Answer: str

When we do a groupby on the Chain column in hotels, this means that the values in the Chain column will be the indices of the DataFrame or Series we get as output, in this case the Series hotels.groupby("Chain")["Number of Rooms"].sum().

Since the values of Chain are strings, and since .idxmax() will return a value from the index of the aforementioned Series, summed is a string.

Answer: summed is the name of the hotel chain with the most total rooms

The result of the .groupby() and .sum() is a Series indexed by the unique Chains, whose values are the total number of rooms in hotels owned by each chain. The idxmax() function gets the index corresponding to the largest value in the Series, which will be the hotel chain name with the most total rooms.

Answer: hotel room

curious gets the most common value of Chain in the DataFrame frame. We already know that summed is the hotel chain with the most rooms in San Diego, so curious only equals summed if the most common Chain in frame is the hotel chain with the most total rooms; this occurs when each row of frame is a single hotel room.

Answers:

1. df.shape[0] >= 5 and df["Number of Rooms"].sum() >= 1000
2. "Location"
3. filter(f)
4. ["Location"].unique() or equivalent

We’d like to only consider certain neighborhoods according to group characteristics (having >= 5 hotels and >= 1000 hotel rooms), and .filter() allows us to do that by excluding groups not meeting those criteria. So, we can write a function that evaluates those criteria on one group at a time (the df of input to f is the subset of hotels containing just one Location value), and calling filter(f) means that the only remaining rows are hotels in neighborhoods that match those criteria. Finally, all we have to do is get the unique neighborhoods from this DataFrame, which are the neighborhoods for which f returned True.

Answers:

1. (cond1 & cond2).sum()
2. cond1.sum() + cond2.sum() - (cond1 & cond2).sum()

Note that cond1 and cond2 are boolean Series, and hotels[cond1] and hotels[cond2] are the subsets of hotels where Chain == "Marriott and "Location" == "Coronado", respectively.

1. When we perform an inner merge, we’re selecting every row where a Hotel Name appears in both hotels[cond1] and hotels[cond2]. This is the same set of indices (and therefore hotel names, since those are unique) as where (cond1 & cond2) == True. So, the length of combined will be the same as the number of Trues in (cond1 & cond2).

2. When we perform an outer merge, we’re selecting every row that appears in either DataFrame, although there will not be repeats for hotels that are both Marriott properties and are in Coronado. So, to find the total number of rows in either DataFrame, we take the sum of the sizes of each, and subtract rows that appear in both, which corresponds to answer cond1.sum() + cond2.sum() - (cond1 & cond2).sum().

Answer: np.arange(1, 76), replace=False, np.count_nonzero(choices == 23) > 0

Here, the idea is to randomly choose 10 different floors repeatedly, and each time, check if floor 23 was selected.

Blank (a): The first argument to np.random.choice needs to be an array/list containing the options we want to choose from, i.e. an array/list containing the values 1, 2, 3, 4, …, 75, since those are the numbers of the floors. np.arange(a, b) returns an array of integers spaced out by 1 starting from a and ending at b-1. As such, the correct call to np.arange is np.arange(1, 76).

Blank (b): Since we want to select 10 different floors, we need to specify replace=False (the default behavior is replace=True).

Blank (c): The if condition needs to check if 23 was one of the 10 numbers that were selected, i.e. if 23 is in choices. It needs to evaluate to a single Boolean value, i.e. True (if 23 was selected) or False (if 23 was not selected). Let’s go through each incorrect option to see why it’s wrong:

• Option 1, choices == 23, does not evaluate to a single Boolean value; rather, it evaluates to an array of length 10, containing multiple Trues and Falses.
• Option 2, choices is 23, does not evaluate to what we want – it checks to see if the array choices is the same Python object as the number 23, which it is not (and will never be, since an array cannot be a single number).
• Option 4, np.count_nonzero(choices) == 23, does evaluate to a single Boolean, however it is not quite correct. np.count_nonzero(choices) will always evaluate to 10, since choices is made up of 10 integers randomly selected from 1, 2, 3, 4, …, 75, none of which are 0. As such, np.count_nonzero(choices) == 23 is the same as 10 == 23, which is always False, regardless of whether or not 23 is in choices.
• Option 5, choices.str.contains(23), errors, since choices is not a Series (and .str can only follow a Series). If choices were a Series, this would still error, since the argument to .str.contains must be a string, not an int.

By process of elimination, Option 3, np.count_nonzero(choices == 23) > 0, must be the correct answer. Let’s look at it piece-by-piece:

• As we saw in Option 1, choices == 23 is a Boolean array that contains True each time the selected floor was floor 23 and False otherwise. (Since we’re sampling without replacement, floor 23 can only be selected at most once, and so choices == 23 can only contain the value True at most once.)
• np.count_nonzero(choices == 23) evaluates to the number of Trues in choices == 23. If it is positive (i.e. 1), it means that floor 23 was selected. If it is 0, it means floor 23 was not selected.
• Thus, np.count_nonzero(choices == 23) > 0 evaluates to True if (and only if) floor 23 was selected.

Answer: \frac{2}{5}

If the five "artist_names" in today and genres are the same, the DataFrame that results from an inner merge will have 15 rows, one for each row in today. This is because there are 3 matches for "harry styles", 3 matches for "olivia rodrigo", 3 matches for "glass animals", and so on.

In the merged DataFrame’s 15 rows, 6 of them will correspond to "Pop" artists — 3 to "harry styles" and 3 to "olivia rodrigo". Thus, the fraction of rows that contain "Pop" in the "genre" column is \frac{6}{15} = \frac{2}{5} (which is the fraction of rows that contained "Pop" in genres["genre"], too).

Answer: \frac{1}{2}

If we perform an inner merge, there will only be 6 rows in the merged DataFrame — 3 for "olivia rodrigo" and 3 for "drake". 3 of those 6 rows will have "Pop" in the "genre" column, hence the answer is \frac{3}{6} = \frac{1}{2}.

Answer: \frac{2}{9}

Since we are performing an outer merge, we can decompose the rows in the merged DataFrame into three groups:

• Rows that are in today that are not in genres. There are 9 of these (3 each for the 3 artists that are in today and not genres). today doesn’t have a "genre" column, and so all of these "genre"s will be NaN upon merging.

• Rows that are in genres that are not in today. There are 3 of these — one for "harry styles", one for "glass animals", and one for "doja cat". 1 of these 3 have "Pop" in the "genre" column.

• Rows that are in both today and genres. There are 6 of these — 3 for "olivia rodrigo" and 3 for "drake" — and 3 of those rows contain "Pop" in the "genre" column.

Tallying things up, we see that there are 9 + 3 + 6 = 18 rows in the merged DataFrame overall, of which 0 + 1 + 3 = 4 have "Pop" in the "genre" column. Hence, the relevant fraction is \frac{4}{18} = \frac{2}{9}.

Answer: Option B

Explanation given in part d) below

Answer: Option D

Explanation given in part d) below

Answer: Option C

Explanation given in part d) below

Answer: Option A

• First, note that in Option B, all three missing values are filled in with the same number, 7. The mean of the observed values in random_10["genre rank"] is 7, so we must have performed unconditional mean imputation in Option B. (Technically, it’s possible for Option B to be the result of unconditional probabilistic imputation, but we stated that each option could only be used once, and there is another option that can only be unconditional probabilistic imputation.)

• Then note that in Option C, the very last missing value (in the "Pop" "genre") is filled in with a 7, which is not the mean of the observed "Pop" values, but rather a value from the "Alternative" "genre". This must mean that unconditional probabilistic imputation was used in Option C, since that’s the only way a value from a different group can be used for imputation (if we are not performing some sort of mean imputation).

• This leaves Option A and Option D. The last two missing values (the two in the "Pop" "genre") are both filled in with the same value, 2 in Option A and 5 in Option D. The mean of the observed values for the "Pop" "genre" is \frac{9+2+4}{3} = 5, so mean imputation conditional on "genre" must have been used in Option D and thus probabilistic imputation conditional on "genre" must have been used in Option A.

Answer: Only product 1 will not be matched; the pattern only looks for “oz” so the “Ounce” is not identified. See this Regex101 link for an illustration.

Answer: 1, 2, and 4 are either not matched or done so incorrectly. This pattern matches either numbers+any character+numbers+[a space]+“oz” OR “Ounce”. However it fails in the following ways:

• For index 1, the pattern matches too little: only “Ounce” when it should match “1.75 Ounce”.
• For index 2, “3 oz” is missed entirely, as the number is a single digit.
• For index 4, the pattern matches too much: “2!6 oz” when it should match of “6 oz”.

See this Regex101 link for an interactive illustration.

Answer: Tree D

Following tree D in the image from top to bottom, we can follow the nesting of tags in the HTML file to verify that the DOM tree D matches the syntax of the HTML file.

Answer: Code Snippet B

Code Snippet B is the only option that finds the tag p with the attribute class being equal to instock availability and then getting the text contained in that tag, which is equal to ‘instock availability’.

Option A will cause an error because of .get('icon-ok') since 'icon-ok' is not the name of the attribute, but is instead the value of the class attribute.

Option C and D will both get the text of the i tag, which is '' and is therefore incorrect.

Answer: Code Snippet C

Code Snippet C finds the first occurence of the tag p, gets the contents of its class attribute as a list, and returns the last element, which is the rating 'Three' as desired.

Option A will error because .get('class') returns ['product_pod'] and strip cannot be used on a list, but also the content of the list does not bring us closer to the desired result.

Option B gets the text contained within the first p tag as a list, which is [''].

Answer: \vec{r}_1 and \vec{r}_2, and \vec{r}_2 and \vec{r}_3

The cosine similarity of two vectors \vec{a} and \vec{b} is \frac{\vec{a} \cdot \vec{b}}{||\vec{a}|| \cdot ||\vec{b}||}.

• The cosine similarity of \vec{r}_1 and \vec{r}_2 is: \frac{2 \cdot 4 + 3 \cdot 4}{\sqrt{13} \cdot \sqrt{32}} = \frac{20}{4 \cdot \sqrt{26}} = \frac{5}{\sqrt{26}}
• The cosine similarity of \vec{r}_1 and \vec{r}_3 is: \frac{2 \cdot 6 + 3 \cdot 4}{\sqrt{13} \cdot \sqrt{52}} = \frac{24}{26} = \frac{12}{13}
• The cosine similarity of \vec{r}_2 and \vec{r}_3 is: \frac{4 \cdot 6 + 4 \cdot 4}{\sqrt{32} \cdot \sqrt{52}} = \frac{40}{8 \cdot \sqrt{26}} = \frac{5}{\sqrt{26}}

\frac{12}{13} \approx 0.9231 and \frac{5}{\sqrt{26}} \approx 0.9806. Since larger cosine similarities mean more similar vectors, our answer is vector pairs \vec{r}_1, \vec{r}_2 AND \vec{r}_2, \vec{r}_3.

Note that we could’ve answered the question without finding the cosine similarity between \vec{r}_1 and \vec{r}_3. Remember, the cosine similarity between two vectors is the cosine of the angle between the two vectors.

• The angle between \vec{r}_1 and \vec{r}_3 is clearly bigger than the angle between \vec{r}_1 and \vec{r}_2, because \vec{r}_2 is in between \vec{r}_1 and \vec{r}_3, so we can rule out \vec{r}_1 and \vec{r}_3.
• The question then boils down to comparing the angle between \vec{r}_1, \vec{r}_2 and the angle between \vec{r}_2, \vec{r}_3. We can compute the cosine similarities of both pairs, as we did above. But, there’s a slightly easier way!
• Specifically, note that \vec{r}_3 = \begin{bmatrix} 6 \\ 4 \end{bmatrix} = 2 \begin{bmatrix} 3 \\ 2 \end{bmatrix}, i.e. it is a scalar multiple of \begin{bmatrix} 3 \\ 2 \end{bmatrix}, meaning \vec{r}_3 points in the same direction as \begin{bmatrix} 3 \\ 2 \end{bmatrix}, just is twice the length. This means that the angle between \vec{r}_2 and \vec{r}_3 is the same as the angle between \vec{r}_2 and \begin{bmatrix} 3 \\ 2 \end{bmatrix} (and remember, if two pairs of vectors have the same angle between them, they have the same cosine similarity).
• Why does that matter? Because, now we’re comparing the cosine similarity between: \vec{r}_1 = \begin{bmatrix} 2 \\ 3 \end{bmatrix}, \vec{r}_2 = \begin{bmatrix} 4 \\ 4 \end{bmatrix} \\ \\ \text{ and } \\ \\ \vec{r_2} = \begin{bmatrix} 4 \\ 4 \end{bmatrix}, \frac{1}{2}\vec{r}_3 = \begin{bmatrix} 3 \\ 2 \end{bmatrix}
• Because of symmetry (work this out yourself!), the two pairs of vectors have the same cosine similarity, and thus the same angle!

Answer: 0

First, it’s worth discussing what information we have.

1. bow_df.sum(axis=0) computes the sum of each column of bow_df. Each column of bow_df refers to a specific word, so the sum of a column in bow_df tells us the number of occurrences of a particular word across the entire corpus (all documents).
2. bow_df.sum(axis=1) computes the sum of each row of bow_df. Each row of bow_df refers to a specific document, so the sum of a row in bow_df tells us the number of words in a particular document.
3. bow_df.loc[0, 'pur'] being 0 tells us that the word "pur" appears 0 times in document 0.
4. (bow_df["paperboard"] > 0).sum() being 20 means that there are 20 documents that contain the word "paperboard".
5. bow_df["gum"].sum() being 41 means that "gum" appears 41 times across all documents.

Now, back to the subpart at hand. The TF-IDF of “pur” in document 0 is 0, because bullet point 3 above tells us that “pur” doesn’t occur at all in document 0. This means that the term frequency of “pur” in document 0 is 0, which means the TF-IDF (which is the product of the term frequency and inverse document frequency) of “pur” in document 0 is also 0, because 0 multiplied by the IDF of “pur” must be 0.

Answer: Need more information

Let’s try and compute the TF-IDF of “gum” in document 0. The formula is as follows:

\text{tfidf}(\text{gum}, \text{document 0}) = \text{tf}(\text{gum}, \text{document 0}) \cdot \text{idf}(\text{gum})

= \frac{\# \text{ of words in document 0 equal to gum}}{\# \text{ of words in document 0}} \cdot \log \left( \frac{\text{ total \# of documents}}{\text{total \# of documents containing gum}} \right)

= \frac{1}{21} \cdot \log \left( \frac{40}{???} \right)

We don’t know the number of documents containing “gum” – all we know (from bullet point 5 in the previous solution) is that “gum” appears 41 times across all documents, but we don’t know how many unique documents contain “gum”. So, we need more information.

Answer: \frac{1}{22}

Let’s try and compute the TF-IDF of “paperboard” in document 1. The formula is as follows (assuming a base-2 logarithm):

\text{tfidf}(\text{"paperboard"}, \text{document 1}) = \text{tf}(\text{"paperboard"}, \text{document 1}) \cdot \text{idf}(\text{"paperboard"})

= \frac{\text{\# of words in document 1 equal to "paperboard"}}{\text{\# of words in document 1}} \cdot \log \left( \frac{\text{total \# of documents}}{\text{total \# of documents containing "paperboard"}} \right)

= \frac{\text{1}}{\text{22}} \cdot \log \left( \frac{\text{40}}{\text{20}} \right) = \frac{\text{1}}{\text{22}} \cdot \log(\text{2}) = \frac{\text{1}}{\text{22}} \cdot \text{1} = \frac{\text{1}}{\text{22}}

30.

The minimizer of empirical risk for the constant model when using zero-one loss is the mode.

7.

The minimizer of empirical risk for the constant model when using absolute loss is the median. If the bar at 30 wasn’t there, the median would be 6, but the existence of that bar drags the “halfway” point up slightly, to 7.

11.

The minimizer of empirical risk for the constant model when using squared loss is the mean. The mean is heavily influenced by the presence of outliers, of which there are many at 30, dragging the mean up to 11. While you can’t calculate the mean here, given the large right tail, this question can be answered by understanding that the mean must be larger than the median, which is 7, and 11 is the next biggest option.

15.

The minimizer of empirical risk for the constant model when using infinity loss is the midrange, i.e. halfway between the min and max.

R_{\text{midterm}}(h)=\frac{1}{n}\sum_{i=1}^{n}[(\alpha y_i - h)^2+\lambda h]=[\frac{1}{n}\sum_{i=1}^{n}(\alpha y_i - h)^2] +\lambda h

h^*=\alpha \bar{y} - \frac{\lambda}{2}

---

\begin{align*} \frac{d}{dh}R_{\text{midterm}}(h)&= [\frac{2}{n}\sum_{i=1}^{n}(h- \alpha y_i )] +\lambda \\ &=2 h-2\alpha \bar{y} + \lambda. \end{align*}

By setting \frac{d}{dh}R_{\text{midterm}}(h)=0 we get 2 h^*-2\alpha \bar{y} + \lambda=0 \Rightarrow h^*=\alpha \bar{y} - \frac{\lambda}{2}.