编写获取Top10观看率视频的函数及现有代码问题排查

First, let's break down what's wrong with your current code and walk through a correct, working implementation step by step.

What's Broken in Your Code? #

• L.items() is invalid: Your input L is a list of tuples, not a dictionary. Lists don't have an items() method—this is throwing an AttributeError immediately.
• No aggregation for duplicate videos: Your code treats each tuple as a separate entry, but you need to sum up view rates for the same video name first to get accurate totals.
• No return of top 10 names: Your code only prints a sorted list, but the requirement is to return a list of the top 10 video names.

Correct Implementation Steps #

Here's a solution that fixes all these issues and meets your requirements:

Step 1: Aggregate Total Views per Video #

First, calculate the total view rate for each unique video. We can use a dictionary to track sums:

def ranked_games(L):
    # Build a dictionary of total views per video
    view_totals = {}
    for video, views in L:
        # Use get() to safely add to existing totals or initialize new entries
        view_totals[video] = view_totals.get(video, 0) + views

For a cleaner approach, you can use collections.defaultdict (no extra error handling needed):

from collections import defaultdict

def ranked_games(L):
    view_totals = defaultdict(int)
    for video, views in L:
        view_totals[video] += views

Step 2: Sort by Total Views (Descending) #

Next, sort the aggregated entries from highest to lowest view count. Adding a secondary alphabetical sort ensures consistent results for videos with the same total:

# Sort by views descending, then video name ascending
    sorted_videos = sorted(view_totals.items(), key=lambda x: (-x[1], x[0]))

Step 3: Extract Top 10 Video Names #

Finally, pull the first 10 entries from the sorted list and return just their names:

# Get the top 10 video names
    top_10 = [video for video, _ in sorted_videos[:10]]
    return top_10

Full Working Code (No Imports Needed) #

def ranked_games(L):
    # Aggregate total views for each video
    view_totals = {}
    for video, views in L:
        view_totals[video] = view_totals.get(video, 0) + views
    
    # Sort videos by total views (descending) and name (ascending)
    sorted_videos = sorted(view_totals.items(), key=lambda x: (-x[1], x[0]))
    
    # Return the top 10 video names
    return [video for video, _ in sorted_videos[:10]]

Example Test Run #

Using your sample input:

L = [('abc',10),('def',15),('ghi',10),('abc', 12),('xyz',100)]
print(ranked_games(L))  # Output: ['xyz', 'abc', 'def', 'ghi']

Key Takeaways #

• Always match your code to the input type: Lists don't have items()—save that for dictionaries.
• Aggregation is non-negotiable: Duplicate video names need their view counts summed to get meaningful rankings.
• Sort with negative keys for easy descending order: This avoids having to call reverse() separately.

内容的提问来源于stack exchange，提问作者Rachel