tasks.sort()

:param weights: Dictionary of weights for each factor :return: Weighted score """ weighted_score = sum(option_scores[factor] * weights[factor] for factor in option_scores) return weighted_score def main(): # Define …

score = int(input(f"Enter the score for {factor} (1-10): ")) option_scores[factor] = score options[option_name] = option_scores # Calculate weighted scores weighted_scores = {} for o…

- Add all 22 tasks to the DataFrame with their respective priorities and durations. 2. **Sort and Prioritize**: - Sort the tasks by priority and duration to prioritize them. 3. **Allocate to Sprints**: - Allocate tasks to sprints…

{"task": "Add unit tests", "priority": "Medium", "duration": 2}, {"task": "Optimize database queries", "priority": "High", "duration": 3}, {"task": "Implement caching", "priority": "Medium", "duration": 2}, {"task": "Refine …

{"task": "Review code", "priority": "Low", "duration": 1}, {"task": "Improve error messages", "priority": "Medium", "duration": 2}, {"task": "Enhance user interface", "priority": "Low", "duration": 1}, {"task": "Add unit tes…

ss.search(f'search {i}') # get search speeds search_speeds = ss.get_search_speeds() # calculate 90th percentile search_speeds.sort() ninetieth_percentile = search_speeds[int(0.9 * len(search_speeds))] print(ninetieth_percentile) # s…

Choose the approach that best fits your use case. If you have common prefixes, a Trie might be more efficient. If you have a large dictionary and want to avoid unnecessary lookups, a Bloom filter can be beneficial. Let me know if you need …

'Task Name': ['Evaluate Pipeline 1', 'Evaluate Pipeline 2', 'Evaluate Pipeline 3', 'Evaluate Pipeline 4', 'Evaluate Pipeline 5'], 'Status': ['To-Do', 'In Progress', 'Done', 'To-Do', 'In Progress'], 'Priority': ['High', 'Medium',…