如何在Python中对比两个列表并生成字典，计算每人平均成绩？

Let's walk through fixing this problem step by step. Your current code has a couple of key issues (like the incorrect if name == names check, and not pairing names with their corresponding grades) so we'll address those first, then build a clean solution.

The Core Problem #

We need to:

1. Pair each name in names with its matching grade in grades (noting that grades is longer, so we only use the first len(names) entries)
2. Track the total score and number of occurrences for each name
3. Calculate the average by dividing total score by occurrence count

Solution 1: Basic Dictionary Implementation #

This is a straightforward approach that uses a regular dictionary to track totals and counts:

def average_grades(names, grades):
    # Initialize a dictionary to store total scores and counts for each name
    grade_tracker = {}
    # Truncate grades to match the length of names (since grades is longer)
    matched_grades = grades[:len(names)]
    
    # Iterate over paired names and grades
    for name, grade in zip(names, matched_grades):
        if name in grade_tracker:
            # If the name exists, add to total score and increment count
            grade_tracker[name]["total"] += grade
            grade_tracker[name]["count"] += 1
        else:
            # If it's the first time seeing the name, initialize the entry
            grade_tracker[name] = {"total": grade, "count": 1}
    
    # Convert the tracker into a dictionary of averages
    return {name: data["total"] / data["count"] for name, data in grade_tracker.items()}

# Test with your sample data
names = ['Mary', 'Jack', 'Rose', 'Mary', 'Carl', 'Fred', 'Meg', 'Phil', 'Carl', 'Jack', 'Fred', 'Mary', 'Phil', 'Jack', 'Mary', 'Fred', 'Meg']
grades = [80, 88, 53, 80, 64, 61, 75, 80, 91, 82, 68, 76, 95, 58, 89, 51, 81, 78]

print(average_grades(names, grades))

Solution 2: Simplified with collections.defaultdict #

For cleaner code, we can use defaultdict from the collections module to avoid checking if a name exists in the dictionary:

from collections import defaultdict

def average_grades(names, grades):
    totals = defaultdict(int)
    counts = defaultdict(int)
    matched_grades = grades[:len(names)]
    
    for name, grade in zip(names, matched_grades):
        totals[name] += grade
        counts[name] += 1
    
    return {name: totals[name] / counts[name] for name in counts}

Solution 3: Using collections.Counter for Occurrences #

If you prefer, we can use Counter to get the occurrence count of each name, then calculate totals separately:

from collections import Counter, defaultdict

def average_grades(names, grades):
    name_counts = Counter(names)
    totals = defaultdict(int)
    matched_grades = grades[:len(names)]
    
    for name, grade in zip(names, matched_grades):
        totals[name] += grade
    
    return {name: totals[name] / name_counts[name] for name in name_counts}

Key Fixes from Your Original Code #

• Replaced the invalid if name == names check with proper dictionary existence checks
• Paired each name with its corresponding grade using zip() (after truncating grades to match names length)
• Added tracking for total scores and occurrence counts to calculate averages correctly

内容的提问来源于stack exchange，提问作者Ethan Chung