Due at 11:59pm on 02/24/2017.

## Starter Files

Download lab05.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the OK autograder.

## Submission

By the end of this lab, you should have submitted the lab with `python3 ok --submit`. You may submit more than once before the deadline; only the final submission will be graded.

• To receive credit for this lab, you must complete Questions 1-5 in lab05.py and submit through OK.
• Questions 6-9 are extra practice. They can also be found in the lab05_extra.py file. It is recommended that you complete these problems on your own time.

## List Mutation

### Question 1: Map

Write a function that maps a function on the given list. Be sure to mutate the original list.

This function should NOT return anything. This is to emphasize that this function should utilize mutability.

``````def map(fn, lst):
"""Maps fn onto lst using mutation.
>>> original_list = [5, -1, 2, 0]
>>> map(lambda x: x * x, original_list)
>>> original_list
[25, 1, 4, 0]
"""
# Iterative solution
for i in range(len(lst)):
lst[i] = fn(lst[i])

# Recursive solution
def map(fn, lst):
"""Maps fn onto lst using mutation.
>>> original_list = [5, -1, 2, 0]
>>> map(lambda x: x * x, original_list)
>>> original_list
[25, 1, 4, 0]
"""
if lst:  # True when lst != []
temp = lst.pop(0)
map(fn, lst)
lst.insert(0, fn(temp))``````

Use OK to test your code:

``python3 ok -q map``

### Question 2: Over 9000

Define `over_nine_thousand`, which takes a list and modifies that list by adding 9000 to each element.

``````def over_nine_thousand(original_list):
"""
>>> original_list = [1, 2, 3, 4, 5]
>>> over_nine_thousand(original_list)
>>> original_list
[9001, 9002, 9003, 9004, 9005]
"""
index = 0
while index < len(original_list):
original_list[index] += 9000
index += 1

# Alternate solution using map
def over_nine_thousand(original_list):
"""
>>> original_list = [1, 2, 3, 4, 5]
>>> over_nine_thousand(original_list)
>>> original_list
[9001, 9002, 9003, 9004, 9005]
"""
map(lambda x: x + 9000, original_list)``````

Use OK to test your code:

``python3 ok -q over_nine_thousand``

## Dictionaries

Dictionaries are unordered sets of key-value pairs. Keys can only be immutable types (strings, numbers, tuples), but their corresponding value can be anything! To create a dictionary, use the following syntax:

``>>> singers = { 'Adele': 'Hello', 1975: 'Chocolate', 'The Weeknd': ['The Hills', 'Earned It'] }``

The curly braces denote the key-value pairs in your dictionary. Each key-value pair is separated by a comma. For each pair, the key appears to the left of the colon and the value appears to the right of the colon. Note keys/values do not all have to be the same type, as you can see we have strings, integers and lists! Each key only appears once in a dictionary. You can retrieve values from your dictionary by "indexing" using the key:

``````>>> singers
'Chocolate'
>>> songs = singers['The Weeknd']
>>> songs
'The Hills'``````

You can add an entry or update an entry for an existing key in the dictionary using the following syntax.

Note they are identical syntax, so be careful! You might end updating (and overwriting an old value) even if you intended to add, and vice versa.

``````>>> singers['Adele'] = 'Rolling in the Deep'
'Rolling in the Deep'
>>> singers['Kanye West'] = 'Real Friends' # New entry!
>>> singers['Kanye West']
'Real Friends'``````

You can also check for membership of keys!

``````>>> 'Adele' in singers
True``````

Finally, here are some useful dictionary functions:

• `dict.keys()` will return a sequence of keys.

``````>>> list(singers.keys()) # We use list() to turn the sequence into a list
• `dict.values()` will return a sequence of values.

``````>>> list(singers.values())
['Chocolate', ['The Hills', 'Earned It'], 'Hello']``````
• `dict.items()` will return a sequences of keys-value pairs.

``````>>> list(singers.items())
[(1975, 'Chocolate'), ('The Weeknd', ['The Hills', 'Earned It']), ('Adele', 'Hello')]``````

### Question 3: WWPD: Dictionaries

What would Python display? Type it in the intepreter if you're stuck!

``python3 ok -q dicts -u``
``````>>> pokemon = {'pikachu': 25, 'dragonair': 148, 'mew': 151}
>>> pokemon['pikachu']
______25
>>> len(pokemon)
______3
>>> pokemon['jolteon'] = 135
>>> pokemon['ditto'] = 25
>>> len(pokemon)
______5
>>> sorted(list(pokemon.keys()))  # Alphabetically sorted list of pokemon's keys
______['ditto', 'dragonair', 'jolteon', 'mew', 'pikachu']
>>> 'mewtwo' in pokemon
______False
>>> pokemon['ditto'] = pokemon['jolteon']
>>> sorted(list(pokemon.keys()))  # Alphabetically sorted list of pokemon's keys
______['ditto', 'dragonair', 'jolteon', 'mew', 'pikachu']
>>> pokemon['ditto']
______135``````
``````>>> letters = {'a': 1, 'b': 2, 'c': 3}
>>> 'a' in letters
______True
>>> 2 in letters
______False``````
``````>>> food = {'bulgogi': 10, 'falafel': 4, 'ceviche': 7}
>>> food['ultimate'] = food['bulgogi'] + food['ceviche']
>>> food['ultimate']
______17
>>> len(food)
______4
>>> food['ultimate'] += food['falafel']
>>> food['ultimate']
______21
>>> sorted(list(food.keys()))  # Alphabetically sorted list of food's keys
______['bulgogi', 'ceviche', 'falafel', 'ultimate']
>>> food['bulgogi'] = food['falafel']
>>> len(food)
______4
>>> 'gogi' in food
______False``````

### Question 4: Replace All

Given a dictionary `d`, replace all occurrences of `x` as a value (not a key) with `y`.

Hint: To loop through the keys of a dictionary use `for key in d:`.

``````def replace_all(d, x, y):
"""Replace all occurrences of x as a value (not a key) in d with y.
>>> d = {3: '3', 'foo': 2, 'bar': 3, 'garply': 3, 'xyzzy': 99}
>>> replace_all(d, 3, 'poof')
>>> d == {3: '3', 'foo': 2, 'bar': 'poof', 'garply': 'poof', 'xyzzy': 99}
True
"""
for key in d:
if d[key] == x:
d[key] = y``````

Use OK to test your code:

``python3 ok -q replace_all``

## Nonlocal

Consider the following function:

``````def make_counter():
"""Makes a counter function.

>>> counter = make_counter()
>>> counter()
1
>>> counter()
2
"""
count = 0
def counter():
count = count + 1
return count
return counter``````

Running this function's doctests, we find that it causes the following error:

``UnboundLocalError: local variable 'count' referenced before assignment``

Why does this happen? When we execute an assignment statement, remember that we are either creating a new binding in our current frame or we are updating an old one in the current frame. For example, the line `count = ...` in `counter`, is creating the local variable `count` inside `counter`'s frame. This assignment statement tells Python to expect a variable called count inside `counter`'s frame, so Python will not look in parent frames for this variable. However, notice that we tried to compute `count + 1` before the local variable was created! That's why we get the `UnboundLocalError`.

To avoid this problem, we introduce the `nonlocal` keyword. It allows us to update a variable in a parent frame! Note we cannot use `nonlocal` to modify variables in the global frame. Consider this improved example:

`````` def make_counter():
"""Makes a counter function.

>>> counter = make_counter()
>>> counter()
1
>>> counter()
2
"""
count = 0
def counter():
nonlocal count
count = count + 1
return count
return counter``````

The line `nonlocal count` tells Python that `count` will not be local to this frame, so it will look for it in parent frames. Now we can update `count` without running into problems.

### Question 5: Count calls

When testing software, it can be useful to count the number of times that a function is called. Define a higher-order function `count_calls` that returns two functions:

1. A counted version of the input function `f` that counts the number of times it has been called, but otherwise behaves identically to the original function, and
2. A function of zero arguments that returns the number of times that the counted function has been called.

Your implementation should not include any lists or dictionaries.

Hint: Remember that you can use the `*args` keyword to define an arbitrary number of parameters for a function.

``````def count_calls(f):
"""A function that returns a version of f that counts calls to f and can
report that count to how_many_calls.

The returned function responds to a special string argument,
'how many calls?'
to return the number of calls.
0
3
1
>>> add(3, 4)  # Doesn't count
7
1
11
2
"""
calls = 0
def counted(*args):
nonlocal calls
calls = calls + 1
return f(*args)
return counted, lambda: calls``````

Use OK to test your code:

``python3 ok -q count_calls``

## Extra Questions

Questions in this section are also not required for submission. We encourage you to try them out on your own time for extra practice. -- they can be found in the the lab05_extra.py file. It is recommended that you complete these problems on your own time.

Python's `list` class that contains many useful methods. Using the builtin `dir()` function will show you all of them, like so:

``dir(list)``

Some of the most common methods include `append()`, `extend()`, and `pop()`.

``````>>> l = [3, 5, 6]
>>> l.append(10) # Adds an element to the end
>>> l
[3, 5, 6, 10]
>>> l.extend([-1, -6]) # Concatenates another list to the end
>>> l
[3, 5, 6, 10, -1, -6]
>>> l.pop() # Removes and returns the last element
-6
>>> l
[3, 5, 6, 10, -1]
>>> l.pop(2) # Removes and returns the element at the index given
6
>>> l
[3, 5, 10, -1]``````

### Question 6: Filter

Write a function that filters a list, only keeping elements that satisfy the predicate. Be sure to mutate the original list.

This function should NOT return anything. This is to emphasize that this function should utilize mutability.

``````def filter(pred, lst):
"""Filters lst with pred using mutation.
>>> original_list = [5, -1, 2, 0]
>>> filter(lambda x: x % 2 == 0, original_list)
>>> original_list
[2, 0]
"""
i = len(lst) - 1
while i >= 0:
if not pred(lst[i]):
lst.pop(i)
i -= 1

def filter(pred, lst):
"""Filters lst with pred using mutation.
>>> original_list = [5, -1, 2, 0]
>>> filter(lambda x: x % 2 == 0, original_list)
>>> original_list
[2, 0]
"""
if lst:
temp = lst.pop(0)
filter(pred, lst)
if pred(temp):
lst.insert(0, temp)``````

Use OK to test your code:

``python3 ok -q filter``

### Question 7: Reverse

Write a function that reverses the given list. Be sure to mutate the original list.

This function should NOT return anything. This is to emphasize that this function should utilize mutability.

``````def reverse(lst):
"""Reverses lst using mutation.

>>> original_list = [5, -1, 29, 0]
>>> reverse(original_list)
>>> original_list
[0, 29, -1, 5]
>>> odd_list = [42, 72, -8]
>>> reverse(odd_list)
>>> odd_list
[-8, 72, 42]
"""
# iterative solution
midpoint = len(lst) // 2
last = len(lst) - 1
for i in range(midpoint):
lst[i], lst[last - i] = lst[last - i], lst[i]

# Recursive solution
def reverse(lst):
"""Reverses lst using mutation.

>>> original_list = [5, -1, 29, 0]
>>> reverse(original_list)
>>> original_list
[0, 29, -1, 5]
>>> odd_list = [42, 72, -8]
>>> reverse(odd_list)
>>> odd_list
[-8, 72, 42]
"""
if len(lst) > 1:
temp = lst.pop()
reverse(lst)
lst.insert(0, temp)

# Alternative recursive solution
def reverse(lst):
"""Reverses lst using mutation.

>>> original_list = [5, -1, 29, 0]
>>> reverse(original_list)
>>> original_list
[0, 29, -1, 5]
>>> odd_list = [42, 72, -8]
>>> reverse(odd_list)
>>> odd_list
[-8, 72, 42]
"""
midpoint = len(lst) // 2
last = len(lst) - 1
def helper(i):
if i == midpoint:
return
lst[i], lst[last - i] = lst[last - i], lst[i]
helper(i + 1)
helper(0)``````

Use OK to test your code:

``python3 ok -q reverse``

### Question 8: Counter

Implement the function `counter` which takes in a string of words, and returns a dictionary where each key is a word in the message, and each value is the number of times that word is present in the original string.

``````def counter(message):
""" Returns a dictionary of each word in message mapped
to the number of times it appears in the input string.

>>> x = counter('to be or not to be')
>>> x['to']
2
>>> x['be']
2
>>> x['not']
1
>>> y = counter('run forrest run')
>>> y['run']
2
>>> y['forrest']
1
"""
word_list = message.split() # .split() returns a list of the words in the string. Try printing it!
result_dict = {}
for word in word_list:
if word in result_dict:
result_dict[word] += 1
else:
result_dict[word] = 1
return result_dict``````

Use OK to test your code:

``python3 ok -q counter``

### Question 9: Next Fibonacci

Write a function `make_fib` that returns a function that returns the next Fibonacci number each time it is called. (The Fibonacci sequence begins with 0 and then 1, after which each element is the sum of the preceding two.) Use a `nonlocal` statement!

``````def make_fib():
"""Returns a function that returns the next Fibonacci number
every time it is called.

>>> fib = make_fib()
>>> fib()
0
>>> fib()
1
>>> fib()
1
>>> fib()
2
>>> fib()
3
>>> fib2 = make_fib()
>>> fib() + sum([fib2() for _ in range(5)])
12
"""
``python3 ok -q make_fib``