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.

Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: [evs.get("Brand") == "Audi"].get("TopSpeed")

There are two parts to this problem:

1. Querying, to make sure that we only keep the rows corresponding to Audis. This is accomplished by:

   • Using evs.get("Brand") == "Audi" to create a Boolean Series, with Trues for the rows we want to keep and Falses for the other rows.
   • Using Boolean indexing to keep only the rows in which the aforementioned Series is True. This is accomplished by evs[evs.get("Brand") == "Audi"] (though the evs part at the front was already provided).

2. 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.

Difficulty: ⭐️⭐️

The average score on this problem was 77%.

Answer:

• (i): Series
• (ii): 32

The .apply method applies a function on every element of a Series. Here, evs.get("Brand").apply(max) applies the max function on every element of the "Brand" column of evs, producing a new Series with the same length as evs.

While not necessary to answer the question, if s is a string, then max(s) evaluates to the single character in s that is last in the alphabet. For instance, max("zebra") evaluates to "z". As such, evs.get("Brand").apply(max) is a Series of 32 elements, each of which is a single character, corresponding to the latest character in the alphabet for each entry in evs.get("Brand").

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 65%.

Answer:

• (i): string
• (ii): 5

The .max() method will find the “largest” element in the Series it is called in, which in this case is evs.get("Brand"). The way that strings are ordered is alphabetically, so evs.get("Brand").max() will be the last value of "Brand" alphabetically. Since we were told that the only values in the "Brand" column are "Tesla", "BMW", "Audi", and "Nissan", the “maximum” is "Tesla", which has a length of 5.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 54%.

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.

Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: [evs.get("Brand") == "Audi"].get("TopSpeed")

There are two parts to this problem:

1. Querying, to make sure that we only keep the rows corresponding to Audis. This is accomplished by:

   • Using evs.get("Brand") == "Audi" to create a Boolean Series, with Trues for the rows we want to keep and Falses for the other rows.
   • Using Boolean indexing to keep only the rows in which the aforementioned Series is True. This is accomplished by evs[evs.get("Brand") == "Audi"] (though the evs part at the front was already provided).

2. 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.

Difficulty: ⭐️⭐️

The average score on this problem was 77%.

Answer: .get("TopSpeed").apply(np.log)

The .apply method is used to apply a function on every element of a Series. The relevant Series here is the column containing the "TopSpeed" of each EV, i.e. evs.get("TopSpeed") (the evs part was already provided to us).

After we get that Series, we need to use the function np.log on every element of it. This is accomplished by using .apply(np.log). Putting our steps so far together, we have evs.get("TopSpeed").apply(np.log), which is a Series containing the natural logarithm of the "TopSpeed" of all EVs in evs.

The number we were asked for was the average of the natural logarithm of the "TopSpeed" of all EVs in evs; all we need to do now is use the .mean() method at the end, which was already done for us.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 61%.

Answer: Options 1, 3, 4, 6

Option 1: In order to get the number of people within a family, we can look at the character at position 0 (for the number of adults) and the character at position 2 (for the number of children). Converting each character into an int and adding these ints yields the correct results.

Option 2: This is similar to Option 1, however, the key difference is that the separate strings are concatenated first, then converted into an integer afterwards. Remember that the plus sign between two strings concatenates the strings together, and does not add mathematically. For example, on a family type of "1a2c", "1" and "2" will be extracted and concatenated together as "12", then converted to the int 12. This is returned instead of the value 3 that we are looking for.

Option 3: This option is similar to Option 2, however, it includes an extra step after concatenation. int(x/10) gets the value in the tens place, taking advantage of the fact that the int() function always rounds down. At the same time, x % 10 gets the value in the ones place by calculating the remainder upon division by ten. Looking at the example of "1a2c", the first line will set x = 12 and then int(12/10) will yield 1 while 12 % 10 yields 2. Adding these together achieves the correct answer of 3.

Option 4: This option is similar to Option 1, but includes the initial step of removing "c" from the string and separating by "a". After this, x is a list of two elements, the first of which represents the number of adults in the family, and the second of which represents the number of children in the family. These are separately converted to ints then added up in the last line.

Option 5: This option iterates through the input string, where i represents each individual character in the string. For example, on an input of "1a2c", i is first set to 1, then a, then 2, then c. However, calculating the remainder when we divide by two (i % 2) only makes sense when i is a number, and results in an error when i is a string.

Option 6: This is a similar approach to Option 5, except this time, i represents each of the numbers 0, 1, 2, and 3, since len(fam) is always 4. For each such i, which we can think of as the position number, the code will check if the position number is even (i % 2 == 0). This is only true for position 0 and 2, which are the positions that contain the numbers of adults and children in the family. When this condition is met, we add the value at that position onto our running total, x, which at the end, equals the total number of adults and children in the family.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

(a). Answer: split(",")[0]

The function find_day takes in a string that is a date with the format "Day of week, Month Day, Year", and we want the "Day of week" part. .split(",") separates the given string by "," and gives us a list containing the separated substrings, which looks something like ["Saturday", " December 7", " 2024"]. We want the first element in the list, so we use [0] to obtain the first element of the list.

Difficulty: ⭐️⭐️

The average score on this problem was 89%.

(b). Answer: 0

This blank follows a call of find_day(date), which returns a string that is a day of the week, such as "Saturday". The blank is inside a pair of [], hinting that we are performing string indexing to pull out a specific character from the string. We want to determine whether the string is "Saturday" or "Sunday". The only thing in common for these two days and is not common for the other five days of the week is that the string starts with the letter "S". Thus, we get the first character using [0] in this blank.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

(c). Answer: "S"

After getting the first character in the string, to check whether it is the letter "S", we use == to evaluate this character against "S". This expression will give us a boolean, either True or False, which is then returned.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 50%.

Answer:

• sales[sales_day.get("weekend")].shape[0] / sales_day.shape[0]
• sales_day.get("weekend").mean()
• np.count_nonzero(sales_day.get("weekend") > 0.5) / sales_day.shape[0]

First of all, we add a new column called "weekend" to the sales DataFrame, and this new DataFrame is called sales_day. The new column contains boolean values that indicate whether the value in the "date" column is "Saturday" or "Sunday", which are obtained by applying is_weekend from the previous question to the "date" column.

Now, we want to find the proportion of book purchases in sales that were made on a weekend, which is basically the proportion of rows in the "weekend" column that are True. Let’s look at each of the options:

• sales_day[sales_day.get("weekend")].count() / sales_day.shape[0] (incorrect): You can only use .count() on a groupby object. The correct alternative is to use .shape[0].

• sales[sales_day.get("weekend")].shape[0] / sales_day.shape[0] (correct): sales_day.get("weekend") gets the "weekend column from the sales_day DataFrame. We then use this Series to query the sales DataFrame and keep only the rows corresponding to weekends. Lastly, we count the number of rows in this filtered DataFrame with .shape[0] and divide by the total number of sales to get the proportion.

• sales_day.get("weekend").median() (incorrect): A boolean value can be seen as 1 (True) or 0 (False). Since the column "weekend" contains all booleans, the median falls on either 1 or 0, which is not the proportion that we want.

• sales_day.get("weekend").mean() (correct): A boolean value can be seen as 1 (True) or 0 (False). The mean of the "weekend" column is equal to the sum of the column divided the total number of rows in the column. The sum of the column is the sum of all the 1’s and 0’s, which is just the number of Trues. The number of Trues divided by the total number of rows is the proportion of sales on weekend.

• np.count_nonzero(sales_day.get("weekend") > 0.5) / sales_day.shape[0] (correct): A boolean value can be seen as 1 (True) or 0 (False). sales_day.get("weekend") > 0.5 gives us a Series of booleans indicating whether the value in "weekend" is greater than 0.5. If the value is True, then it’s greater than 0.5. np.count_nonzero counts the number of values that are not zero, which is the number of Trues. We then divide the number of Trues by the total number of sales to get the proportion.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer:

• (a): var = 3, var = -1 or alternate solution var = 1

• (b): index_value

• (c): split(", ")[var] or alternate solution split(", San Diego, CA, ")[var]

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 58%.

Answer: treat.get("address").apply(zip_as_int)

Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: takes a string as input, returns a string

To use the Series method .apply, we first need a Series, containing values of any type. We pass in the name of a function to .apply and essentially, .apply calls the given function on each value of the Series, producing a Series with the resulting outputs of those function calls. In this case, .apply(extract_product_line) is called on the Series ikea.get('product'), which contains string values. This means the function extract_product_line must take strings as inputs. We’re told that the code assigns a new column to the ikea DataFrame containing the product line associated with each product, and we know that the product line is a string, as it’s the first word of the product name. This means the function extract_product_line must output a string.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: x.split(' ')[0]

This function should take as input a string x, representing a product name, and return the first word of that string, representing the product line. Since words are separated by spaces, we want to split the string on the space character ' '.

It’s also correct to answer x.split()[0] without specifying to split on spaces, because the default behavior of the string .split method is to split on any whitespace, which includes any number of spaces, tabs, newlines, etc. Since we’re only extracting the first word, which will be separated from the rest of the product name by a single space, it’s equivalent to split using single spaces and using the default of any whitespace.

Difficulty: ⭐️⭐️

The average score on this problem was 84%.

Answer: int(first_split[0])*60+int(second_split[0])

As the last subpart demonstrated, if we want to compare times, it doesn’t make sense to do so when times are represented as strings. In the to_minutes function, we convert a time string into an integer number of minutes.

The first step is to understand the logic. Every hour contains 60 minutes, so for a time string formatted like x hr, y min' the total number of minutes comes from multiplying the value of x by 60 and adding y.

The second step is to understand how to extract the x and y values from the time string using the string methods .split. The string method .split takes as input some separator string and breaks the string into pieces at each instance of the separator string. It then returns a list of all those pieces. The first line of code, therefore, creates a list called first_split containing two elements. The first element, accessed by first_split[0] contains the part of the time string that comes before ' hr, '. That is, first_split[0] evaluates to the string x.

Similarly, first_split[1] contains the part of the time string that comes after ' hr, '. So it is formatted like 'y min'. If we split this string again using the separator of ' min', the result will be a list whose first element is the string 'y'. This list is saved as second_split so second_split[0] evaluates to the string y.

Now we have the pieces we need to compute the number of minutes, using the idea of multiplying the value of x by 60 and adding y. We have to be careful with data types here, as the bolded instructions warn us that the function must return an integer. Right now, first_split[0] evaluates to the string x and second_split[0] evaluates to the string y. We need to convert these strings to integers before we can multiply and add. Once we convert using the int function, then we can multiply the number of hours by 60 and add the number of minutes. Therefore, the solution is int(first_split[0])*60+int(second_split[0]).

Note that failure to convert strings to integers using the int function would lead to very different behavior. Let’s take the example time string of '3 hr, 5 min' as input to our function. With the return statement as int(first_split[0])*60+int(second_split[0]), the function would return 185 on this input, as desired. With the return statement as first_split[0]*60+second_split[0], the function would return a string of length 61, looking something like this '3333...33335'. That’s because the * and + symbols do have meaning for strings, they’re just different meanings than when used with integers.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 71%.

Answer: False

Even though people must have distinct usernames, one person can build multiple different IKEA products and log their time for each build. So we don’t expect every row of app_data to have a distinct username associated with it, and therefore username would not be suitable as an index, since the index should have distinct values.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 52%.

Answer: '1 hr, 45 min'

The code says that we should take the first four rows of app_data (which we can see in the preview at the start of this exam), sort the rows in ascending order of 'assembly_time', and extract the very first entry of the 'assembly_time' column. In other words, we have to find the 'assembly_time' value that would be first when sorted in ascending order. As given in the hint, the 'assembly_time' column contains strings, and strings get sorted alphabetically, so we are looking for the assembly time, among the first four, that would come first alphabetically. This is '1 hr, 45 min'.

Note that first alphabetically does not correspond to the least amount of time. '1 hr, 5 min' represents less time than '1 hr, 45 min', but in alphabetical order '1 hr, 45 min' comes before '1 hr, 5 min' because both start with the same characters '1 hr, ' and comparing the next character, '4' comes before '5'.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 46%.

Answer: False

Even though people must have distinct usernames, one person can build multiple different IKEA products and log their time for each build. So we don’t expect every row of app_data to have a distinct username associated with it, and therefore username would not be suitable as an index, since the index should have distinct values.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 52%.

Answer: False

Even though people must have distinct usernames, one person can build multiple different IKEA products and log their time for each build. So we don’t expect every row of app_data to have a distinct username associated with it, and therefore username would not be suitable as an index, since the index should have distinct values.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 52%.

Answer: '1 hr, 45 min'

The code says that we should take the first four rows of app_data (which we can see in the preview at the start of this exam), sort the rows in ascending order of 'assembly_time', and extract the very first entry of the 'assembly_time' column. In other words, we have to find the 'assembly_time' value that would be first when sorted in ascending order. As given in the hint, the 'assembly_time' column contains strings, and strings get sorted alphabetically, so we are looking for the assembly time, among the first four, that would come first alphabetically. This is '1 hr, 45 min'.

Note that first alphabetically does not correspond to the least amount of time. '1 hr, 5 min' represents less time than '1 hr, 45 min', but in alphabetical order '1 hr, 45 min' comes before '1 hr, 5 min' because both start with the same characters '1 hr, ' and comparing the next character, '4' comes before '5'.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 46%.

Answer: int(first_split[0])*60+int(second_split[0])

As the last subpart demonstrated, if we want to compare times, it doesn’t make sense to do so when times are represented as strings. In the to_minutes function, we convert a time string into an integer number of minutes.

The first step is to understand the logic. Every hour contains 60 minutes, so for a time string formatted like x hr, y min' the total number of minutes comes from multiplying the value of x by 60 and adding y.

The second step is to understand how to extract the x and y values from the time string using the string methods .split. The string method .split takes as input some separator string and breaks the string into pieces at each instance of the separator string. It then returns a list of all those pieces. The first line of code, therefore, creates a list called first_split containing two elements. The first element, accessed by first_split[0] contains the part of the time string that comes before ' hr, '. That is, first_split[0] evaluates to the string x.

Similarly, first_split[1] contains the part of the time string that comes after ' hr, '. So it is formatted like 'y min'. If we split this string again using the separator of ' min', the result will be a list whose first element is the string 'y'. This list is saved as second_split so second_split[0] evaluates to the string y.

Now we have the pieces we need to compute the number of minutes, using the idea of multiplying the value of x by 60 and adding y. We have to be careful with data types here, as the bolded instructions warn us that the function must return an integer. Right now, first_split[0] evaluates to the string x and second_split[0] evaluates to the string y. We need to convert these strings to integers before we can multiply and add. Once we convert using the int function, then we can multiply the number of hours by 60 and add the number of minutes. Therefore, the solution is int(first_split[0])*60+int(second_split[0]).

Note that failure to convert strings to integers using the int function would lead to very different behavior. Let’s take the example time string of '3 hr, 5 min' as input to our function. With the return statement as int(first_split[0])*60+int(second_split[0]), the function would return 185 on this input, as desired. With the return statement as first_split[0]*60+second_split[0], the function would return a string of length 61, looking something like this '3333...33335'. That’s because the * and + symbols do have meaning for strings, they’re just different meanings than when used with integers.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 71%.

Answer: app_data.get('assembly_time').apply(to_minutes)

We want to create a Series of times in minutes, since it’s to be added to the app_data DataFrame using .assign. The correct way to do that is to use .apply so that our function, which works for one input time string, can be applied separately to every time string in the 'assembly_time' column of app_data. Remember that .apply takes as input one parameter, the name of the function to apply, with no arguments. The correct syntax is app_data.get('assembly_time').apply(to_minutes).

Difficulty: ⭐️

The average score on this problem was 98%.

Answer: takes a string as input, returns a string

To use the Series method .apply, we first need a Series, containing values of any type. We pass in the name of a function to .apply and essentially, .apply calls the given function on each value of the Series, producing a Series with the resulting outputs of those function calls. In this case, .apply(extract_product_line) is called on the Series ikea.get('product'), which contains string values. This means the function extract_product_line must take strings as inputs. We’re told that the code assigns a new column to the ikea DataFrame containing the product line associated with each product, and we know that the product line is a string, as it’s the first word of the product name. This means the function extract_product_line must output a string.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: x.split(' ')[0]

This function should take as input a string x, representing a product name, and return the first word of that string, representing the product line. Since words are separated by spaces, we want to split the string on the space character ' '.

It’s also correct to answer x.split()[0] without specifying to split on spaces, because the default behavior of the string .split method is to split on any whitespace, which includes any number of spaces, tabs, newlines, etc. Since we’re only extracting the first word, which will be separated from the rest of the product name by a single space, it’s equivalent to split using single spaces and using the default of any whitespace.

Difficulty: ⭐️⭐️

The average score on this problem was 84%.

Answer: survey.get("IG Followers") == 0

survey.get("IG Followers") is a Series containing the values in the "IG Followers" column. survey.get("IG Followers") == 0, then, is a Series of Trues and Falses, containing True for rows where the number of "IG Followers" is 0 and False for all other rows.

Put together, survey[survey.get("IG Followers") == 0] evaluates to a new DataFrame, which only has the rows in survey where the student’s number of "IG Followers" is 0.

Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer: Options 2 and 3

Let’s look at each option.

• Option 1: get("Unread Emails").apply(max) This option is invalid because .apply(max) will produce a new Series containing the maximums of each individual element in "Unread Emails" rather than the maximum of the entire column. Since the values in the "Unread Emails" columns are numbers, this will try to call max on a number individually (many times), and that errors, since e.g. max(2) is invalid.
• Option 2: get("Unread Emails").max() This option is valid because it directly fetches the "Unread Emails" column and uses the max method to retrieve the highest value within it, aligning with the goal of identifying the most unread emails.
• Option 3: sort_values("Unread Emails", ascending=False).get("Unread Emails").iloc[0] This option is valid. The sort_values("Unread Emails", ascending=False) method call sorts the "Unread Emails" column in descending order. Following this, .get("Unread Emails") retrieves the sorted column, and .iloc[0] selects the first element, which, given the sorting, represents the maximum number of unread emails.
• Option 4: sort_values("Unread Emails", ascending=False).get("Unread Emails").iloc[-1] This option is invalid because, although it sorts the "Unread Emails" column in descending order, it selects the last element with .iloc[-1], which represents the smallest number of unread emails due to the descending sort order.

Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answers: (temp.get("College") == "Revelle") & (temp.get("IG Followers") == 0)

To identify the students in "Revelle" with 0 Instagram followers, we filter the temp DataFrame using two conditions: (temp["IG Followers"] == 0) ensures students have no Instagram followers, and (temp["College"] == "Revelle") ensures the student is from "Revelle". The & operator combines these conditions, resulting in a subset of students from "Revelle" with 0 Instagram followers.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 54%.

Answer: iloc[0], iloc[-1], sum(), max(), mean(), or any other method that when called on a Series with a single value outputs that same value

To create temp, we grouped on both "IG Followers" and "College" and used the max aggregation method. This means that temp has exactly one row where "College" is "Revelle" and "IG Followers" is 0, since when we group on both "IG Followers" and "College", the resulting DataFrame has exactly one row for every unique combination of "IG Followers" and "College" name. We’ve already identified that before blank (d) is

temp[(temp.get("College") == "Revelle") & (temp.get("IG Followers") == 0)].get("Unread Emails")

Since temp[(temp.get("College") == "Revelle") & (temp.get("IG Followers") == 0)] is a DataFrame with just a single row, the entire expression is a Series with just a single value, being the maximum value we ever saw in the "Unread Emails" column when "College" was "Revelle" and "IG Followers" was 0. To extract the one element out of a Series that only has one element, we can use many aggregation methods (max, min, mean, median) or use .iloc[0] or .iloc[-1].

Difficulty: ⭐️⭐️

The average score on this problem was 79%.

Answer: No

When the survey DataFrame is grouped by both "IG Followers" and "College" using the .max aggregation method, it retrieves the maximum value for each column based on alphabetical order. Even if a student in "Revelle" with 0 Instagram followers has the most unread emails and is a "HI25" major, the "HI25" value would only be the maximum for the "Major" column if no other major code in the same group alphabetically surpasses it. Therefore, if there’s another major code like "HI26" or "MA30" in that group, it would be chosen as the maximum instead of "HI25".

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 59%.

Answer: n_baths = n_baths + 0.5

The behavior that we want this line of code to have is to work regardless if the bath string contains "One", "Two", or "Three". This means we need to have some way of taking the value that n_baths is already assigned and adding 0.5 to it. So, our code should read n_baths = n_baths + 0.5.

Difficulty: ⭐️

The average score on this problem was 92%.

Answer:

• (b): 12
• (c): int(s[0])

The code in blank (b) will only be run if the last letter of s is "r", which only happens when s = "1 year". So, blank (b) should return 12.

The code in blank (c) will run when s has any value other than "1 year". This includes only two options: 1 month, and 6 months. In order to get the corresponding number of the months for these two string values, we just need to take the first character of the string and convert it from a str type to an int type. So, blank (c) should return int(s[0]).

Difficulty: ⭐️⭐️

The average score on this problem was 84%.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answer: apts = apts.assign(Bed = apts.get("Bed").apply(int_bed))

The code above takes the “Bed” column, apts.get("Bed"), and uses .apply(int_bed), which runs each entry through the int_bed function that we have defined above. All that is left is to save the result back to the dataframe; this can be done with .assign().

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 68%.

Answer:

• (a) digit != "-"
• (b) total = total + int(digit)

1. We only care if the current digit is a number, and the only character in the string of the phone_number is the hyphen (-). For instance, in the given phone number of "501-800-3002", we want to extract the actual numbers in the string, without the hyphens. So, because we cannot establish a numerical value to a hyphen, we exclude it.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 34%.

2. This converts the current character to an integer and adds it to the total sum if the character is a digit.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 71%.

Answer:

• (a): int(dob[0])
• (b): int(dob[1])
• (c): int(dob[2])
• (d): (month < 5) or (month == 5 and day <= 3)

.split() method divides a string into a list based on delimiter. In this probelm, dob.split("-") returns a list of substrings that were separated by hyphens in the original string formatted like "MM-DD-YYYY". Thus, after using the .split("-"), the resulting list will be formatted like[“MM”, “DD”, “YYYY”]. Thus, we can access month, day, year by its position in the list dob.

Note that the comment asks month, day, year as int, do we need to convert them from string to int datatype, by using int(). Thus, we have month = int(dob[0]), day = int(dob[1]), year = int(dob[2]). This is for us to calculate age later by comparing month and day numebrs.

Then to calculate the age of the person given today is “05-03-2024”, we can use 2024-year. But if a person is born after May 3rd, then they are techinically 1 year younger than that, so we should use 2024-year-1. Thus in the if statment, we use (month < 5) or (month == 5 and day <= 3) to compare if the person is born in January, February, March, April or one of the first three days of May.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 59%.

Answer:

contacts.get("DOB").apply(age_today).mean()

contacts.get("DOB") gets the column ‘DOB’ column in contacts DataFrame as a series.

.apply(age_today) applies the function age_today to each element in the series to calculate age, and returns a series of ages of all contacts.

.mean() calculates the average of the series, giving the average age of all contacts.

Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer (a): str(x)+"%"

In order to add a percent symbol to the end, we first need to convert the variable x from an integer into a string by casting it. From there we can add a "%" sign to x through concatenation.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 70%.

Answer (b): int(x.replace("%", "")) or int(x.strip("%"))

In this case, the variable x is now a string with "%" at the end of it. Our first step is to get rid of the "%" at the end by either replacing it with an empty string "", or stripping it from the end. Note that the .strip("%") approach works because "%" is at the end of the string. If it were in the middle instead of the start or end, .strip("%") would not work. After we are left with only the number to cast to a string value for our output.

Difficulty: ⭐️⭐️

The average score on this problem was 79%.

Answer: z = tariffs.get("Reciprocal Tariff").apply(str) + "%"

The Reciprocal Tariff column consists of integer values. In order to add a percent sign (“%”) to the end of each of these values, we must first convert the Series of integers into a Series of strings before we are able to concatenate each with the percent sign (“%”).

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 33%.

Answer: "32% tax on goods from Taiwan"

y.iloc[3] consists of the Reciprocal Tariff (as a string) corresponding to the fourth entry of the Series y. From the given DataFrame, the fourth row corresponds to a Reciprocal Tariff 32. Because y is a Series of strings that added the “%” sign to each Reciprocal Tariff, y.iloc[3] evaluates to the string “32%”. tariffs.get("Country").loc[3] evaluates to the country at the fourth position from the tariffs DataFrame, or “Taiwan”.

Difficulty: ⭐️⭐️

The average score on this problem was 88%.

Answer: "Team"

Recall a choice for a good index would be a column wherein the values would act as a unique identifier for a particular observation. The only column that suits this description would be “Team” since each row represents a unique team.

Difficulty: ⭐️⭐️

The average score on this problem was 86%.

Answer: kart = kart.set_index("Team")

We use set_index(“Team”) to set the DataFrame’s index to “Team” and set this equal to kart to save this change.

Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer:

• (i) return
• (ii) int
• (iii) split() or split(" ")
• (iv) -1 or 1

Using the argument division, we just want to retrieve the number. To accomplish this, we can use split() on the division to separate the string in order to access the number. Note that the default behavior of split() is to split by the space.

The output of division.split() is now a list of “Division” and the number. To retrieve the number in this list, we can index for it with either -1 (since it is the last element of the list) or 1 (since it is the second element in the list). Because we want the output to be an integer type, we use int to cast the value to an integer. Finally, to have the function output the desired value, we start with a return statement.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 71%.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 52%.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 52%.

Answer: kart.assign(Division=kart.get("Division").apply(division_to_int))

First let’s start by getting the information we want for the new column. We get the column for transformation with kart.get(“Division”) and use .apply(division_to_int) in order to apply the function to this column. In order to update this transformed Series as a column “Division”, we use the .assign method on the DataFrame and set the transformed Series to the column name “Division”. Note that when using .assign to add a column and the chosen column name already exists, .assign will update the information in this column with the new input information.

Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: input: function, output: Series

It helps to think of an example of how we typically use .apply. Consider a DataFrame called books and a function called year_to_century that converts a year to the century it belongs to. We might use .apply as follows:

books.assign(publication_century = books.get('publication_year').apply(year_to_century))

.apply is called a Series method because we use it on a Series. In this case that Series is books.get('publication_year'). .apply takes one input, which is the name of the function we wish to apply to each element of the Series. In the example above, that function is year_to_century. The result is a Series containing the centuries for each book in the books DataFrame, which we can then assign back as a new column to the DataFrame. So .apply therefore takes as input a function and outputs a Series.

Difficulty: ⭐️

The average score on this problem was 98%.

Answer: concrete_city.shape[0] / all_city.shape[0]

Let’s first understand the code that is already provided for us. Note that city is a string corresponding to the name of a city.

all_city contains only the rows for the passed in city. Note that all_city.shape[0] or len(all_city) is the number of rows in all_city, i.e. it is the number of skyscrapers in city. Then, concrete_city contains only the rows in all_city corresponding to 'concrete' skyscrapers, i.e. it contains only the rows corresponding to 'concrete' skyscrapers in city. Note that concrete_city.shape[0] or len(concrete_city) is the number of skyscrapers in city that are made of 'concrete'.

We want to return True only if at least 50% of the skyscrapers in city are made of concrete. The last line in the function, return proportion >= 0.5, is already provided for us, so all we need to do is compute the proportion of skyscrapers in city that are made of concrete. This is concrete_city.shape[0] / all_city.shape[0].

Another possible answer is len(concrete_city) / len(all_city).

Difficulty: ⭐️⭐️

The average score on this problem was 85%.

Answer: by_city.get('city').apply(majority_concrete)

We are told to add a column to by_city. Recall, the way that .assign works is that the name of the new column comes before the = symbol, and a Series (or array) containing the values for the new column comes after the = symbol. As such, a Series needs to go in the blank.

majority_concrete takes in the name of a single city and returns either True or False accordingly. All we need to do here then is use the majority_concrete function on every element in the 'city' column. After accessing the 'city' column using by_city.get('city'), we need to use the .apply method using the argument majority_concrete. Putting it all together yields by_city.get('city').apply(majority_concrete), which is a Series.

Note: Here, by_city.get('city') only works because .reset_index() was used in the line where by_city was defined. If we did not reset the index, 'city' would not be a column!

Difficulty: ⭐️⭐️

The average score on this problem was 86%.

Answer: concrete_city.shape[0] / all_city.shape[0]

Let’s first understand the code that is already provided for us. Note that city is a string corresponding to the name of a city.

all_city contains only the rows for the passed in city. Note that all_city.shape[0] or len(all_city) is the number of rows in all_city, i.e. it is the number of skyscrapers in city. Then, concrete_city contains only the rows in all_city corresponding to 'concrete' skyscrapers, i.e. it contains only the rows corresponding to 'concrete' skyscrapers in city. Note that concrete_city.shape[0] or len(concrete_city) is the number of skyscrapers in city that are made of 'concrete'.

We want to return True only if at least 50% of the skyscrapers in city are made of concrete. The last line in the function, return proportion >= 0.5, is already provided for us, so all we need to do is compute the proportion of skyscrapers in city that are made of concrete. This is concrete_city.shape[0] / all_city.shape[0].

Another possible answer is len(concrete_city) / len(all_city).

Difficulty: ⭐️⭐️

The average score on this problem was 85%.

Answer: by_city.get('city').apply(majority_concrete)

We are told to add a column to by_city. Recall, the way that .assign works is that the name of the new column comes before the = symbol, and a Series (or array) containing the values for the new column comes after the = symbol. As such, a Series needs to go in the blank.

majority_concrete takes in the name of a single city and returns either True or False accordingly. All we need to do here then is use the majority_concrete function on every element in the 'city' column. After accessing the 'city' column using by_city.get('city'), we need to use the .apply method using the argument majority_concrete. Putting it all together yields by_city.get('city').apply(majority_concrete), which is a Series.

Note: Here, by_city.get('city') only works because .reset_index() was used in the line where by_city was defined. If we did not reset the index, 'city' would not be a column!

Difficulty: ⭐️⭐️

The average score on this problem was 86%.

Answer: 2

Recall, the 'is_majority' column we created in the previous subpart contains only two possible values – True and False. When we group by_city by 'is_majority', we create two “groups” – one for True and one for False. As such, no matter what aggregation method we use (here we happened to use .count()), the resulting DataFrame will only have 2 rows (again, one for True and one for False).

Note: The question asked for the “largest possible value” that mystery.shape[0], because it is possible that mystery only has 1 row. This can only happen in two cases:

1. It is true in all cities that the majority of skyscrapers are made of 'concrete'.
2. It is true in no cities that the majority of skyscrapers are made of 'concrete'.

Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: True

In the solution to the previous subpart, we noted that mystery contains at most 2 rows, one corresponding to cities where 'is_majority' is True and one corresponding to cities where 'is_majority' is False. Furthermore, recall that we used the .count() aggregation method, which means that the entries in each column of mystery contain the number of cities where 'is_majority' is True and the number of cities where 'is_majority' is False.

If mystery.get('city').iloc[0] == mystery.get('city').iloc[1], it must be the case that the number of cities where 'is_majority' is True and False are equal. This must mean that in exactly half of the cities in sky, it is true that the majority of skyscrapers are made of 'concrete'.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 69%.

Answer: Option 2

Option 1: Let’s look at the code piece by piece to understand why this does not work. games.get("Rating") gives you the column "Rating" as a Series. As per the note, .str.replace(",", ".") can be used on a Series, and will replace all commas with periods. The issue is with the use of the float function; the float function can convert a single string to a float, like float("3.14"), but not an entire Series of floats. This will cause an error making this option wrong.

Option 2: Once again we are getting "Rating" as a Series and replacing the commas with periods. We then apply float() to the Series, which will successfully convert all of the values into floats.

Option 3: This piece of code attempts to replace commas with nothing, which is correct for values using commas as decimal separators. However, this approach ignores values that use dots as decimal separators. Something like games.get("Rating").str.replace(",", "").str.replace(".", "").apply(int)/100 could correct this mistake.

Option 4: Again, we are getting "Rating" as a Series and replacing the commas with an empty string. The values inside of "Rating" are then converted to floats, which is fine, but remember, that the numbers are 100 times too big. This means we have altered the actual value inappropriately, which makes this option incorrect.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answer: int(date/10000)

The problem is asking us to find the code for blank (a) and get_year is asking us to find the year. Let’s use 19500812 as an example, so we need to convert 19500812 to 1950. If we plug in Option 2 for (a) we will get 1950. This is because \frac{19500812}{10000} = 1950.0812, and when int() is applied to 1950.0812 then it will drop the decimal, which returns 1950.

Option A: date / 10000 If we plugged in Option 1 into blank (a) we would get: \frac{19500812}{10000} = 1950.0812. This is a float and is not equal to 1950.

Option C: int(date % 10000) If we plugged in Option 3 into blank (a) we would get: 812. Remember, % is the operation to find the remainder. We can manually find the remainder by doing \frac{19500812}{10000} = 1950.0812, then looking at the decimal, and noticing 812 cannot be divided by 10000 evenly. This is an int, but not the year.

Option D: int((date % 10000) / 10000) If we plugged in Option 4 into blank (a) we would get: 0.0812. We get this number by once again looking at the remainder of \frac{19500812}{10000} = 1950.0812, which is 812, and then dividing 812 by 10000. This is a float and is not equal to 1950.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answer: date % 100

The problem is asking us to find the code for blank (b) and get_day is asking us to get the day. Let’s use 19500812 as the example again, so we need to convert 19500812 to 12. Remember, % is the operation to find the remainder. If we plug in Option 4 for (b) we will get 12. This is because \frac{19500812}{100} = 195008.12 and by looking at the decimal place we notice 12 cannot be divided by 100 evenly, making the remainder 12.

Option 1: int(date / 100) If we plugged in Option 1 into blank (b) we would get 195008. This is because Python would do: \frac{19500812}{100} = 195008.12 then would drop the decimal due to the int() function. This is not the day.

Option 2: int(date / 1000000) If we plugged in Option 2 into blank (b) we would get 19. This is because Python would do: \frac{19500812}{1000000} = 19.500812 then would drop the decimal due to the int() function. This is a day, but not the one we are looking for.

Option 3: int((date % 100) / 10000) If we plugged in Option 3 into blank (b) we would get 0. This is because Python works from the inside out, so it will first evaluate the remainder: \frac{19500812}{100} = 195008.12, by looking at the decimal place we notice 12 cannot be divided by 100 evenly, making the remainder 12. Python then does \frac{12}{10000} = 0.0012. Remember that int() drops the decimal, so by plugging a date into this code it will return 0, which is not a day.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 56%.

Answer: int(date/10000)

The problem is asking us to find the code for blank (a) and get_year is asking us to find the year. Let’s use 19500812 as an example, so we need to convert 19500812 to 1950. If we plug in Option 2 for (a) we will get 1950. This is because \frac{19500812}{10000} = 1950.0812, and when int() is applied to 1950.0812 then it will drop the decimal, which returns 1950.

Option A: date / 10000 If we plugged in Option 1 into blank (a) we would get: \frac{19500812}{10000} = 1950.0812. This is a float and is not equal to 1950.

Option C: int(date % 10000) If we plugged in Option 3 into blank (a) we would get: 812. Remember, % is the operation to find the remainder. We can manually find the remainder by doing \frac{19500812}{10000} = 1950.0812, then looking at the decimal, and noticing 812 cannot be divided by 10000 evenly. This is an int, but not the year.

Option D: int((date % 10000) / 10000) If we plugged in Option 4 into blank (a) we would get: 0.0812. We get this number by once again looking at the remainder of \frac{19500812}{10000} = 1950.0812, which is 812, and then dividing 812 by 10000. This is a float and is not equal to 1950.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answer: date % 100

The problem is asking us to find the code for blank (b) and get_day is asking us to get the day. Let’s use 19500812 as the example again, so we need to convert 19500812 to 12. Remember, % is the operation to find the remainder. If we plug in Option 4 for (b) we will get 12. This is because \frac{19500812}{100} = 195008.12 and by looking at the decimal place we notice 12 cannot be divided by 100 evenly, making the remainder 12.

Option 1: int(date / 100) If we plugged in Option 1 into blank (b) we would get 195008. This is because Python would do: \frac{19500812}{100} = 195008.12 then would drop the decimal due to the int() function. This is not the day.

Option 2: int(date / 1000000) If we plugged in Option 2 into blank (b) we would get 19. This is because Python would do: \frac{19500812}{1000000} = 19.500812 then would drop the decimal due to the int() function. This is a day, but not the one we are looking for.

Option 3: int((date % 100) / 10000) If we plugged in Option 3 into blank (b) we would get 0. This is because Python works from the inside out, so it will first evaluate the remainder: \frac{19500812}{100} = 195008.12, by looking at the decimal place we notice 12 cannot be divided by 100 evenly, making the remainder 12. Python then does \frac{12}{10000} = 0.0012. Remember that int() drops the decimal, so by plugging a date into this code it will return 0, which is not a day.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 56%.

Answer: float(lat_long.strip("NW"))

According to the hint, .strip() takes as input of a string of characters and removes all instances of those characters at the beginning or the end of the string. The input we are given, lat_long is the latitude and longitude, so we able to take it and use .strip() to remove the "N" and "W". However, it is important to mention the number we now have is still a string, so we put float() around it to convert the string to a float.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 70%.

Answer: Option 1 and Option 3

Option 1 is correct because it applies the function lat_long_numerical to the 'Latitude' and 'Longitude' columns. It then drops the old 'Latitude' and 'Longitude' columns. Then creates new 'Latitude' and 'Longitude' columns that contain the float versions.

Option 3 is correct because it applies the function lat_long_numerical to the 'Latitude' and 'Longitude' columns. It then re-assigns the 'Latitude' and 'Longitude' columns to the float versions.

Option 2 is incorrect because it re-assigns the 'Latitude' and 'Longitude' columns to the float versions and then drops the 'Latitude' and 'Longitude' columns from the DataFrame.

Difficulty: ⭐️⭐️

The average score on this problem was 86%.

Answer: Way 1 and Way 4

• Way 1: This function first checks if the year is divisible by 4, and returns “Summer” if it is. If the year isn’t divisible by 4, then the code inside that if statement won’t execute, and we move on to the next line of code, which just returns “Winter”, as we wanted. So, way 1 is correct.
• Way 2 looks similar to way 1, but has one key difference: the return "Winter" line is before the if statement. Since nothing after a return statement gets executed (assuming that the return statement gets executed), no matter what the year is this function will always return “Winter”. So, way 2 is incorrect.
• Way 3 doesn’t account for the fact that all years which are multiples of 4 are also multiples of 2. So even though it gets the 2022 Winter Olympics correct, it will also return “Winter” for 2020 since 2020 % 2 evaluates to 0. So, way 3 is incorrect.
• Way 4 uses similar logic to way 1, just using a different method to check for divisibility. Instead of using the modulo operator, we check if casting year / 4 to an integer using int changes its value. If the two aren’t equal, then we know that year / 4 wasn’t an integer before casting, which means that the year isn’t divisible by 4 and we should return "Winter". If the code inside the if statement doesn’t execute, then we know that the year is divisible by 4, so we return “Summer”. So, way 4 is correct.

Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer: We would need four parameters:

• df, the DataFrame to change.

• row, the row label or row number of the entry to change.

• col, the column label of the entry to change.

• val, the value that we want to store at that location.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 43%.

Answer: 0

First inspect the function parameters. With the example logwarts(["RQ", "GQ", "RB", "GS"], "Gryffindor", "Ravenclaw"), we observe game_log will be a list of strings, and team1 and team2 will be the full name of the respective competing houses. We can infer from the given structure that our code will

1. Initialize two scores variables for the two houses,

2. Run a for loop through all the entries in the list, update score based on given conditions,

3. Return the scores calculated by the loop.

To set up score_1 and score_2 so we can accumulate them in the for loop, we first set both equal to 0. So blank (a) will be 0.

Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer: action[0]

We observe the for loop iterates over the list of actions, where each action is a two letter string in the game_log array. Recall question statement- “Each element of a game log is a two-letter string where the first letter represents the house that performed the action”. Therefore, to get the house, we will want to get the first letter of each action string. This is accessed by action[0].

Note: A common mistake here is using action.split()[0]. Recall what split() does/) - it takes in a string, splits it according to the delimiter given (if not given, it separates by blank space), and returns a list of separated strings. This means that action.split()[0] will actually return itself. Example: if action is "RQ", action.split() will split "RQ" by blank space, in this case, there are none; so it will return the one-element list ["RQ"]. Accessing the zero-th index of this list by action.split()[0] gives us "RQ" back, instead of what we actually want ("R").

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 61%.

Answer: action[1]

Following the same logic as blank (b), we get the type of ball used in the action by accessing the second character of the string, which is action[1].

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 62%.

Answer: house == team1[0]

Now enter the fun part: to figure out the correct conditions of the if-statments, we must observe the code inside our conditional blocks carefully.

Recall question statement:

• For each entry of the game log, the house is represented by the first letter. “G” for Griffindor, “H” for Hufflepuff, “R” for Ravenclaw, “S” for Slytherin.

• Quaffle (“Q”) gets 10 points,

• Snitch (“S”) gets 150 points,

• Bludger (“B”) gets no point.

• score1 is the score of the first team, and score2 is the score of the second.

We have two conditions to take care of, the house and the type of ball. How on earth do we know which one is nested and which one is on the outside? Observe in the first big if statment, we are only updating score1. This means this block takes care of the score of the first house. Therefore, blank (d) should set the condition for the first house.

Now careful! team1 and team2 are given as full house names. We can match the house variable by getting the first letter of the each team string (e.g. if team1 is Griffindor, we will get “G” to match with “G”). We want to match house with team1[0], so our final answer is house == team1[0]. Since there are only two houses here, the following else block will take care of calculating score for the second house using the same scoring scheme as we do for the first house.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

Answer: if ball == "Q"

After gracefully handling the outer conditional statement, the rest is simple. We now simply condition the scores. Here, we see score1 increments by 10, so we know this is accounting for a Quaffle. Recall ball variable represents the type of ball we have. In this case, we use if ball == "Q" to filter for Quaffles.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 69%.

Answer: elif ball == "S" or if ball == "S"

Using the same logic as blank (e), since the score is incremented by 150 here, we know this is a snitch. Using elif ball == "S" or if ball == "S" will serve the purpose. We do not need to worry about bludgers, since those do not add to the score.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 64%.

Answer (a): split(":")

We first want to separate the time string into hours and minutes, so we use split(":") to turn a string like "HH:MM" into a list of the form ["HH", "MM"].

Difficulty: ⭐️⭐️

The average score on this problem was 84%.

Answer (b): separate[0] or int(separate[0])

The hour will be the first element of the split list, so we can access it using separate[0]. This is a string, which we can convert to an int now, or later in blank (d).

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer (c): separate[1] or int(separate[1])

The minute will be the second element of the split list, so we can access it using separate[1] or separate[-1]. Likewise, this is also a string, which we can convert to an int now, or later in blank (d).

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer (d): int(hour) * 60 + int(minute) or hour * 60 + minute (depending on the answers to (b) and (c))

In order to convert hours and minutes into minutes since midnight, we have to multiply the hours by 60 and add the number of minutes. These operations must be done with ints, not strings.

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer (e): apply(to_minutes)

In order to find the time between when an order is placed and completed, we first need to convert the "Start" and "End" times to minutes since midnight, so that we can subtract them. We apply the function we wrote in Problem 3.1 to get the values of "Start" and "End" in minutes elapsed since midnight.

Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answer (f): end_min - start_min

Now that we have the values of "Start" and "End" columns in minutes, we simply need to subtract them to find the wait time. Note that we are subtracting two Series here, so the subtraction happens element-wise by matching up corresponding elements.

Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answer (g): orders.shape[0]

Since "Wait" was already calculated in the previous question by subtracting "End" minus "Start", we know that positive values indicate where "End" is after "Start". When we evaluate orders.get("Wait") = 0, we get a Boolean Series where True indicates rows where "End" is after "Start". Taking the .sum() of this Boolean Series counts how many rows satisfy this condition, using the fact that True is 1 and False is 0 in Python. If "End" is always after "Start", this sum should equal the total number of rows in the DataFrame, which is orders.shape[0]. Therefore, comparing the sum to orders.shape[0] will return True exactly when "End" is after "Start" for all rows.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 36%.

Answer (h): groupby('Dining Hall').mean()

Since we want to compare the average wait time for each Dining Hall we know we need to aggregate each Dining Hall and find the mean of the "Wait" column. This can be done using groupby('Dining Hall').mean().

Difficulty: ⭐️⭐️⭐️

The average score on this problem was 62%.

Answer (i): index or reset_index().get('Dining Hall')

The desired output is an array with the names of each dining hall. After grouping and sorting, the dining hall names become the index of our DataFrame. Since we want an array of these names in descending order of wait times, we can either use .index to get them directly, or reset_index().get("Dining Hall") to convert the index back to a column and select it.

Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 43%.

Answer: bar chart

"Dining Hall" is a categorical variable and the average wait time is a numerical variable, so a bar chart should be used. Our visualization would have a bar for each dining hall, and the length of that bar would be proportional to the average wait time.

Difficulty: ⭐️⭐️

The average score on this problem was 89%.