Answer: a) streams['date'] == day_str, b) (200 - n), c) day_only['track_name']

The first line in the function gives us an idea that maybe later on in the function we’re going to filter for all the days that match the given data. Indeed, in blank a, we filter for all the rows in which the 'date' column matches day_str. In blank b, we could access directly access the row with the n-th most stream using iloc. (Remember, the image above shows us that the streams are sorted by most streamed in ascending order, so to find the n-th most popular song of a day, we simply do 200-n). Finally, to return the track name, we could simply do day_only['track_name'].

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 63%.

Answer: See below.

pivoted.apply(lambda s: s.isna().sum(), axis=1) computes the number of days that a song was not on the Top 200, so 31 - pivoted.apply(lambda s: s.isna().sum(), axis=1) computes the number of days the song was in the Top 200. As such, the correct interpretation is that "pepas" was in the Top 200 for 23 days in May.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 68%.

Answer: Option A, B and D

For each combination of "track_name" and "date", there is just a single value — the number of streams that song received on that date. As such, the aggfunc needs to take in a Series containing a single number and return that same number.

• The mean and median of a Series containing a single number is equal to that number, so the first two options are correct.

• The length of a Series containing a single number is 1, no matter what that number is, so the third option is not correct.

• lambda df: df.iloc[0] takes in a Series and returns the first element in the Series, which is the only element in the Series. This option is correct as well. (The parameter name df is irrelevant.)

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Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer: See below.

1in Since streams is sorted by "date" in descending order and, within each "date", by "streams" in ascending order, streams.groupby("date").last() is a DataFrame containing the song with the most "streams" on each day in May. In other words, we found the “top song" for each day. (The DataFrame we created has 31 rows.)

When we then execute .groupby(["artist_names", "track_name"]).count(), we create one row for every unique combination of song and artist, amongst the “top songs". (If no two artists have a song with the same name, this is the same as creating one row for every unique song.) Since there are 5 rows in this new DataFrame (resetting the index doesn’t do anything here), it means that there were only 5 unique songs that were ever the “top song" in a day in May; this is the correct interpretation.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 50%.

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).

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Difficulty: ⭐️

The average score on this problem was 97%.

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}.

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Difficulty: ⭐️⭐️

The average score on this problem was 86%.

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}.

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 29%.

Answer: \frac{1}{2}

First, we must normalize their distributions so they sum to 1. \text{billy} = \begin{bmatrix} \frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6} \end{bmatrix}, \: \: \: \text{daisy} = \begin{bmatrix} \frac{1}{2}, 0, 0, \frac{1}{4}, 0, \frac{1}{4} \end{bmatrix}

Then, recall that the TVD is the sum of the absolute differences in proportions, all divided by 2:

\begin{aligned} \text{TVD} &= \frac{1}{2} \Bigg( \Big| \frac{1}{2} - \frac{1}{6} \Big| + \Big| 0 - \frac{1}{6} \Big| + \Big| 0 - \frac{1}{6} \Big| + \Big| \frac{1}{4} - \frac{1}{6} \Big| + \Big| 0 - \frac{1}{6} \Big| + \Big| \frac{1}{4} - \frac{1}{6} \Big| \Bigg) \\ &= \frac{1}{2} \Bigg( \frac{2}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{12} + \frac{1}{6} + \frac{1}{12} \Bigg) \\ &= \frac{1}{2} \cdot 1 \\ &= \frac{1}{2}\end{aligned}

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 42%.

Answer: Smallest: 0, Largest: 1

There is an absolute value in the definition of TVD, so it must be at least 0. The TVD does not scale with the number of categories; its maximum possible value is 1. We will not provide a rigorous proof here, but intuitively, the most “different" two distributions can be is if they both have 100% of their values in different categories. In such a case, the TVD is 1.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 74%.

Answer: Option C

• These are two quantitative distributions, and the total variation distance only applies for categorical distributions.

• These two distributions have similar means, so the difference in means and absolute difference in means won’t be able to tell them apart.

• As such, the correct answer is the Kolmogorov-Smirnov statistic, which roughly measures the largest “gap" between two cumulative distribution functions.

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Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answer: Option C: "track_name"

The first thing to notice is that first line in the function filters for all the rap songs in df. Next, since the helper function is supposed to return the number of unique Hip-Hop/Rap songs in a given DataFrame, we group by "track_name" so that each group/category is a unique song. Grouping by "year", "artist_names", or "genre" won’t help us compute the number of unique songs.

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Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer: Option A: .shape

Note that after grouping by "track_name", each individual row in the resulting dataframe is a unique song within the rap genre. Thus the number of rows in the resulting dataframe is just the number of unique rap songs in a dataframe. To get the number of rows of a dataframe, we simply do .shape[0]

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Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer: Option D: 200 * 365

Consider the following statement from the problem: “To generate data under the null, you decide to treat the songs that were in the Top 200 in 2021 as a random sample from the songs that were in the Top 200 in all ten years.” Thus it follows that we need to generate 200 * 365 songs since there are 200 * 365 songs in the Top 200 in 2021.

Alternatively, it makes sense to sample 200 * 365 songs from all_years because we use the number of unique rap songs as a test statistic for each sample. Thus our sample should be the same size as the number of songs in the Top 200 in 2021.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 52%.

Answer: (counts >= count_2021)

Note that counts is an array that contains the number of Hip-Hop/Rap songs from each sample. Because our alternative hypothesis is that “The number of unique Hip-Hop/Rap songs that were popular in 2021 is greater than the average number of unique Hip-Hop/Rap songs that were popular each year between 2012 and 2021”, we’re interested in the proportion of simulated test statistics that were equal to our observed or greater. counts >= count_2021 is an array of Booleans, denoting whether or not count_2021 was greater than the number of rap songs in each sample, and (counts >= count_2021).mean() is the proportion (p-value) we’re looking for.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 53%.

Answer: Option B

Explanation given in part d) below

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Difficulty: ⭐️

The average score on this problem was 99%.

Answer: Option D

Explanation given in part d) below

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Difficulty: ⭐️

The average score on this problem was 96%.

Answer: Option C

Explanation given in part d) below

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Difficulty: ⭐️

The average score on this problem was 92%.

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.

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Difficulty: ⭐️

The average score on this problem was 92%.

Answer: Total Variation Distance (TVD)

We are comparing two distributions:

• The distribution of "genre" when "genre rank" is missing.

• The distribution of "genre" when "genre rank" is not missing.

Since the distribution of "genre" is categorical, the above two distributions are also categorical. The only test statistic we have for comparing categorical distributions is the total variation distance (TVD).

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 63%.

Answer: Option D: Probabilistic imputation conditional on "genre"

Mean imputation does not preserve the variance of the imputed values — since it fills in all missing numbers with the same number (either overall, or within each group), the variance of the imputed dataset is less than the pre-imputed dataset. To preserve the variance of the imputed values, we must use probabilistic imputation of some sort. Since the missingness of "genre rank" was found to be dependent on "genre", we perform probabilistic imputation conditional on "genre", i.e. impute "genre rank"s randomly within each "genre".

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Difficulty: ⭐️⭐️

The average score on this problem was 75%.

Answer: Option B: NMAR or Option C: MAR

We accepted two answers here — not missing at random and missing at random.

• MCAR is ruled out right away, since there is some “pattern" to the missingness, i.e. some sort of relationship between "day_before_yest" and the missingness of "yest".

• One could argue not missing at random, because stream counts are more likely to be missing from "yest" if they are smaller. A song is missing from "yest" but present in "day_before_yest" if its number of streams was in the Top 200 yesterday but not today; if this is true, this must mean that its number of streams is less than any of the songs whose stream counts are actually in "yest".

• One could also argue missing at random, because the missingness of "yest" does indeed depend on "day_before_yest".

• Missing by design is not a valid answer here. While it is true that "day_before_yest" tells you something about the missingness of "yest", it is not the case that you can 100% of the time predict if "yest" will be missing just by looking at "day_before_yest"; this would need to be the case if "yest" were missing by design.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 64%.

Answer: 14

There’s 1 <title>, 1 <h1>, 3 <th>s, and 9 <td>s, adding up to 14.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 64%.

Answer: 9

As mentioned in the solution to the part above, there are 9 <td> nodes, and soup.find_all finds them all.

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Difficulty: ⭐️

The average score on this problem was 95%.

Answer: ["heading"] or "heading"

soup.find("tr") finds the first occurrence of a <tr> node, and get("class") accesses the value of its "class" attribute.

Note that technically the answer is ["heading"], but "heading" received full credit too.

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Difficulty: ⭐️

The average score on this problem was 96%.

Answer: a) {'class' : n} b) [2].text or [-1].text

The logic is to find the <tr> node with the correct class attribute (which we do by setting attr to {'class' : 2}), then access the text of the node’s last <td> child (since that’s where the song titles are stored).

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 66%.

Answer: Option B: False

A status code of 200 means that the request has succeeded. Hence, the response could be JSON, it is not necessarily HTML.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 44%.

Answer: One solution is given below.

Click this link to interact with the solution on regex101.

While grading, we were not particular about students’ treatment of uppercase characters in country names.

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: Option B, Option C, and Option E

Let’s first dissect the regular expression into manageable groups:

• "^" matches the regex to its right at the start of a given string
• "\w{2,5}" matches alphanumeric characters (a-Z, 0-9 and _) 2 to 5 times inclusively. (Note the that it does indeed match with the underscore)
• "." is a basic wildcard
• "\d*" matches digits (0-9), at least 0 times
• "\/" matches the "/" character
• "[^A-Z5]{1,}" matches any character that isn’t (A-Z or 5) at least once.

Thus using these rules, it’s not hard to verify that Options B, C and E are matches.

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Difficulty: ⭐️⭐️

The average score on this problem was 85%.

Answer: See below

1. "\b" matches “word boundaries", which are any locations that separate words. As such, there are 4 matches — ["doja", "cat", "you", "right"]. Thus the answer is 4.

2. The 3 matches are [" cat", " you", " right"]. Thus the answer is 3.

3. This was quite tricky! The key is remembering that re.findall only finds non-overlapping matches (if you look at the solutions to the above two parts, none of the matches overlapped). Reading from left to right, there is only a single non-overlapping match: "cat". Sure, " you " also matches the pattern, but since the space after "cat" was already “found" by re.findall, it cannot be included in any future matches. Thus the answer is 1.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 60%.

Answer: \frac{1}{6}

There are 12 words in Song 0’s title, and 2 of them are "hate", so the term frequency of "hate" in Song 0’s title is \frac{2}{12} = \frac{1}{6}.

There are 4 documents total, and 2 of them contain "hate" (Song 0’s title and Song 3’s title), so the inverse document frequency of "hate" in the corpus is \log_2 \left( \frac{4}{2} \right) = \log_2 (2) = 1.

Then, the TF-IDF of "hate" in Song 0’s title is

\text{TF-IDF} = \frac{1}{6} \cdot 1 = \frac{1}{6}

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Difficulty: ⭐️⭐️

The average score on this problem was 86%.

Answer: Option A: "i"

It was not necessary to compute the TF-IDFs of all words in Song 0’s title to determine the answer. \text{tfidf}(t, d) is high when t occurs often in d but rarely overall. That is the case with "i" — it is the most common word in Song 0’s title (with 4 appearances), but it does not appear in any other document. As such, it must be the word with the highest TF-IDF in Song 0’s title.

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Difficulty: ⭐️⭐️

The average score on this problem was 84%.

Answer: Option B

Recall that \text{tfidf}(t, d) = \text{tf}(t, d) \cdot \text{idf}(t), and note that \text{tf}(t, d) is just $ (t, d) $. Thus, \text{tfidf}(t, d) is 0 is if either \text{bow}(t, d) = 0 or \text{idf}(t) = 0.

So, if \text{bow}(t, d) = 0, then \text{tf}(t, d) = 0 and \text{tfidf}(t, d) = 0, so the second option is true. However, if \text{tfidf}(t, d) = 0, it could be the case that \text{bow}(t, d) > 0 and \text{idf}(t) = 0 (which happens when term t is in every document), so the first option is not necessarily true.

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Difficulty: ⭐️

The average score on this problem was 91%.

Answer: Option B: Song 2

Recall, the cosine similarity between two vectors \vec{a}, \vec{b} is computed as

\cos \theta = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b}|}

We are told that each row vector is already normalized to have a length of 1, so to compute the similarity between two songs’ titles, we can compute dot products directly.

• Song 0 and Song 1: 0.47 \cdot 0.76

• Song 0 and Song 2: 0.47 \cdot 0.58 + 0.71 \cdot 0.58

• Song 0 and Song 3: 0.47 \cdot 0.71

Without using a calculator (which students did not have access to during the exam), it is clear that the dot product between Song 0’s title and Song 2’s title is the highest, hence Song 2’s title is the most similar to Song 0’s.

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Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answer: \frac{1}{12}

Since we are using a trigram model, to compute the conditional probability of a token, we must condition on the prior two tokens. For the first and second tokens, "name" and "is", there aren’t two prior tokens to condition on, so we instead look at unigrams and bigrams instead.

First, we decompose P(\text{name is what my}):

\begin{aligned} P(\text{name is what my}) &= P(\text{name}) \cdot P(\text{is $|$ name}) \cdot P(\text{what $|$ name is}) \cdot P(\text{my $|$ is what}) \end{aligned}

• P(\text{name}) is \frac{2}{12}, since there are 12 total tokens and 2 of them are equal to "name".

• P(\text{is $|$ name}) is \frac{2}{2} = 1, because of the 2 times "name" appears, "is" always follows it.

• P(\text{what $|$ name is}) is \frac{1}{2}, because of the 2 times "name is" appears, "what" follows it once ("who" follows it the other time).

• P(\text{my $|$ is what}) is \frac{1}{1} = 1, because "is what" only appeared once, and "my" appeared right after it.

Thus: \begin{aligned} P(\text{name is what my}) &= P(\text{name}) \cdot P(\text{is $|$ name}) \cdot P(\text{what $|$ name is}) \cdot P(\text{my $|$ is what}) \\ &= \frac{2}{12} \cdot 1 \cdot \frac{1}{2} \cdot 1 \\ &= \frac{1}{12} \end{aligned}

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

Answer: False

For regression models, the score method computes R^2. A higher R^2 indicates a better linear fit. If the training R^2 is much greater than the testing R^2, that is indication of overfitting. However, that is not the case here — here, we were asked what happens if the training R^2 is much less than the testing R^2, which is not indication of overfitting (it is an indication that your model, luckily, performs much better on unseen data than it does on observed data!).

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Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: 100

We are given that there are n=50 data points, and that the sum of squared errors \sum_{i = 1}^n (y_i - H(x_i))^2 is 500{,}000. Then:

\begin{aligned} \text{RMSE} &= \sqrt{\frac{1}{n} \sum_{i = 1}^n (y_i - H(x_i))^2} \\ &= \sqrt{\frac{1}{50} \cdot 500{,}000} \\ &= \sqrt{10{,}000} \\ &= 100\end{aligned}

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Difficulty: ⭐️⭐️

The average score on this problem was 87%.

Answer: 7 and 9

There are 5 unique values of "genre" and 4 unique values of "rec_label", so if we create a single one-hot encoded column for each one, there would be 5 + 4 = 9 one-hot encoded columns (which there are in lr_no_drop).

If we drop one one-hot-encoded column per category, which is what drop="first" does, then we only have (5 - 1) + (4 - 1) = 7 one-hot encoded columns (which there are in lr_drop).

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Difficulty: ⭐️⭐️

The average score on this problem was 75%.

Answer: Option B: roughly equal to

Multicollinearity does not impact a linear model’s ability to make predictions (even on unseen data), it only impacts the interpretability of its coefficients. As such, the testing RMSE of both Model 3 and Model 4 will be roughly the same.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 70%.

Answer: \frac{1}{1000}

“My favorite problem on the exam!" -Suraj

Model 4 is made up of 11 features, i.e. 11 columns.

• 4 of the columns correspond to the different values of "rec_label". Since Daisy and Billy have the same "rec_label", their values in these four columns are all the same.

• One of the columns corresponds to "speechiness". Since Daisy’s song and Billy’s song have the same "speechiness", their values in this column are the same.

• 5 of the columns correspond to the different values of "genre". Daisy is a "Pop" artist, so she has a 1 in the "genre_Pop" column and a 0 in the other four "genre_" columns, and similarly Billy has a 1 in the "genre_Country" column and 0s in the others.

• One of the columns corresponds to "danceability", and Daisy and Billy have different quantitative values in this column.

Let d_1 and d_2.

The key is in recognizing that all features in Daisy’s prediction and Billy’s prediction are the same, other than the coefficients on "genre_Pop", "genre_Country", and "danceability". Let’s let d_1 be Daisy’s song’s "danceability", and let d_2 be Billy’s song’s "danceability". Then:

\begin{aligned} 2000 + 10^{6} \cdot d_1 = 1000 + 10^{6} \cdot d_2 \\ 1000 &= 10^{6} (d_2 - d_1) \\ \frac{1}{1000} &= d_2 - d_1\end{aligned}

Thus, the absolute difference between their songs’ "danceability"s is \frac{1}{1000}.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 70%.

Answer: 2400

There are 12 \cdot 10 \cdot 5 = 600 combinations of hyperparameters. For each combination of hyperparameters, we will train a BillyClassifier with that combination of hyperparameters k = 4 times. So, the total number of BillyClassifier instances that will be trained is 600 \cdot 4 = 2400.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer: 600, 200

Since we performed k=4 cross-validation, we must divide the training set into four disjoint groups each of the same size. \frac{800}{4} = 200, so each group is of size 200. Each time we perform cross validation, one group is used for validation, and the other three are used for training, so the validation set size is 200 and the training set size is 200 \cdot 3 = 600.

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Difficulty: ⭐️⭐️

The average score on this problem was 77%.

Answer: Option B

When performing cross validation, we select the combination of hyperparameters that had the highest average validation accuracy across all four folds of the data. That is, by definition, how best_params_ came to be. None of the other options are guaranteed to be true.

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Difficulty: ⭐️⭐️

The average score on this problem was 82%.

Answer: \frac{35}{57}

There are 105 true positives and 66 false negatives. Hence, the recall is \frac{105}{105 + 66} = \frac{105}{171} = \frac{35}{57}.

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Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer: 33

Let x be the number of true negatives. The number of correctly classified data points is 105 + x, and the total number of data points is 105 + 30 + 66 + x = 201 + x. Hence, this boils down to solving for x in \frac{69}{117} = \frac{105 + x}{201 + x}.

It may be tempting to cross-multiply here, but that’s not necessary (in fact, we picked the numbers specifically so you would not have to)! Multiply \frac{69}{117} by \frac{2}{2} to yield \frac{138}{234}. Then, conveniently, setting x = 33 in \frac{105 + x}{201 + x} also yields \frac{138}{234}, so x = 33 and hence the number of true negatives our classifier has is 33.

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Difficulty: ⭐️⭐️

The average score on this problem was 84%.

Answer: True

Remember that \text{precision} = \frac{TP}{TP + FP} and \text{recall} = \frac{TP}{TP + FN}. In order for precision to be the same as recall, it must be the case that FP = FN, i.e. that our classifier makes the same number of false positives and false negatives. The only kinds of “errors" or”mistakes" a classifier can make are false positives and false negatives; thus, we must have

\text{mistakes} = FP + FN = FP + FP = 2 \cdot FP

2 times any integer must be an even integer, so the number of mistakes must be even.

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Difficulty: ⭐️

The average score on this problem was 100%.

Answer: Option B: low precision and high recall.

Our classifier is good at identifying when the input stream is “Kiss Me thru the Phone", i.e. it is good at identifying true positives amongst all positives. This means it has high recall.

Since our classifier makes many false positive predictions — in other words, it often incorrectly predicts “Kiss Me thru the Phone" when that’s not what the input stream is — it has many false positives, so its precision is low.

Thus, our classifier has low precision and high recall.

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Difficulty: ⭐️

The average score on this problem was 91%.