Quiz 3: Solutions
June 2nd, 2021
Below is a historgram of the results.
Key Statistics
- Median: 62
- Mean: 57
- Maximum: 70
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1. One-liners (20 points)
Do not need to write a def for these questions. For each part, write a 1-line expression to compute the indicated value with: map() or sorted() or min() or max() or a comprehension. You do not need to call list() for these.
# a. Given a list of numbers. # Call map() to produce a result where each number is multiplied by 2. # (or equivalently write a comprehension) >>> nums = [3, 0, -2, 5] # yields [6, 0, -4, 10] # your expression:map(lambda x: x*2, nums)or equivalently[num*2 for num in nums]# b. Given a list of numbers. # Call map() to produce a result where each number is converted # to string form with a '!' added at its end. # (or equivalently write a comprehension) >>> nums = [3, 0, -2, 5] # yields ['3!', '0!', '-2!', '5!'] # your expression:map(lambda x: str(x)+'!', nums)or equivalently[str(num)+'!' for num in nums]# c. Given a list of city (name, population) tuples like this: # cities = [('palo alto', 51202), ('kensington', 12300), ...] # Call sorted() to produce a list of these tuples sorted in increasing # order by population # your expression:sorted(cities, key=lambda x: x[1])# d. Given a non-empty list of food (name, sour-score, sweet-score) tuples # like the following, scores are in the range 1 .. 10 # foods = [('banana', 1, 8), ('orange', 9, 6), ... ] # Call max() or min() to return the food tuple with the highest sweet-score. # your expression:max(foods, key=lambda food: food[2])
2. Dict (25 points)
Suppose we have a "foods" dict, where each key is a food, and its value is a len-2 tuple (tasty-score, healthy-score), where each score is in the range 1 .. 10, like this:
foods = {
'coffee': (3, 4),
'oatmeal': (6, 9),
'donut': (10, 1),
}
We have a "meals" dict where each key is a meal, and its value is a list of foods for that meal:
meals = {
'breakfast': ['coffee', 'oatmeal'],
'lunch': ['donut'],
'dinner': ['beets', 'donut']
}
Write code that, given the foods and meals dicts, creates and returns a new "healthy" dict which gathers the healthy scores for each planned meal. The healthy dict has the same keys as the meals dict, and the value for each key is a list of the healthy scores of that meal's foods. If a particular food is not present in the foods dict, assume it has a healthy score of 5, e.g. 'beets' in the example below.
# Input foods and meals dicts:
foods = {
'coffee': (3, 4),
'oatmeal': (6, 9),
'donut': (10, 1),
}
meals = {
'breakfast': ['coffee', 'oatmeal'],
'lunch': ['donut'],
'dinner': ['beets', 'donut']
}
Computed healthy dict:
{
'breakfast': [4, 9],
'lunch': [1],
'dinner': [5, 1]
}
def healthy_dict(foods, meals):
healthy = {}
for meal in meals.keys():
scores = []
for food in meals[meal]:
if food in foods:
tup = foods[food]
scores.append(tup[1])
else:
scores.append(5)
healthy[meal] = scores
return result
3. Dict-File (25 points)
Suppose we are working on a dumbed-down Tweet technology called Bleet. In a Bleet, all hashtags look like '#omg' - the hashtag is made of a '#' char and the 3 following chars. No logic or loop is needed to find the end of the hashtag; it's simply the 3 chars after the '#'. We have a text file, where each line is made of 2 or more bleets, separated by plus chars '+'.
hello #omgthere+hi and#idk as if#omg+#idkwhat
Each bleet, e.g. 'hello #omgthere', contains exactly 1 hashtag, and other than the hashtag, does not contain any '#' or '+' chars.
Process all the lines in the file, building and and returning a dict that has a key for each hashtag string without the leading '#', and its value is a list of all the bleets that include that hashtag.
{
'omg': ['hello #omgthere', 'as if#omg']
'idk': ['hi and#idk', '#idkwhat']
}
def read_posts(filename):
posts = {}
with open(filename) as f:
for line in f:
parts = line.split('+')
for bleet in parts:
hash = bleet.find('#')
tag = bleet[hash + 1:hash + 4]
if tag not in bleets:
bleets[tag] = []
bleets[tag].append(bleet)
return posts