Python Puzzlers

Python Puzzlers

Tendayi Mawushe
PyCon Ireland 2010

Introduction
Eight Python Puzzles
Short Python program with curious behaviour
What is the output? (multiple choice)
The correct answer given
How to fix the problem (if there was one)
The moral of the story
What will be covered
Language and core libraries
Python 2.6 & 3.x (some puzzles apply to 2.6 only)

1. Exceptional Circumstances

try:
raise NameError('some_name')
except TypeError, NameError:
print ('caught exception NameError')
except Exception:
pass

1. What is the output?

try:
except Exception:
pass

(a) caught exception NameError
(b) SyntaxError: invalid syntax
(c) <no output>
(d) caught exception TypeError


(a) caught exception NameError
(b) SyntaxError: invalid syntax
(c) <no output>
(d) caught exception TypeError

1. A closer look

try:
except Exception:
pass

1. How do you fix it?

try:
except (TypeError, NameError):
except Exception:
pass

>>>
caught exception NameError

1. The moral of the story

When catching multiple exceptions in a single
except clause you must surround them in
parentheses
This problem is non-existent problem in Python
3.x because the problematic syntax is not
permitted:
except SomeException, variable # not valid
3.x syntax
except SomeException as variable

2. Final Countdown

seconds = 10
for i in range(10):
--seconds
if seconds:
print('Wait for it.', seconds)
else:
print('Happy New Year!', seconds)


seconds = 10
for i in range(10):
--seconds
if seconds:
else:

(a) ('Wait for it.', 10)
(b) -10
(c) SyntaxError: invalid syntax
(d) ('Happy New Year!', 0)


(a) ('Wait for it.', 10)
(b) -10
(c) SyntaxError: invalid syntax
(d) ('Happy New Year!', 0)

2. A closer look

seconds = 10
for i in range(10):
--seconds
if seconds:
else:


seconds = 10
for i in range(10):
seconds -= 1
if seconds:
else:

>>>
('Happy New Year!', 0)


There is no -- or ++ operator in Python to
achieve that effect use -= 1 and += 1
--seconds is actually the same as -(-seconds)

3. Local News

def news(headline):
sports = 'Soccer'
for story in locals():
print(locals()[story])
news('Politics')


def news(headline):
sports = 'Soccer'
news('Politics')

(a) Politics
Soccer
(b) {'sports': 'Soccer'}
(c) Soccer
(d) RuntimeError: dictionary changed size during
iteration


(a) Politics
Soccer
(b) {'sports': 'Soccer'}
(c) Soccer
(d) RuntimeError: dictionary changed
size during iteration

3. A closer look

def news(headline):
sports = 'Soccer'
news('Politics')


def news(headline):
sports = 'Soccer'
stories = locals()
for story in stories:
print(stories[story])
news('Politics')

>>>
Politics
Soccer


When locals() is invoked it updates and returns
a dictionary representing the current local
symbol table
You should never attempt to update the locals
dictionary, however if you need to access it's
contents in a loop assign it to another name
first

4. TGIF

days = ['Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat',
'Sun']
weekend = enumerate(days)[5:]
for day in weekend:
print(day[0], day[1])


'Sun']
for day in weekend:

(a) (5, 'Sat') (6, 'Sun')
(b) ('Sat', 'Sun')
(c) TypeError: object is unsubscriptable
(d) (5, 6)


(a) (5, 'Sat') (6, 'Sun')
(b) ('Sat', 'Sun')
(c) TypeError: object is unsubscriptable
(d) (5, 6)

4. A closer look

'Sun']
for day in weekend:


'Sun']
weekend = list(enumerate(days))[5:]
for day in weekend:

>>>
(5, 'Sat')
(6, 'Sun')


The enumerate built-in function is a generator,
that is it returns an iterator
Iterators are not sequences therefore they
cannot be indexed or sliced:
If you need to index or slice an iterator
you must first convert it to a list, this
loads the entire dataset into memory
Generators can represent infinite chain of
values for example itertools.count(), these
cannot be meaningfully sliced in reverse

5. Rabbits everywhere
a, b = 0, 1
def fibonacci(n):
for i in range(n):
a, b = b, a + b
return a
fib8 = fibonacci(8)
print(fib8)


a, b = 0, 1
def fibonacci(n):
for i in range(n):
a, b = b, a + b
return a
fib8 = fibonacci(8)
print(fib8)

(a) UnboundLocalError: local variable
(b) 21
(c) 1
(d) 0


(a) UnboundLocalError: local variable
(b) 21
(c) 1
(d) 0

5. A closer look

a, b = 0, 1
def fibonacci(n):
for i in range(n):
a, b = b, a + b
return a
fib8 = fibonacci(8)
print(fib8)


a, b = 0, 1
def fibonacci(n):
global a, b
for i in range(n):
a, b = b, a + b
return a
fib8 = fibonacci(8)
print(fib8)

>>>
21


The issue is local variable optimisation.
If a variable is assigned in a function it is a local
variable, the bytecode generated to access it is
different to that for global variables.
A variable in a function can either be local or
global, but not both.
Do not mix up global and local names in this way, it
is confusing and problematic.

6. The Whole Truth

w = False
h = []
o = 0,
l = None
e = {}
print(any((w, h, o, l, e)))


w = False
h = []
o = 0,
l = None
e = {}

(a) True
(b) (w, h, o, l, e)
(c) (False, [], 0, None, {})
(d) False


(a) True
(b) (w, h, o, l, e)
(c) (False, [], 0, None, {})
(d) False

6. A closer look

w = False
h = []
O = 0,
l = None
e = {}


w = False
h = []
O = 0
l = None
e = {}

>>> False


The comma is the tuple constructor, not the
parentheses
Though it is not required it is generally
considered good style to use parentheses when
creating a tuple:
(0,) is better than 0,

7. Double or Nothing

def double(items, doubles=[]):
for item in items:
doubles.append(item * 2)
return doubles
numbers = double([1, 2, 3])
words = double(['one', 'two', 'three'])
print(words)


for item in items:
return doubles
print(words)

(a) [2, 4, 6, 'oneone', 'twotwo', 'threethree']
(b) ['oneone', 'twotwo', 'threethree']
(c) TypeError: unsupported operand type(s) for *
(d) [2, 4, 6]


(a) [2, 4, 6, 'oneone', 'twotwo', 'threethree']
(b) ['oneone', 'twotwo', 'threethree']
(c) TypeError: unsupported operand type(s) for *
(d) [2, 4, 6]

7. A closer look

for item in items:
return doubles
print(words)


def double(items, doubles=None):
if doubles is None: doubles = []
for item in items:
return doubles
print(words)

>>> ['oneone', 'twotwo', 'threethree']


Do not use mutable types as default arguments
Default arguments are evaluated when the function
is defined not when the function is called
If you want to use a mutable type as a default
argument, set the default to None and initialise it
properly inside the function

8. Evening Out the Odds

nums = [01, 02, 03, 04, 05, 06, 07, 08, 09, 10]
evens = []
for num in nums:
if num % 2 != 0: # is the number odd
evens.append(num + 1)
print(evens)


nums = [01, 02, 03, 04, 05, 06, 07, 08, 09, 10]
evens = []
for num in nums:
print(evens)

(a) [2, 4, 6, 8, 10]
(b) SyntaxError: invalid token
(c) [02, 04, 06, 08, 10]
(d) [2, 2, 4, 4, 6, 6, 8, 8, 10, 10]


(a) [2, 4, 6, 8, 10]
(b) SyntaxError: invalid token
(c) [02, 04, 06, 08, 10]
(d) [1, 2, 3, 4, 5]

8. A closer look

nums = [01, 02, 03, 04, 05, 06, 07, 08, 09, 10]
evens = []
for num in nums:
print(evens)


nums = [01, 02, 03, 04, 05, 06, 07, 010, 011, 012]
evens = []
for num in nums:
evens.append(oct(num + 1))
print(evens)

>>> ['02', '04', '06', '010', '012']


In Python 2.x a leading 0 specifies an octal
literal
If you want to work with octal numbers
remember the valid digits are 0 though 7
In Python 3.x octal literals are specified using
0o, which removes the ambiguity
01 # not valid 3.x syntax
0o1

Links
Slides:
insmallportions.com

Q & A:
stackoverflow.com
Inspiration:
javapuzzlers.com

Python Puzzlers

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