Re-ranking retrieved results in Prompt Engineering / GenAI - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the retrieved results by their scores in descending order.

Prompt Engineering / GenAI

sorted_results = sorted(results, key=lambda x: x['score'], reverse=[1])

Drag options to blanks, or click blank then click option'

A0

BFalse

CTrue

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Using reverse=False which sorts ascending instead of descending.

Using None or 0 which are invalid for reverse parameter.

Setting reverse=True sorts the list from highest to lowest score, which is needed for re-ranking.

2fill in blank

medium

Complete the code to select the top 5 results after re-ranking.

Prompt Engineering / GenAI

top_results = sorted_results[:[1]]

Drag options to blanks, or click blank then click option'

A3

B5

C10

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 3 or 10 which selects wrong number of results.

Using 0 which results in an empty list.

We want the top 5 results, so slicing up to index 5 selects the first five items.

3fill in blank

hard

Fix the error in the code to compute new scores by multiplying original scores by a re-ranker score.

Prompt Engineering / GenAI

for item in results:
    item['new_score'] = item['score'] * item[[1]]

Drag options to blanks, or click blank then click option'

A'rank_score'

B'score'

C'new_score'

D're_rank_score'

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'score' which multiplies the score by itself.

Using 'new_score' which is not yet defined.

Using 'rank_score' which is not a valid key.

The re-ranker score is stored under the key 're_rank_score', so we multiply by that.

4fill in blank

hard

Fill both blanks to create a dictionary of item IDs mapped to their new scores, filtering only items with new_score above 0.5.

Prompt Engineering / GenAI

filtered_scores = {item['id']: item[[1]] for item in results if item[[2]] > 0.5}

Drag options to blanks, or click blank then click option'

A'new_score'

B'score'

D're_rank_score'

Attempts:

3 left

💡 Hint

Common Mistakes

Using different keys for mapping and filtering causing errors.

Using 'score' or 're_rank_score' which are not the updated scores.

We want to map IDs to new_score and filter by new_score greater than 0.5.

5fill in blank

hard

Fill all three blanks to sort the filtered scores dictionary by score descending and get a list of top 3 item IDs.

Prompt Engineering / GenAI

top_ids = [k for k, v in sorted(filtered_scores.items(), key=lambda item: item[[1]], reverse=[2])[:[3]]]

Drag options to blanks, or click blank then click option'

A1

BTrue

C3

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by index 0 which is the key, not the score.

Using reverse=False which sorts ascending.

Slicing with wrong number like 0 or 1.

We sort by the value at index 1 (the score), descending (reverse=True), and take the top 3 IDs.

Practice

1.

What is the main purpose of re-ranking retrieved results in a search system?

easy

A. To sort the initial search results again using a better scoring method

B. To remove duplicate results from the search output

C. To speed up the initial search query processing

D. To translate results into different languages

Solution

1. Step 1: Understand the role of re-ranking

   Re-ranking means sorting results again after the first search to improve order.

2. Step 2: Identify the goal of re-ranking

   The goal is to use a smarter scoring method to show the most relevant results at the top.

3. Final Answer:

   To sort the initial search results again using a better scoring method -> Option A

4. Quick Check:

   Re-ranking = better sorting [OK]

Hint: Re-ranking means sorting results again for better relevance [OK]

Common Mistakes:

• Confusing re-ranking with removing duplicates
• Thinking re-ranking speeds up initial search
• Assuming re-ranking translates results

2.

Which of the following code snippets correctly represents a simple re-ranking step that sorts a list of results by their score in descending order?

results = [{'id': 1, 'score': 0.5}, {'id': 2, 'score': 0.9}, {'id': 3, 'score': 0.7}]
# Re-rank results here

easy

A. results.sort(reverse=True)

B. results.sort(key=lambda x: x['id'])

C. results.sort(key=lambda x: x['score'])

D. results.sort(key=lambda x: x['score'], reverse=True)

Solution

1. Step 1: Identify sorting by score descending

   We want to sort by 'score' in descending order, so reverse=True is needed.

2. Step 2: Check each option

   results.sort(key=lambda x: x['score'], reverse=True) sorts by 'score' with reverse=True, which is correct. Others either sort by 'id' or ascending score or missing key.

3. Final Answer:

   results.sort(key=lambda x: x['score'], reverse=True) -> Option D

4. Quick Check:

   Sort by score descending = results.sort(key=lambda x: x['score'], reverse=True) [OK]

Hint: Sort with key and reverse=True for descending order [OK]

Common Mistakes:

• Forgetting reverse=True for descending sort
• Sorting by wrong key like 'id'
• Using sort without key causing error

3.

Given the following code that re-ranks search results by a new score, what will be the output after re-ranking?

results = [
  {'id': 'a', 'score': 0.3},
  {'id': 'b', 'score': 0.8},
  {'id': 'c', 'score': 0.5}
]

# New scores from a re-ranker
new_scores = {'a': 0.9, 'b': 0.4, 'c': 0.7}

for r in results:
    r['score'] = new_scores[r['id']]

results.sort(key=lambda x: x['score'], reverse=True)
print([r['id'] for r in results])

medium

A. ['b', 'c', 'a']

B. ['a', 'c', 'b']

C. ['c', 'a', 'b']

D. ['a', 'b', 'c']

Solution

1. Step 1: Update scores with new_scores

   Results get scores: 'a' = 0.9, 'b' = 0.4, 'c' = 0.7.

2. Step 2: Sort results by updated score descending

   Sorted order by score: 0.9 ('a'), 0.7 ('c'), 0.4 ('b').

3. Final Answer:

   ['a', 'c', 'b'] -> Option B

4. Quick Check:

   Sort by new scores descending = ['a', 'c', 'b'] [OK]

Hint: Replace scores then sort descending by score [OK]

Common Mistakes:

• Sorting by old scores instead of new
• Sorting ascending instead of descending
• Mixing up ids and scores

4.

Identify the error in this re-ranking code snippet and select the fix:

results = [{'id': 1, 'score': 0.2}, {'id': 2, 'score': 0.5}]
new_scores = {1: 0.7, 2: 0.9}

for r in results:
    r['score'] = new_scores[r['id']]

results.sort(key=lambda x: x['score'], reverse=True)
print(results)

medium

A. Use sorted() instead of sort() to avoid in-place sorting

B. Change new_scores keys to strings to match 'id' type

C. No error; code runs correctly and sorts results

D. Add a try-except block to handle missing keys

Solution

1. Step 1: Check key types in new_scores and results

   Both use integer keys for 'id', so lookup works correctly.

2. Step 2: Verify sorting and printing

   Sorting by updated 'score' descending is valid and prints sorted list.

3. Final Answer:

   No error; code runs correctly and sorts results -> Option C

4. Quick Check:

   Matching key types = no error [OK]

Hint: Check key types match for dictionary lookups [OK]

Common Mistakes:

• Assuming string keys when they are integers
• Thinking sort() causes error without reason
• Adding unnecessary try-except blocks

5.

You have a list of 5 retrieved documents with initial scores. You want to re-rank them using a machine learning model that outputs a relevance score. Which approach best improves the final ranking?

1. Use the model scores to replace initial scores and sort descending.
2. Combine initial and model scores by averaging, then sort descending.
3. Sort only by initial scores, ignoring model scores.
4. Randomly shuffle results to avoid bias.

hard

A. Combine initial and model scores by averaging, then sort descending

B. Use the model scores to replace initial scores and sort descending

C. Sort only by initial scores, ignoring model scores

D. Randomly shuffle results to avoid bias

Solution

1. Step 1: Understand re-ranking with model scores

   Replacing scores fully may ignore useful initial info; combining scores balances both.

2. Step 2: Evaluate options for best ranking

   Averaging initial and model scores uses all info, improving relevance and stability.

3. Final Answer:

   Combine initial and model scores by averaging, then sort descending -> Option A

4. Quick Check:

   Combine scores for best re-ranking [OK]

Hint: Blend initial and model scores for better ranking [OK]

Common Mistakes:

• Replacing scores blindly losing initial info
• Ignoring model scores completely
• Random shuffling breaks relevance