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)] > > (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)] :) -- Duncan Booth http://kupuguy.blogspot.com -- http://mail.python.org/mailman/listinfo/python-list
