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.

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Difficulty: ⭐️

The average score on this problem was 97%.

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.

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Difficulty: ⭐️

The average score on this problem was 94%.

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.

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Difficulty: ⭐️

The average score on this problem was 93%.

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.

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Difficulty: ⭐️

The average score on this problem was 94%.

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'.

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Difficulty: ⭐️

The average score on this problem was 91%.

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')

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Difficulty: ⭐️⭐️

The average score on this problem was 86%.

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.

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Difficulty: ⭐️⭐️

The average score on this problem was 81%.

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].

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Difficulty: ⭐️⭐️

The average score on this problem was 86%.

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!)

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Difficulty: ⭐️

The average score on this problem was 94%.

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.

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Difficulty: ⭐️

The average score on this problem was 95%.

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:

• Among all books in books written by Charles Dickens, Oliver Twist is the title that is alphabetically last.
• Among all books in books written by Charles Dickens, 53 is the greatest number of chapters.
• Among all books in 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.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

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.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

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.

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Difficulty: ⭐️

The average score on this problem was 95%.

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.

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Difficulty: ⭐️⭐️

The average score on this problem was 89%.

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.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 70%.

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.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 56%.

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".

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 33%.

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.

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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.

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Difficulty: ⭐️⭐️

The average score on this problem was 77%.

Answer:

To draw the histogram, we first need to bin the data and figure out how many data values fall into each bin. The code includes bins=np.arange(0, 7, 2) which means the bin endpoints are 0, 2, 4, 6. This gives us three bins: [0, 2), [2, 4), and [4, 6]. Remember that all bins, except for the last one, include the left endpoint but not the right. The last bin includes both endpoints.

Now that we know what the bins are, we can count up the number of values in each bin. We need to look at the 'Appearance_Order' column of sungod.take(np.arange(5)), or the first five rows of sungod. The values there are 1, 4, 3, 1, 3. The two 1s fall into the first bin [0, 2). The two 3s fall into the second bin [2, 4), and the one 4 falls into the last bin [4, 6]. This means the proportion of values in each bin are \frac{2}{5}, \frac{2}{5}, \frac{1}{5} from left to right.

To figure out the height of each bar in the histogram, we use the fact that the area of a bar in a density histogram should equal the proportion of values in that bin. The area of a rectangle is height times width, so height is area divided by width.

For the bin [0, 2), the area is \frac{2}{5} = 0.4 and the width is 2, so the height is \frac{0.4}{2} = 0.2.

For the bin [2, 4), the area is \frac{2}{5} = 0.4 and the width is 2, so the height is \frac{0.4}{2} = 0.2.

For the bin [4, 6], the area is \frac{1}{5} = 0.2 and the width is 2, so the height is \frac{0.2}{2} = 0.1.

Since the bins are all the same width, the fact that there an equal number of values in the first two bins and half as many in the third bin means the first two bars should be equally tall and the third should be half as tall. We can use this to draw the rest of the histogram quickly once we’ve drawn the first bar.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 45%.

Answer: 4 grocery stores

We are given that the bins start at 0.8 and have a width of 0.1, which means one of the bins has endpoints 1.3 and 1.4. This bin (the tallest bar) includes all grocery stores that sold broccoli for a price greater than or equal to $1.30 per pound, but less than $1.40 per pound.

This bar has a width of 0.1 and we’d estimate the height to be around 2.2, though we can’t say exactly. Multiplying these values, the area of the bar is about 0.22, which means about 22 percent of the grocery stores fall into this bin. There are 18 grocery stores in total, as we are told in the introduction to this question. We can compute using a calculator that 22 percent of 18 is 3.96. Since the actual number of grocery stores this represents must be a whole number, this bin must represent 4 grocery stores.

The reason for the slight discrepancy between 3.96 and 4 is that we used 2.2 for the height of the bar, a number that we determined by eye. We don’t know the exact height of the bar. It is reassuring to do the calculation and get a value that’s very close to an integer, since we know the final answer must be an integer.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 71%.

Answer: 1.25

First, we need to figure out how many grocery stores the bin [\$1.10, \$1.50) contains. We already know from the previous subpart that there are four grocery stores in the bin [\$1.30, \$1.40). We could do similar calculations to find the number of grocery stores in each of these bins:

• [\$1.10, \$1.20)
• [\$1.20, \$1.30)
• [\$1.40, \$1.50)

However, it’s much simpler and faster to use the fact that when the bins are all equally wide, the height of a bar is proportional to the number of data values it contains. So looking at the histogram in the previous subpart, since we know the [\$1.30, \$1.40) bin contains 4 grocery stores, then the [\$1.10, \$1.20) bin must contain 1 grocery store, since it’s only a quarter as tall. Again, we’re taking advantage of the fact that there must be an integer number of grocery stores in each bin when we say it’s 1/4 as tall. Our only options are 1/4, 1/2, or 3/4 as tall, and among those choices, it’s clear.

Therefore, by looking at the relative heights of the bars, we can quickly determine the number of grocery stores in each bin:

• [\$1.10, \$1.20): 1 grocery store
• [\$1.20, \$1.30): 3 grocery stores
• [\$1.30, \$1.40): 4 grocery stores
• [\$1.40, \$1.50): 1 grocery store

Adding these numbers together, this means there are 9 grocery stores whose broccoli prices fall in the interval [\$1.10, \$1.50). In the new histogram, these 9 grocery stores will be represented by a bar of width 1.50-1.10 = 0.4. The area of the bar should be \frac{9}{18} = 0.5. Therefore the height must be \frac{0.5}{0.4} = 1.25.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 33%.

Answer: Yes

When we use .plot without specifying a y column, it uses every column in the DataFrame as a y column and creates an overlaid plot. Since we first use get with the list ['broccoli', 'ice_cream'], this keeps the 'broccoli' and 'ice_cream' columns from prices, so our bar chart will overlay broccoli prices with ice cream prices. Notice that this get is unnecessary because prices only has these two columns, so it would have been the same to just use prices directly. The resulting bar chart will look something like this:

Each grocery store has its broccoli price represented by the length of the blue bar and its ice cream price represented by the length of the red bar. We can therefore answer the question by simply counting the number of red bars that are shorter than their corresponding blue bars.

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Difficulty: ⭐️⭐️

The average score on this problem was 78%.

Answer: No

This will create an overlaid histogram of broccoli prices and ice cream prices. So we will be able to see the distribution of broccoli prices together with the distribution of ice cream prices, but we won’t be able to pair up particular broccoli prices with ice cream prices at the same store. This means we won’t be able to answer the question. The overlaid histogram would look something like this:

This tells us that broadly, ice cream tends to be more expensive than broccoli, but we can’t say anything about the number of stores where ice cream is cheaper than broccoli.

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Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: Yes, and there are 2 such stores.

In this scatterplot, each grocery store is represented as one dot. The x-coordinate of that dot tells the price of broccoli at that store, and the y-coordinate tells the price of ice cream. If a grocery store’s ice cream price is cheaper than its broccoli price, the dot in the scatterplot will have y<x. To identify such dots in the scatterplot, imagine drawing the line y=x. Any dot below this line corresponds to a point with y<x, which is a grocery store where ice cream is cheaper than broccoli. As we can see, there are two such stores.

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Difficulty: ⭐️⭐️

The average score on this problem was 78%.

Answer: 80%

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer: scatterplot

The number of customers is given by 'num_diners' which is an integer, and 'income' is a float. Since both are numerical variables, neither of which represents time, it is most appropriate to use a scatterplot.

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Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answer: line plot

Since we want to plot a trend of a numerical quantity ('income') over time, it is best to use a line plot.

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Difficulty: ⭐️

The average score on this problem was 95%.

Answer: 90

Remember, the key property of histograms is that the proportion of values in a bin is equal to the area of the corresponding bar. To find the number of values in the range 2-8 (the x-axis is measured in hundreds), we’ll need to find the proportion of values in the range 2-8 and multiply that by 200, which is the total number of students in DSC 10. To find the proportion of values in the range 2-8, we’ll need to find the areas of the 2-4, 4-6, and 6-8 bars.

Area of the 2-4 bar: \text{width} \cdot \text{height} = 2 \cdot 0.1 = 0.2

Area of the 4-6 bar: \text{width} \cdot \text{height} = 2 \cdot 0.0625 = 0.125.

Area of the 6-8 bar: \text{width} \cdot \text{height} = 2 \cdot 0.0625 = 0.125.

Then, the total proportion of values in the range 2-8 is 0.2 + 0.125 + 0.125 = 0.45, so the total number of students with between 200 and 800 Instagram followers is 0.45 \cdot 200 = 90.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 49%.

Answer: 400 \cdot h

As we said at the start of the last solution, the key property of histograms is that the proportion of values in a bin is equal to the area of the corresponding bar. Then, the number of students represented bar a bar is the total number of students in DSC 10 (200) multiplied by the area of the bar.

Since all bars in this histogram have a width of 2, the area of a bar in this histogram is \text{width} \cdot \text{height} = 2 \cdot h. If there are 200 students in total, then the number of students represented in a bar with height h is 200 \cdot 2 \cdot h = 400 \cdot h.

To verify our answer, we can check to see if it makes sense in the context of the previous subpart. The 2-4 bin has a height of 0.1, and 400 \cdot 0.1 = 40. The total number of students in the range 2-8 was 90, so it makes sense that 40 of them came from the 2-4 bar, since the 2-4 bar takes up about half of the area of the 2-8 range.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 36%.