← return to practice.dsc10.com
Welcome! The problems shown below should be worked on on
paper, since the quizzes and exams you take in this course will
also be on paper. You do not need to submit your solutions anywhere.
We encourage you to complete this worksheet in groups during an
extra practice session on Friday, January 19. Solutions will be posted
after all sessions have finished. This problem set is not designed to
take any particular amount of time - focus on understanding concepts,
not on getting through all the questions.
Consider the following assignment statement.
puffin = np.array([5, 9, 13, 17, 21])Provide arguments to call np.arange with so that the
array penguin is identical to the array
puffin.
penguin = np.arange(____)Answer: We need to provide np.arange
with three arguments: 5, anything in (21,
25], 4. For instance, something line
penguin = np.arange(5, 25, 4) would work.
The average score on this problem was 90%.
Fill in the blanks so that the array parrot is also
identical to the array puffin.
Hint: Start by choosing y so that
parrot has length 5.
parrot = __(x)__ * np.arange(0, __(y)__, 2) + __(z)__Answer:
x: 2y: anything in (8,
10]z: 5
The average score on this problem was 74%.
Suppose students is a DataFrame of all students who took
DSC 10 last quarter. students has one row per student,
where:
The index contains students’ PIDs as strings starting with
"A".
The "Overall" column contains students’ overall
percentage grades as floats.
The "Animal" column contains students’ favorite
animals as strings.
What type is students.get("Overall")? If this expression
errors, select “this errors."
float
string
array
Series
this errors
Answer: Series
The average score on this problem was 73%.
What type is students.get("PID")? If this expression
errors, select “this errors."
float
string
array
Series
this errors
Answer: this errors
The average score on this problem was 67%.
Vanessa is one student who took DSC 10 last quarter. Her PID is A12345678, she earned the sixth-highest overall percentage grade in the class, and her favorite animal is the giraffe.
Supposing that students is already sorted by
"Overall" in descending order, fill in the
blanks so that animal_one and animal_two
both evaluate to "giraffe".
animal_one = students.get(__(x)__).loc[__(y)__]
animal_two = students.get(__(x)__).iloc[__(z)__]Answer:
x: "Animal"y: "A12345678"z: 5
The average score on this problem was 69%.
If students wasn’t already sorted by
"Overall" in descending order, which of your answers would
need to change?
Neither y nor z would need to change
Both y and z would need to change
y only
z only
Answer: z only
The average score on this problem was 82%.
You are given a DataFrame called sports, indexed by
'Sport' containing one column,
'PlayersPerTeam'. The first few rows of the DataFrame are
shown below:
| Sport | PlayersPerTeam |
|---|---|
| baseball | 9 |
| basketball | 5 |
| field hockey | 11 |
Which of the following evaluates to
'basketball'?
sports.loc[1]
sports.iloc[1]
sports.index[1]
sports.get('Sport').iloc[1]
Answer: sports.index[1]
We are told that the DataFrame is indexed by 'Sport' and
'basketball' is one of the elements of the index. To access
an element of the index, we use .index to extract the index
and square brackets to extract an element at a certain position.
Therefore, sports.index[1] will evaluate to
'basketball'.
The first two answer choices attempt to use .loc or
.iloc directly on a DataFrame. We typically use
.loc or .iloc on a Series that results from
using .get on some column. Although we don’t typically do
it this way, it is possible to use .loc or
.iloc directly on a DataFrame, but doing so would produce
an entire row of the DataFrame. Since we want just one word,
'basketball', the first two answer choices must be
incorrect.
The last answer choice is incorrect because we can’t use
.get with the index, only with a column. The index is never
considered a column.
The average score on this problem was 88%.
Suppose you are given a DataFrame of employees for a given company.
The DataFrame, called employees, is indexed by
'employee_id' (string) with a column called
'years' (int) that contains the number of years each
employee has worked for the company.
Suppose that the code
employees.sort_values(by='years', ascending=False).index[0]outputs '2476'.
True or False: The number of years that employee 2476 has worked for the company is greater than the number of years that any other employee has worked for the company.
True
False
Answer: False
This is false because there could be other employees who worked at the company equally long as employee 2476.
The code says that when the employees DataFrame is
sorted in descending order of 'years', employee 2476 is in
the first row. There might, however, be a tie among several employees
for their value of 'years'. In that case, employee 2476 may
wind up in the first row of the sorted DataFrame, but we cannot say that
the number of years employee 2476 has worked for the company is greater
than the number of years that any other employee has worked for the
company.
If the statement had said greater than or equal to instead of greater than, the statement would have been true.
The average score on this problem was 29%.
What will be the output of the following code?
employees.assign(start=2021-employees.get('years'))
employees.sort_values(by='start').index.iloc[-1]the employee id of an employee who has worked there for the most years
the employee id of an employee who has worked there for the fewest years
an error message complaining about iloc[-1]
an error message complaining about something else
Answer: an error message complaining about something else
The problem is that the first line of code does not actually add a
new column to the employees DataFrame because the
expression is not saved. So the second line tries to sort by a column,
'start', that doesn’t exist in the employees
DataFrame and runs into an error when it can’t find a column by that
name.
This code also has a problem with iloc[-1], since
iloc cannot be used on the index, but since the problem
with the missing 'start' column is encountered first, that
will be the error message displayed.
The average score on this problem was 27%.
Suppose df is a DataFrame and b is any
boolean array whose length is the same as the number of rows of
df.
True or False: For any such boolean array b,
df[b].shape[0] is less than or equal to
df.shape[0].
True
False
Answer: True
The brackets in df[b] perform a query, or filter, to
keep only the rows of df for which b has a
True entry. Typically, b will come from some
condition, such as the entry in a certain column of df
equaling a certain value. Regardless, df[b] contains a
subset of the rows of df, and .shape[0] counts
the number of rows, so df[b].shape[0] must be less than or
equal to df.shape[0].
The average score on this problem was 86%.
You are given a DataFrame called books that contains
columns 'author' (string), 'title' (string),
'num_chapters' (int), and 'publication_year'
(int).
Suppose that after doing books.groupby('Author').max(),
one row says
| author | title | num_chapters | publication_year |
|---|---|---|---|
| Charles Dickens | Oliver Twist | 53 | 1838 |
Based on this data, can you conclude that Charles Dickens is the alphabetically last of all author names in this dataset?
Yes
No
Answer: No
When we group by 'Author', all books by the same author
get aggregated together into a single row. The aggregation function is
applied separately to each other column besides the column we’re
grouping by. Since we’re grouping by 'Author' here, the
'Author'column never has the max() function
applied to it. Instead, each unique value in the 'Author'
column becomes a value in the index of the grouped DataFrame. We are
told that the Charles Dickens row is just one row of the output, but we
don’t know anything about the other rows of the output, or the other
authors. We can’t say anything about where Charles Dickens falls when
authors are ordered alphabetically (but it’s probably not last!)
The average score on this problem was 94%.
Based on this data, can you conclude that Charles Dickens wrote Oliver Twist?
Yes
No
Answer: Yes
Grouping by 'Author' collapses all books written by the
same author into a single row. Since we’re applying the
max() function to aggregate these books, we can conclude
that Oliver Twist is alphabetically last among all books in the
books DataFrame written by Charles Dickens. So Charles
Dickens did write Oliver Twist based on this data.
The average score on this problem was 95%.
Based on this data, can you conclude that Oliver Twist has 53 chapters?
Yes
No
Answer: No
The key to this problem is that groupby applies the
aggregation function, max() in this case, independently to
each column. The output should be interpreted as follows:
books written by Charles Dickens,
Oliver Twist is the title that is alphabetically last.books written by Charles Dickens, 53
is the greatest number of chapters.books written by Charles Dickens,
1838 is the latest year of publication.However, the book titled Oliver Twist, the book with 53 chapters, and the book published in 1838 are not necessarily all the same book. We cannot conclude, based on this data, that Oliver Twist has 53 chapters.
The average score on this problem was 74%.
Based on this data, can you conclude that Charles Dickens wrote a book with 53 chapters that was published in 1838?
Yes
No
Answer: No
As explained in the previous question, the max()
function is applied separately to each column, so the book written by
Charles Dickens with 53 chapters may not be the same book as the book
written by Charles Dickens published in 1838.
The average score on this problem was 73%.
King Triton, UCSD’s mascot, is quite the traveler! For this question,
we will be working with the flights DataFrame, which
details several facts about each of the flights that King Triton has
been on over the past few years. The first few rows of
flights are shown below.
Here’s a description of the columns in flights:
'DATE': the date on which the flight occurred. Assume
that there were no “redeye” flights that spanned multiple days.'FLIGHT': the flight number. Note that this is not
unique; airlines reuse flight numbers on a daily basis.'FROM' and 'TO': the 3-letter airport code
for the departure and arrival airports, respectively. Note that it’s not
possible to have a flight from and to the same airport.'DIST': the distance of the flight, in miles.'HOURS': the length of the flight, in hours.'SEAT': the kind of seat King Triton sat in on the
flight; the only possible values are 'WINDOW',
'MIDDLE', and 'AISLE'. Which of these correctly evaluates to the number of flights King
Triton took to San Diego (airport code 'SAN')?
flights.loc['SAN'].shape[0]
flights[flights.get('TO') == 'SAN'].shape[0]
flights[flights.get('TO') == 'SAN'].shape[1]
len(flights.sort_values('TO', ascending=False).loc['SAN'])
Answer:
flights[flights.get('TO') == 'SAN'].shape[0]
The strategy is to create a DataFrame with only the flights that went
to San Diego, then count the number of rows. The first step is to query
with the condition flights.get('TO') == 'SAN' and the
second step is to extract the number of rows with
.shape[0].
Some of the other answer choices use .loc['SAN'] but
.loc only works with the index, and flights
does not have airport codes in its index.
The average score on this problem was 95%.
Fill in the blanks below so that the result also evaluates to the
number of flights King Triton took to San Diego (airport code
'SAN').
flights.groupby(__(a)__).count().get('FLIGHT').__(b)__ What goes in blank (a)?
'DATE'
'FLIGHT'
'FROM'
'TO'
What goes in blank (b)?
.index[0]
.index[-1]
.loc['SAN']
.iloc['SAN']
.iloc[0]
True or False: If we change .get('FLIGHT') to
.get('SEAT'), the results of the above code block will not
change. (You may assume you answered the previous two subparts
correctly.)
True
False
Answer: 'TO',
.loc['SAN'], True
The strategy here is to group all of King Triton’s flights according
to where they landed, and count up the number that landed in San Diego.
The expression flights.groupby('TO').count() evaluates to a
DataFrame indexed by arrival airport where, for any arrival airport,
each column has a count of the number of King Triton’s flights that
landed at that airport. To get the count for San Diego, we need the
entry in any column for the row corresponding to San Diego. The code
.get('FLIGHT') says we’ll use the 'FLIGHT'
column, but any other column would be equivalent. To access the entry of
this column corresponding to San Diego, we have to use .loc
because we know the name of the value in the index should be
'SAN', but we don’t know the row number or integer
position.
The average score on this problem was 89%.
Consider the DataFrame san, defined below.
san = flights[(flights.get('FROM') == 'SAN') & (flights.get('TO') == 'SAN')]Which of these DataFrames must have the same number
of rows as san?
flights[(flights.get('FROM') == 'SAN') and (flights.get('TO') == 'SAN')]
flights[(flights.get('FROM') == 'SAN') | (flights.get('TO') == 'SAN')]
flights[(flights.get('FROM') == 'LAX') & (flights.get('TO') == 'SAN')]
flights[(flights.get('FROM') == 'LAX') & (flights.get('TO') == 'LAX')]
Answer:
flights[(flights.get('FROM') == 'LAX') & (flights.get('TO') == 'LAX')]
The DataFrame san contains all rows of
flights that have a departure airport of 'SAN'
and an arrival airport of 'SAN'. But as you may know, and
as you’re told in the data description, there are no flights from an
airport to itself. So san is actually an empty DataFrame
with no rows!
We just need to find which of the other DataFrames would necessarily
be empty, and we can see that
flights[(flights.get('FROM') == 'LAX') & (flights.get('TO') == 'LAX')]
will be empty for the same reason.
Note that none of the other answer choices are correct. The first
option uses the Python keyword and instead of the symbol
&, which behaves unexpectedly but does not give an
empty DataFrame. The second option will be non-empty because it will
contain all flights that have San Diego as the departure airport or
arrival airport, and we already know from the first few rows of
flight that there are some of these. The third option will
contain all the flights that King Triton has taken from
'LAX' to 'SAN'. Perhaps he’s never flown this
route, or perhaps he has. This DataFrame could be empty, but it’s not
necessarily going to be empty, as the question requires.
The average score on this problem was 70%.
The American Kennel Club (AKC) organizes information about dog
breeds. We’ve loaded their dataset into a DataFrame called
df. The index of df contains the dog breed
names as str values.
The columns are:
'kind' (str): the kind of dog (herding, hound, toy,
etc.). There are six total kinds.'size' (str): small, medium, or large.'longevity' (float): typical lifetime (years).'price' (float): average purchase price (dollars).'kids' (int): suitability for children. A value of
1 means high suitability, 2 means medium, and
3 means low.'weight' (float): typical weight (kg).'height' (float): typical height (cm).The rows of df are arranged in no particular
order. The first five rows of df are shown below
(though df has many more rows than
pictured here).
Assume we have already run import babypandas as bpd and
import numpy as np.
The following code computes the breed of the cheapest toy dog.
df[__(a)__].__(b)__.__(c)__Fill in part (a).
Answer: df.get('kind') == 'toy'
To find the cheapest toy dog, we can start by narrowing down our
dataframe to only include dogs that are of kind toy. We do this by
constructing the following boolean condition:
df.get('kind') == 'toy', which will check whether a dog is
of kind toy (i.e. whether or not a given row’s 'kind' value
is equal to 'toy'). As a result,
df[df.get('kind') == 'toy'] will retrieve all rows for
which the 'kind' column is equal to 'toy'.
The average score on this problem was 91%.
Fill in part (b).
Answer: .sort_values('price')
Next, we can sort the resulting dataframe by price, which will make
the minimum price (i.e. the cheapest toy dog) easily accessible to us
later on. To sort the dataframe, simply use .sort_values(),
with parameter 'price' as follows:
.sort_values('price')
The average score on this problem was 86%.
Which of the following can fill in blank (c)? Select all that apply.
loc[0]
iloc[0]
index[0]
min()
Answer: index[0]
loc[0]: loc retrieves an element by the
row label, which in this case is by 'breed', not by index
value. Furthermore, loc actually returns the entire row,
which is not what we are looking for. (Note that we are trying to find
the singular 'breed' of the cheapest toy dog.)iloc[0]: While iloc does retrieve elements
by index position, iloc actually returns the entire row,
which is not what we are looking for.index[0]: Note that since 'breed' is the
index column of our dataframe, and since we have already filtered and
sorted the dataframe, simply taking the 'breed' at index 0,
or index[0] will return the 'breed' of the
cheapest toy dog.min(): min() is a method used to find the
smallest value on a series not a dataframe.
The average score on this problem was 81%.
The following code computes an array containing the unique kinds of dogs that are heavier than 20 kg or taller than 40 cm on average.
foo = df.__(a)__.__(b)__
np.array(foo[__(c)__].__d__)Fill in blank (a).
Answer: groupby('kind')
We start this problem by grouping the dataframe by
'kind' since we’re only interested in whether each unique
'kind' of dog fits some sort of constraint. We don’t quite
perform querying yet since we need to group the DataFrame first. In
other words, we first need to group the DataFrame into each
'kind' before we could apply any sort of boolean
conditionals.
The average score on this problem was 97%.
Fill in blank (b).
Answer: .mean()
After we do .groupby('kind'), we need to apply
.mean() since the problem asks if each unique
'kind' of dog satisfies certain constraints on
average. .mean() calculates the average of each
column of each group which is what we want.
The average score on this problem was 94%.
Fill in blank (c).
Answer:
(foo.get('weight') > 20 | foo.get(height) > 40)
Once we have grouped the dogs by 'kind' and have
calculated the average stats of each kind of dog, we can do some
querying with two conditionals: foo.get('weight') > 20
gets the kinds of dogs that are heavier than 20 kg on average and
foo.get('height') > 40) gets the kinds of dogs that are
taller than 40 cm on average. We combine these two conditions with the
| operator since we want the kind of dogs that satisfy
either condition.
The average score on this problem was 93%.
Which of the following should fill in blank (d)?
.index
.unique()
.get('kind')
.get(['kind'])
Answer: .index
Note that earlier, we did groupby('kind'), which
automatically sets each unique 'kind' as the index. Since
this is what we want anyways, simply doing .index will give
us all the kinds of dogs that satisfy the given conditions.
The average score on this problem was 94%.
In September 2020, Governor Gavin Newsom announced that by 2035, all new vehicles sold in California must be zero-emissions vehicles. Electric vehicles (EVs) are among the most popular zero-emissions vehicles (though other examples include plug-in hybrids and hydrogen fuel cell vehicles).
The DataFrame evs consists of 32 rows,
each of which contains information about a different EV model.
"Brand" (str): The vehicle’s manufacturer."Model" (str): The vehicle’s model name."BodyStyle" (str): The vehicle’s body style."Seats" (int): The vehicle’s number of seats."TopSpeed" (int): The vehicle’s top speed, in
kilometers per hour."Range" (int): The vehicle’s range, or distance it can
travel on a single charge, in kilometers.The first few rows of evs are shown below (though
remember, evs has 32 rows total).
Assume that:
"Brand" column are
"Tesla", "BMW", "Audi", and
"Nissan".import babypandas as bpd and
import numpy as np.Suppose we’ve run the following line of code.
counts = evs.groupby("Brand").count()What value does counts.get("Range").sum() evaluate
to?
Answer: 32
counts is a DataFrame with one row per
"Brand", since we grouped by "Brand". Since we
used the .count() aggregation method, the columns in
counts will all be the same – they will all contain the
number of rows in evs for each "Brand"
(i.e. they will all contain the distribution of "Brand").
If we sum up the values in any one of the columns in
counts, then, the result will be the total number of rows
in evs, which we know to be 32. Thus,
counts.get("Range").sum() is 32.
The average score on this problem was 56%.
What value does counts.index[3] evaluate to?
Answer: "Tesla"
Since we grouped by "Brand" to create
counts, the index of counts will be
"Brand", sorted alphabetically (this sorting happens
automatically when grouping). This means that counts.index
will be the array-like sequence
["Audi", "BMW", "Nissan", "Tesla"], and
counts.index[3] is "Tesla".
The average score on this problem was 33%.
Consider the following incomplete assignment statement.
result = evs______.mean()In each part, fill in the blank above so that result evaluates to the specified quantity.
A DataFrame, indexed by "Brand", whose
"Seats" column contains the average number of
"Seats" per "Brand". (The DataFrame may have
other columns in it as well.)
Answer: .groupby("Brand")
When we group by a column, the resulting DataFrame contains one row
for every unique value in that column. The question specified that we
wanted some information per "Brand", which implies
that grouping by "Brand" is necessary.
After grouping, we need to use an aggregation method. Here, we wanted
the resulting DataFrame to be such that the "Seats" column
contained the average number of "Seats" per
"Brand"; this is accomplished by using
.mean(), which is already done for us.
Note: With the provided solution, the resulting DataFrame also has
other columns. For instance, it has a "Range" column that
contains the average "Range" for each "Brand".
That’s fine, since we were told that the resulting DataFrame may have
other columns in it as well. If we wanted to ensure that the only column
in the resulting DataFrame was "Seats", we could have used
.get(["Brand", "Seats"]) before grouping, though this was
not necessary.
The average score on this problem was 76%.
A number, corresponding to the average "TopSpeed" of all
EVs manufactured by Audi in evs
Answer:
[evs.get("Brand") == "Audi"].get("TopSpeed")
There are two parts to this problem:
Querying, to make sure that we only keep the rows corresponding to Audis. This is accomplished by:
evs.get("Brand") == "Audi" to create a Boolean
Series, with Trues for the rows we want to keep and
Falses for the other rows.True. This is accomplished by
evs[evs.get("Brand") == "Audi"] (though the
evs part at the front was already provided).Accessing the "TopSpeed" column. This is
accomplished by using .get("TopSpeed").
Then, evs[evs.get("Brand") == "Audi"].get("TopSpeed") is
a Series contaning the "TopSpeed"s of all Audis, and mean
of this Series is the result we’re looking for. The call to
.mean() was already provided for us.
The average score on this problem was 77%.