如何在Python中合并列表元组的同名元素并累加对应数值？

这里有几种简洁的方式实现同名数据的合并累加：

方法一：基础字典遍历 #

这是最直观的实现方式，用字典记录每个名字的累计统计值：

# 原始数据
baseball_data = [
    ("John Doe", 2, 3),
    ("Sponge Bob", 4, 5),
    ("John Doe", 1, 5)
]

# 初始化字典存储累计结果
total_stats = {}

for name, hits, runs in baseball_data:
    if name in total_stats:
        # 累加已有数据
        total_stats[name] = (total_stats[name][0] + hits, total_stats[name][1] + runs)
    else:
        # 首次出现，初始化数据
        total_stats[name] = (hits, runs)

# 转换回元组列表格式
result = [(name, *values) for name, values in total_stats.items()]

print(result)
# 输出：[('John Doe', 3, 8), ('Sponge Bob', 4, 5)]

方法二：用collections.defaultdict简化代码 #

defaultdict可以自动处理键不存在的情况，省去判断逻辑：

from collections import defaultdict

baseball_data = [
    ("John Doe", 2, 3),
    ("Sponge Bob", 4, 5),
    ("John Doe", 1, 5)
]

# 默认值设为(0, 0)，避免键不存在的报错
total_stats = defaultdict(lambda: (0, 0))

for name, hits, runs in baseball_data:
    current_hits, current_runs = total_stats[name]
    total_stats[name] = (current_hits + hits, current_runs + runs)

result = [(name, *values) for name, values in total_stats.items()]

print(result)
# 输出：[('John Doe', 3, 8), ('Sponge Bob', 4, 5)]

方法三：使用itertools.groupby（需先排序） #

如果允许先对数据按名字排序，groupby可以快速分组统计：

from itertools import groupby

baseball_data = [
    ("John Doe", 2, 3),
    ("Sponge Bob", 4, 5),
    ("John Doe", 1, 5)
]

# 必须先按名字排序，groupby只能分组连续的相同元素
sorted_data = sorted(baseball_data, key=lambda x: x[0])

result = []
for name, group in groupby(sorted_data, key=lambda x: x[0]):
    # 将分组转为列表，方便多次遍历
    group_items = list(group)
    total_hits = sum(item[1] for item in group_items)
    total_runs = sum(item[2] for item in group_items)
    result.append((name, total_hits, total_runs))

print(result)
# 输出：[('John Doe', 3, 8), ('Sponge Bob', 4, 5)]

内容的提问来源于stack exchange，提问作者Dylan Eye