timeit -s'import operator; string = "abcdefghij1312 3kj12l3k1j23lk1 2j3l12kj3"' \
'reduce(operato r.xor, map(ord, string))'
100000 loops, best of 3: 15.6 usec per loop
--
Nick Craig-Wood <nick@craig-wood.com> -- http://www.craig-wood.com/nick
Comment
Comment