99 Bottles of Beer/Python
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99 Bottles of Beer done in Python.
Normal Code
def sing(b, end):
print(b or 'No more','bottle'+('s' if b-1 else ''), end)
for i in range(99, 0, -1):
sing(i, 'of beer on the wall,')
sing(i, 'of beer,')
print('Take one down, pass it around,')
sing(i-1, 'of beer on the wall.\n')
Using a template
verse = '''\
%i bottles of beer on the wall
%i bottles of beer
Take one down, pass it around
%i bottles of beer on the wall
'''
for bottles in range(99,0,-1):
print verse % (bottles, bottles, bottles-1)
New-style template (Python 2.6)
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
'''
for bottles in range(99,0,-1):
print verse.format(some=bottles, less=bottles-1)
"Clever" generator expression
a, b, c, s = " bottles of beer", " on the wall\n", "Take one down, pass it around\n", str
print "\n".join(s(x)+a+b+s(x)+a+"\n"+c+s(x-1)+a+b for x in xrange(99, 0, -1))
Enhanced "Clever" generator expression using lambda
a = lambda n: "%u bottle%s of beer on the wall\n" % (n, "s"[n==1:])
print "\n".join(a(x)+a(x)[:-13]+"\nTake one down, pass it around\n"+a(x-1) for x in xrange(99, 0, -1))
Using a generator expression (Python 3)
#!/usr/bin/env python3
"""\
{0} {2} of beer on the wall
{0} {2} of beer
Take one down, pass it around
{1} {3} of beer on the wall
"""
print("\n".join(
__doc__.format(
i, i - 1,
"bottle" if i == 1 else "bottles",
"bottle" if i - 1 == 1 else "bottles"
) for i in range(99, 0, -1)
), end="")
A wordy version
ones = (
'', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine'
)
prefixes = ('thir', 'four', 'fif', 'six', 'seven', 'eigh', 'nine')
tens = ['', '', 'twenty' ]
teens = ['ten', 'eleven', 'twelve']
for prefix in prefixes:
tens.append(prefix + 'ty')
teens.append(prefix +'teen')
tens[4] = 'forty'
def number(num):
"get the wordy version of a number"
ten, one = divmod(num, 10)
if ten == 0 and one == 0:
return 'no'
elif ten == 0:
return ones[one]
elif ten == 1:
return teens[one]
elif one == 0:
return tens[ten]
else:
return "%s-%s" % (tens[ten], ones[one])
def bottles(beer):
"our rephrase"
return "%s bottle%s of beer" % (
number(beer).capitalize(), 's' if beer > 1 else ''
)
onthewall = 'on the wall'
takeonedown = 'Take one down, pass it around'
for beer in range(99, 0, -1):
print bottles(beer), onthewall
print bottles(beer)
print takeonedown
print bottles(beer-1), onthewall
print
String Formatting
for n in xrange(99, 0, -1):
## The formatting performs a conditional check on the variable.
## If it formats the first open for False, and the second for True
print n, 'bottle%s of beer on the the wall.' % ('s', '')[n == 1]
print n, 'bottle%s of beer.' % ('s', '')[n == 1]
print 'Take one down, pass it around.'
print n - 1, 'bottle%s of beer on the wall.\n' % ('s', '')[n - 1 == 1]
Python 3 f-strings and walrus operator
bottles = 100
while (bottles:=bottles-1) != 0:
print(f"{bottles} bottles of beer on the wall\n{bottles} bottles of beer\nTake one down, pass it around\n{bottles-1} bottles of beer on the wall\n")
Function recursion
v=99
def s():
global v
print(f"{v} bottles of beer on the wall,")
print(f"{v} bottles of beer.")
print("Take one down, pass it around,")
v = v - 1
print(f"{v} bottles of beer on the wall.")
print()
if v!=0:
s()
s()