Skill Test #2 Practice Questions

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No writeup provided for practice problems. Feel free to collapse this window.

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• skill_test_#2_practice_questions.py

### Skill #1: Updating a dictionary

def update_money(dct, name, money):
    """
    Takes in a dictionary, key, and val if the key already exists
    add val to the current val, key is guaranteed to be in the dct.

>>> bank = {"anish": 0, "carter": 0}
    >>> update_money(bank, "anish", 50)
    {'anish': 50, 'carter': 0}

>>> bank = {"anish": 0, "carter": 0}
    >>> update_money(bank, "carter", 10)
    {'anish': 0, 'carter': 10}
    """
    # YOUR CODE GOES HERE

def change_string(dct, s):
    """
    Takes in a dictionary and a string, s. If the string already exists, 
    square its value, otherwise create the a key-val pair of the form
    s: len(s).

>>> dct = {"hi": 2, "bye": 3}
    >>> change_string(dct, "hi")
    {'hi': 4, 'bye': 3}

>>> dct = {"hi": 2, "bye": 3}
    >>> change_string(dct, "hello")
    {'hi': 2, 'bye': 3, 'hello': 5}
    """
    # YOUR CODE GOES HERE

### Skill #2: Aggregating keys/values

def val_sum(dct):
    """
    Find the sum of a dictionary's keys, values will be ints/floats only.

>>> dct = {1: 2, 3: 4, 5: 6, 7: 8}
    >>> val_sum(dct)
    16
    """
    # YOUR CODE GOES HERE

def avg_val(dct):
    """
    Find the average of a dictionary's keys, keys will be ints/floats only. 
    Don't round.

>>> dct = {1: "hello", 2: "hi", 3: "hey"}
    >>> avg_val(dct)
    2.0

>>> dct = {1: 2, 3: 4, 5: 6}
    >>> avg_val(dct)
    3.0
    """
    # YOUR CODE GOES HERE

### Skill #3: Iterating over a dictionary

def can_drink(dct, drinking_age):
    """
    Given a dictionary of names and ages, return a list of names of people
    who are old enough to drink.

>>> names = {"anish": 20, "carter": 19, "ben": 21}
    >>> can_drink(names, 21)
    ['ben']

>>> names = {"anish": 20, "carter": 19}
    >>> can_drink(names, 18)
    ['anish', 'carter']
    """
    # YOUR CODE GOES HERE

def oldest(dct):
    """
    Given a dictionary of names and ages, return the name of the 
    person who is the oldest, assume unique ages.

>>> names = {"anish": 20, "carter": 19, "ben": 21}
    >>> oldest(names)
    'ben'

>>> names = {"anish": 20, "carter": 19}
    >>> oldest(names)
    'anish'
    """
    # YOUR CODE GOES HERE

### Skill #4: Creating a dictionary

def create_dct1(keys, vals):
    """
    Takes in two lists of equal length, keys and vals 
    and creates a dictionary.

>>> keys = ["anish", "carter", "ty"]
    >>> vals = [20, 19, 19]
    >>> create_dct1(keys, vals)
    {'anish': 20, 'carter': 19, 'ty': 19}
    """
    # YOUR CODE GOES HERE

def create_dct2(tups, key_idx, val_idx):
    """
    Takes in a list of 3-tuples and creates a dictionary 
    of the form tup[key_idx]: tup[val_idx]. You can assume every
    value in every tuple is a valid key and unique.

>>> tups = [("anish", 20, 175), ("carter", 19, 160)]
    >>> create_dct2(tups, 0, 2)
    {'anish': 175, 'carter': 160}

>>> tups = [("anish", 20, 175), ("carter", 19, 160)]
    >>> create_dct2(tups, 0, 1)
    {'anish': 20, 'carter': 19}
    """
    # YOUR CODE GOES HERE

## Skill #5: Extra Credit

# no practice questions for the extra credit, but 
# it will combine/build on many of the topics above

Python UTF-8

Saved Ln 1, Col 1

Terminal

Output

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