Answer:

dogs["district"].value_counts().idxmax()

To find the most common district in the ’dogs’ DataFrame, the solution uses the value_counts() method, which counts the occurrences of each unique value in the ’district’ column. The idxmax() method is then used to identify the index of the maximum value in the resulting series, which corresponds to the most frequent district. Above, we presented one possible solution, but there are many:

• dogs["district"].value_counts().index[0]
  Directly selects the first district in the sorted counts.

• dogs.groupby("district").size().sort_values(ascending=False).index[0]
  Groups by ’district’, counts rows in each group, sorts by count, and retrieves the first district. Additionally, grouping by ’district’ and using .size() does the same thing as using .value_counts() on the ’district’ column; the only difference is that the values are sorted in a different order, which is why we set ascending=False

• dogs.groupby("district").count()["owner_id"].sort_values(ascending=False).index[0]
  Counts non-NaN values of owner_id for each district, sorts by count, and retrieves the first district.

Answer:

loc[dogs["dog_sex"] == "f", "primary_breed"]

Another possible answer is:

query("dog_sex == 'f'")["primary_breed"]

Note that the question didn’t ask for unique primary breeds.

Answers:

• 1. :, "Netflix": or some variation of that
• 2. :, 5: or some variation of that
• 3. axis=1
• 4. -1, 0

In Expression 1, keep in mind that idxmax() is a Series method returns the index of the row with the maximum value. As such, we can infer that Expression 1 sums the service-specific indicator columns (that is, the columns "Netflix", "Hulu", "Prime Video", and "Disney+") for each row and returns the index of the row with the greatest sum. To do this, we need the loc accessor to select all the service-specific indicator columns, which we can do using loc[:, "Netflix":] or loc[:, ["Netflix", "Hulu", "Prime Video", "Disney+"]].

When looking at Expression 2, we can split the problem into two parts: the code inside the assign statement and the code outside of it.

• Glancing at the code inside of the assign statement, (and also noticing the variable num_services), we realize that we, once again, want to sum up the values in the service-specific indicator columns. We do this by first selecting the last four columns, using .iloc[:, 5:] (notice the iloc), and then summing over axis=1. We use axis=1 (different from axis=0 in Expression 1), because unlike Expression 1, we’re summing over each row, instead of each column. If there had not been a .T in the code for Expression 1, we would’ve also used axis=1 in Expression 1.
• Finally, we need to select the "Title" of the last row in DataFrame in Expression 2, because sort_values sorts in ascending order by default. The last row has an integer position of -1, and the "Title" column has an integer position of 0, so we use iloc[-1, 0].

Answer: Series

The .value_counts() method, when called on a Series s, produces a new Series in which

• the index contains all unique values in s.
• the values are the frequencies of the unique values in s.

Since tv["Title"] is a Series, tv["Title"].value_counts() is a Series, and so is tv["Title"].value_counts.value_counts(). We provide an interpretation of each of these Series in the solution to the next subpart.

Answers:

• The only case in which it would make sense to set the index of tv to "Title" is if double_count.loc[1] == tv.shape[0] is True.
• If double_count.loc[2] == 5 is True, there are 5 pairs of 2 TV shows such that each pair shares the same "Title".

To answer, we need to understand what each of tv["Title"], tv["Title"].value_counts(), and tv["Title"].value_counts().value_counts() contain. To illustrate, let’s start with a basic, unrelated example. Suppose tv["Title"] looks like:

0    A
1    B
2    C
3    B
4    D
5    E
6    A
dtype: object

Then, tv["Title"].value_counts() looks like:

A    2
B    2
C    1
D    1
E    1
dtype: int64

and tv["Title"].value_counts().value_counts() looks like:

1    3
2    2
dtype: int64

Back to our actual dataset. tv["Title"], as we know, contains the name of each TV show. tv["Title"].value_counts() is a Series whose index is a sequence of the unique TV show titles in tv["Title"], and whose values are the frequencies of each title. tv["Title"].value_counts() may look something like the following:

Breaking Bad                             1
Fresh Meat                               1
Doctor Thorne                            1
                                       ...
Styling Hollywood                        1
Vai Anitta                               1
Fearless Adventures with Jack Randall    1
Name: Title, Length: 5368, dtype: int64

Then, tv["Title"].value_counts().value_counts() is a Series whose index is a sequence of the unique values in the above Series, and whose values are the frequencies of each value above. In the case where all titles in tv["Title"] are unique, then tv["Title"].value_counts() will only have one unique value, 1, repeated many times. Then, tv["Title"].value_counts().value_counts() will only have one row total, and will look something like:

1    5368
Name: Title, dtype: int64

This allows us to distinguish between the first two answer choices. The key is remembering that in order to set a column to the index, the column should only contain unique values, since the goal of the index is to provide a “name” (more formally, a label) for each row.

• The first answer choice, “The only case in which it would make sense to set the index of tv to "Title" is if double_count.iloc[0] == 1 is True”, is false. As we can see in the example above, all titles are unique, but double_count.iloc[0] is something other than 1.
• The second answer choice, “The only case in which it would make sense to set the index of tv to "Title" is if double_count.loc[1] == tv.shape[0] is True”, is true. If double_count.loc[1] == tv.shape[0], it means that all values in tv["Title"].value_counts() were 1, meaning that tv["Title"] consisted solely of unique values, which is the only case in which it makes sense to set "Title" to the index.

Now, let’s look at the second two answer choices. If double_counts.loc[2] == 5, it would mean that 5 of the values in tv["Title"].value_counts() were 2. This would mean that there were 5 pairs of titles in tv["Title"] that were the same.

• This makes the fourth answer choice, “If double_count.loc[2] == 5 is True, there are 5 pairs of 2 TV shows such that each pair shares the same "Title"”, correct.
• The third answer choice, “If double_count.loc[2] == 5 is True, there are 5 TV shows that all share the same "Title"”, is incorrect; if there were 5 TV shows with the same title, then double_count.loc[5] would be at least 1, but we can’t make any guarantees about double_counts.loc[2].

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.

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. Specifically, sports.loc[1] attempts to locate a row with the index label 1. However, since the DataFrame is indexed by 'Sport', and 1 is not a valid index label, this would result in a KeyError. On the other hand, sports.iloc[1] accesses the second row of the DataFrame (based on zero-based indexing), but it returns the entire row as a Series, including all its columns.

The last answer choice is incorrect because 'Sport' is not a column in the DataFrame. Instead, 'Sport' is the index of the DataFrame, which is never considered a column.

Answer: The first three options are correct.

Explanation:

• sports.loc['basketball', 'PlayersPerTeam']:
  .loc will access the value of a specific row ('basketball') and column ('PlayersPerTeam') by label. This will return 5.

• sports['PlayersPerTeam'].loc['basketball']:
  Selecting the column 'PlayersPerTeam' as a Series and then using .loc['basketball'] accesses the value for the 'basketball' index, returning 5.

• sports['PlayersPerTeam'].iloc[1]:
  Selecting the column 'PlayersPerTeam' as a Series and using .iloc[1] accesses the second value (index 1) in the Series, which corresponds to 'basketball'. This will also return 5.

• sports.loc['PlayersPerTeam']['basketball']:
  This throws a KeyError because 'PlayersPerTeam' is not a valid index label in the DataFrame. The sports DataFrame is indexed by 'Sport', so attempting to locate a row labeled 'PlayersPerTeam' is invalid.

• sports.loc['basketball']: While this syntax is valid, it retrieves the entire row corresponding to the index 'basketball'. The result will be a Series, not just the single value.

• sports.iloc[1]: In this case, sports.iloc[1] retrieves the entire second row of the DataFrame.