Python - for loop ignoring if statement

Question

I'm attempting to write a Python program that will print out the lyrics of 99 Bottles of Beer in chunks, which I've gotten to work. However, I need the last line to say "1 bottle of beer" as opposed to "1 bottles of beer".

I wrapped my code in an if statement, but it appears the if statement is getting ignored. What am I doing wrong?

verse = '''
{some} bottles of beer on the wall
{some} bottles of beer
Take one down, pass it around
{less} bottles of beer on the wall
'''

verse2 = '''
{some} bottles of beer on the wall
{some} bottles of beer
Take one down, pass it around
1 bottle of beer on the wall
'''

for bottles in range(99 , 1, -1):
  if bottles >= 2:
    print(verse.format(some = bottles, less = bottles - 1))
  else:
    print(verse2.format(some = bottles))

Philip Tzou · Accepted Answer · 2018-09-21 19:07:04Z

2

Try this:

print(list(range(99 , 1, -1)))

The result should be:

[99, 98, 97, ..., 5, 4, 3, 2]

Number 1 never appeared.

To Fix this issue, simply changing range(99, 1, -1) to range(99, 0, -1).

answered Sep 21, 2018 at 19:07

Philip Tzou

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1 Comment

Nightviz1on Over a year ago

Ahhh, I can't believe I missed that, that's exactly it. Thank you for being my second pair of eyes!

wim · Accepted Answer · 2018-09-21 19:06:48Z

1

Ranges don't include the endpoint, so the final element of your range is not 1 but 2.

Should you change the if-statement to if bottles > 2: instead, the output will be correct.

answered Sep 21, 2018 at 19:06

wim

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1 Comment

Nightviz1on Over a year ago

Sure enough, that also did the trick. Thank you for the assistance!

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