Answer: ([A-Za-z]+): Available

In the above availability strings, we’re looking for a one-or-more-length sequence of letters (not numbers, as mentioned in the problem), followed by a colon and the word “Available.” However, since we only want to return the room type, and not the colon or the word “Available,” we use a capturing group (the parentheses) to just return the word before the colon.

Answers:

1. ["$350.25"]
2. ["200.75", "402.99", "350.25"]

In the first part, the string that the pattern matches is a dollar sign ($), one or more digits (\d+), a period (\.), two digits (\d{2}), followed by the end of the string ($). Since the string we’re passing into re.findall is prices, which is the concatenation of one + ", " + two, the only option that matches is the last price displayed, "$350.25". (If we had not specified the match ending with the end of the string, "$200.75" would also have been a match.)

In the second part, the beginning of the pattern is "\$?", which means to match zero or one instance of the dollar sign character. In most cases, this means that if there is a dollar sign before the remainder of the pattern, it will be included in the match, but if not, the rest of the pattern will still match. However, the capturing group around the remainder of the pattern means that in either case, only the remainder of the pattern after the dollar sign is captured. (So, in this particular example, the \$? has no effect on the output.)

The rest of the pattern is structured similarly to the previous part, except now, the pattern does not require the end of the string after the price, so we’re just selecting sequences of digits followed by a period and two more digits, which is how we get our solution.

**Answer: Option A, B, C and D

The regex pattern r'a+' searches for any string that contains at least one substring consisting of the character a one or more times. Clearly all of these strings contain a substring og the character a one or more times.

**Answer: Option A and B

The regex pattern r'a+ b+' searches for any string that contains at least one substring consisting of the character a one or more times followed by a space followed by the character b one or more times. The only strings that have this pattern are Options A and B.

**Answer: Option A

The regex pattern r'\b' matches for the boundary of a word (so like the start and end of a word which could seperated by spaces). Thus the regex pattern searches for the substring 'aa' that is its own standalone word. The only string that has this pattern is Option A.

**Answer: Option C

The regex pattern r'(aba){2,}' searches the substring consisting of aba 2 or more times. The only string that has this pattern is Option C.

**Answer: Option C

The regex pattern r'a..a' searches the substring consisting of a followed by any two characters followed by a. The only string that has this pattern is Option C.

**Answer: Option A, B, C and D

The regex pattern r'.*' searches the substring consisting of any character 0 or more times. Clearly all of the strings contain that pattern.

Answer: 'U(m+)[A-Z]([A-z]*)(\?)'

Starting our regular expression, it’s not too difficult to see that we need a 'U' followed by (m+) which will match with a singular capital 'U' followed by at least one lowercase 'm'. Next it is required that we follow that up with any string of letters, where the first letter is capitalized. we do this with '[A-Z]([A-z]*)', where '[A-Z]' will match with any capital letter and '([A-z]*)' will match with lowercase and uppercase letters 0 or more times. Finally we end the regex with '\?' which matches with a question mark.

Answer: Option A, Option B and Option E

Let’s dissect what the regex in the question actually means. First, the '\$' simply matches with any question mark. Next, '\s*' matches with whitespace characters 0 or more times, and '\d{1,3}' will match with any digits 1-3 times inclusive. Finally, '(\.\d{2})?' will match with any expression consisting of a period and any two digits following that period 0 or 1 times (due to the '?' mark). With those rules in mind, it’s not too difficult to check that Options A, B and E work.

Options C, D and F don’t work because none of those expressions have 1-3 digits before the period.

Answer: '(.*)((borough$)|(brough$)|(burgh$))'

We will assume that the question wants us to create a regex that matches with any string that ends with the strings described in the problem. First, '(.*)' will match with anything 0 or more times at the start of the expression. Then '((borough$)|(brough$)|(burgh$))' will match with any of the described strings in the problem , since '$' indicates the end of string and '|' is just or in regex.

Answer: list3

This regex pattern \d+ matches one or more digits anywhere in the string. It doesn’t concern itself with the context of the digits, whether they are inside brackets or not. As a result, it extracts all sequences of digits in s, including '10', '3', '15', '4', '5', '3', '91', '8', and '100', which together form list3. This is because \d+ greedily matches all contiguous digits, capturing both the citation numbers and any other numbers present in the text.

Answer: list5

This pattern [\d+] is slightly misleading because the square brackets are used to define a character class, and the plus sign inside is treated as a literal character, not as a quantifier. However, since there are no plus signs in s, this detail does not affect the outcome. The character class \d matches any digit, so this pattern effectively matches individual digits throughout the string, resulting in list5. This list contains every single digit found in s, separated as individual string elements.

Answer: list2

This pattern is specifically designed to match digits that are enclosed in square brackets. The \[(\d+)\] pattern looks for a sequence of one or more digits \d+ inside square brackets []. The parentheses capture the digits as a group, excluding the brackets from the result. Therefore, it extracts just the citation numbers as they appear in s, matching list2 exactly. This method is precise for extracting citation numbers from a text formatted in the verbose numeric style.

Answer: list4

Similar to the previous explanation but with a key difference: the entire pattern of digits within square brackets is captured, including the brackets themselves. The pattern \[\d+\] specifically searches for sequences of digits surrounded by square brackets, and the parentheses around the entire pattern ensure that the match includes the brackets. This results in list4, which contains all the citation markers found in s, preserving the brackets to clearly denote them as citations.

Answer: 1

Answer: 1, 2, 4

Answer: Options A and C

• Option A: First, we grab everything but 'am' / 'pm' with x[:-2]. Then splitting on ':' means that values such as '1:15' become a list like ['1', '15'], while single numbers such as '12' become a list like ['12']. We then pass in the hour value, the minute value or 0 if there is none, and then the 'am' / 'pm' value by grabbing just the last two indices of x.
• Option B: We use regex to define groups from x and return a list of those groups using .groups(). The regex matches to any number of integers '(\d+)' as the first group, then a colon ':', another group of any number of integers '(\d+)', then a final group that matches exactly twice to characters from “apm”, which can be “am” or “pm” with '([apm]{2})'. The problem with this solution is that while it does correctly match to input such as '1:15pm', it would not match to '12pm' because there is no ':' character.
• Option C: We use regex to define groups from x and return a list of those groups using .groups(). The regex matches to any number of integers '(\d+)' as the first group, then looks for a colon ':' as the second group, and matches another group to any number of integers following it with '(\d+)'. The '?' character makes the previous matches optional, so if there is no ':' followed by integers then it will return None for groups 2 and 3 and still match the rest of the string. Finally, it matches one last group that is any two characters after the previous groups. This solution works because the full regex matches the format '1:15pm', while the '?' checks so if there is no ':' it can still match something formatted like '12pm'. It also passes in the group indices 0, 2, and 3 because group 1 will just be ':' or None, so we want to skip it.
• Option D: We use regex to define groups from x and return a list of those groups using .groups(). The regex matches any character unlimited times with '.+' as the first group, then has another group that takes 3 of any character after the first group from '(.{3})?, and this group is optional because of the ending '?'. Finally, the last two characters are the last group '(..)'. This solution does not work because while the intention is that '.+' grabs the hour and '(.{3})?' grabs the colon and minutes if present, the '.+' just grabs everything because the '+' matches unlimited times. Therefore, '1:15pm' returns the groups ['1:15', None, 'pm'], which doesn’t work with our convert function.

Answer: One solution is given below.

Click this link to interact with the solution on regex101.

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.

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.

Answer: 8

Recall that bag of word creates a new column for each unique word. Thus the problem boils down to “how many unique words are there in the following four sentences”, which we count 8: “this”, “is”, “one”, “two”, “the”, “third”, “and”, “fourth”.

Answer: 0

Recall that TF is calculated as the number of terms that appear in that sentence divided by the total number of terms in the sentence. In this case the TF value of “this” in the first sentence is \frac{1}{3}.

IDF is calculated as the log of the number of sentences divided by the number of sentences in which that term appears in. In this case, the IDF value of “this” in the first sentence is \log_{2}(\frac{4}{4}) = 0.

Thus the TF-IDF is just \frac{1}{3} * 0 = 0

Answer: 0.4

Recall that TF is calculated as the number of terms that appear in that sentence divided by the total number of terms in the sentence. In this case the TF value of “and” in the last sentence is \frac{1}{5}.

IDF is calculated as the log of the number of sentences divided by the number of sentences in which that term appears in. In this case, the IDF value of “and” in the last sentence is \log_{2}(\frac{4}{1}) = 2.

Thus the TF-IDF is just \frac{1}{5} * 2 = 0.4

Answer: \frac{3}{10}

The answer is simply the proportion of words in the sentence that are the word "beach". There are 10 words in the sentence, 3 of which are "beach".

Answer: 12

The TF-IDF is the product of the TF and IDF terms. So if the TF-IDF of this document is \frac{9}{10}, and the TF is \frac{3}{10}, as established in the last part, the IDF of the term "beach" is 3. The IDF for a word is the log of the inverse of the proportion of documents in which the word appears. So, since we know there are 96 total documents.

3 = log_{2}(\frac{96}{\text{\# documents containing "beach"}})

2^{3} = \frac{96}{\text{\# documents containing "beach"}}

\boxed{12} = {\text{\# documents containing "beach"}}

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}}

Answer: Yes

Recall,

\text{tf-idf}(t, d) = \text{term frequency}(t, d) \cdot \text{inverse document frequency}(t)

In the context of TF-IDF, if a word appears in every sentence, its inverse document frequency (IDF) would be \log(\frac{5}{5}) = 0. Since a word’s TF-IDF in a document is its TF (term frequency) in that document multiplied by its IDF, if the word’s IDF is 0, it’s TF-IDF is also 0. Since “the” appears in all five sentences, its IDF is zero, leading to a column of zeros in the TF-IDF matrix for “the”.

Answer: Sentence 4

Remember, the IDF of a word is the same for all documents, since $(t) = ( )$. This means that the sentence where “college” is the word with the highest TF-IDF is the same as the sentence where “college” is the word with the highest TF, or term frequency. Sentence 4 is the only sentence where “college” appears twice; in all other sentences, “college” appears at most once. (Since all of these sentences have the same length, we know that if “college” appears more times in Sentence 4 than it does in other sentences, then “college”’s term frequency in Sentence 4, \frac{2}{5}, is also larger than in any other sentence.) As such, the answer is Sentence 4.

Answer: the smallest DF-IDF in document d

The key idea behind TF-IDF, as we learned in class, is that t is a good summary of d if t occurs commonly in d but rarely across all documents.

When t occurs often in d, then \frac{\text{\# of occurrences of $t$ in $d$}}{\text{total \# of words in $d$}} is large, which means \frac{\text{total \# of words in $d$}}{\text{\# of occurrences of $t$ in $d$}} and hence \log \left( \frac{\text{total \# of words in $d$}}{\text{\# of occurrences of $t$ in $d$}} \right) is small.

Similarly, if t is rare across all documents, then \frac{\text{\# of documents in which $t$ appears}}{\text{total \# of documents}} is small.

Putting the above two pieces together, we have that \text{df-itf}(t, d) is small when t occurs commonly in d but rarely overall, which means that the term t that best summarizes d is the term with the smallest DF-IDF in d.

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}

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.

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.

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.

Answer: Documents 3 and 4

The bag-of-words representation for the documents is:

Document number	yesterday	rainy	today	sunny
1	1	1	1	1
2	1	0	1	2
3	1	1	2	0
4	2	0	2	0

The dot product between documents 3 and 4 is 6, which is the highest among all pairs of documents.

Answer: \frac{1}{2}

The dot product between documents 2 and 3 is: 1 + 0 + 2 + 0 = 3 The magnitude of document 2 is equal to document 3 and is: \sqrt{1^2 + 0 ^2 + 1^2 + 2^2} = \sqrt{6} So, the cosine similarity is \frac{3}{\sqrt{6}\times\sqrt{6}} = \frac{1}{2}

Answer: yesterday and today

Remember,

\text{tf-idf}(t, d) = \text{tf}(t, d) \cdot \text{idf}(t)

where \text{tf}(d, t), the term frequency of term t in document d, is the proportion of terms in d that are equal to t, and \text{idf}(t) = \log \left( \frac{\text{number of documents}}{\text{number of documents containing $t$}} \right).

In order for \text{tf-idf}(t, d) to be 0, either t must not be in document d (meaning \text{tf}(t, d) = 0), or t is in every document (meaning \text{idf}(t) = \log(\frac{n}{n}) = \log(1) = 0). In this case, we’re looking for words that have a \text{tf-idf} of 0 across all documents, which means we’re looking for words that have a \text{idf}(t) of 0, meaning they’re in every document. The words that are in every document are yesterday and today.