Duncan Booth wrote:
Mark Wooding <[EMAIL PROTECTED]> wrote:
This is still inelegant, though. We can glue the results mod 3 and 5
together using the Chinese Remainder Theorem and working mod 15
instead. For example,
[['Fizz', 'FizzBuzz', False, None, 'Buzz'][(pow(i, 4, 15) + 1)%7] or
str(i) for i in xrange(1, 101)]
The lookup table is a constant. If made a tuple, it will be compiled as
a constant (as least in 2.6, maybe 2.5). In any case, it could (and to
me should) be lifted out of the string comp.
(A less mathematical approach would just use i%15 to index a table. But
that's not interesting. ;-) )
Ooh. Doesn't having 5 elements make you shudder? (Even though you did
change one to avoid a repeated value.) You have 4 options for output, so
for elegance that list should also have 4 elements:
[[str(i), 'FizzBuzz', 'Fizz', 'Buzz'][25/(pow(i, 4, 15) + 1)%4] for i in
xrange(1, 101)]
I feel it is even more elegant with the lookup table in its natural order:
[['Fizz', 'Buzz', 'FizzBuzz', str(i)][62/(pow(i, 4, 15) + 1)%4] for i in
xrange(1, 101)]
These make the lookup table variable, so it has to be recalculated for
each i.
tjr
--
http://mail.python.org/mailman/listinfo/python-list