what is the most efficent Python script to the below question?

Question

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What is your solution to the FizzBuzz problem?

Write a program that prints the numbers from 1 to 100. But for multiples of three print "Fizz" instead of the number and for the multiples of five print "Buzz". For numbers which are multiples of both three and five print "FizzBuzz".

Here is what I have right now:

#!/usr/bin/python
for i in range(101):
    if i % 3 == 0:
        i = 'fizz'
        print i
    elif i % 5 == 0:
        i = 'buzz'
        print i
    elif i % 3 == 0 and i % 5 == 0:
        i = 'fizz buzz'
        print i
    else:
        print i

Answer 1

#!/usr/bin/python 
for i in range(101): 
 if i % 15 == 0:
  print 'FizzBuzz'
 elif i % 3 == 0: 
  print 'Fizz'
 elif i % 5 == 0: 
  print 'Buzz' 
 else:
  print i

Please, send the credits to my univerzity, I need them.

Answer 2

Well it's working now, i swapped two lines (the 'if i % 3 == 0 & i % 5 == 0' line and one 'elif' line.)

Working script is:

#!/usr/bin/python

for i in range(1,101):
 if i % 3 and i % 5 == 0:
  i = 'fizz'
  print i
 elif i % 5 == 0:
  i = 'buzz'
  print i
 elif i % 3 == 0:
  i = 'fizz buzz'
  print i
 else:
  print i

Answer 3

The following is probably more efficient:

print "1"
print "2"
print "Fizz"
print "4"
print "Buzz"
...
print "98"
print "Fizz"
print "Buzz"

Answer 4

#!/usr/bin/python

for i in range(1,101):
  out = ''
  if i % 3 == 0: out = 'Fizz '
  if i % 5 == 0: out = out + 'Buzz'
  print (out or i)

This avoids a third check for multiples of 15

Answer 5

for i in xrange(1,101):
    if not i % 15:
        print 'FizzBuzz'
    elif not i % 5:
        print 'Buzz'
    elif not i % 3:
        print 'Fizz'
    else:
        print i

It's silly to assign a new value to i if all you're doing is printing it.

If you want to be clever:

for i in xrange(1, 101):
   print 'FuzzBuzz' if not i % 15 else \
         'Buzz' if not i % 5 else \
         'Fuzz' if not i % 3 else \
         i

---

After calling timeit.timeit with 100,000 repetitions, on each method from 1-100, and 1-1000, there doesn't seem to be a whole lot of difference in the two methods:

Clever to 100:  9.38496558538
Clever to 1000:  79.0384339106
Straight to 100:  9.45068505041
Straight to 1000:  78.4853689489

Answer 6

for i in range(1, 101):
  s = ''
  if not i % 3:
    s += 'Fizz'
  if not i % 5:
    s += 'Buzz
  print(s or i)