Python对字典列表按多键统计通过率等指标的高效实现方法

你原来的实现存在两个明显问题：

• 统计逻辑错误：sum(["Passed" and "name" for d in input if 'Status' in d and name in d])中"Passed" and "name"的返回值固定为字符串name，求和时会被转为布尔值True（对应数值1），实际统计的是匹配name的所有条目总数，无法区分Passed和Failed的数量
• 效率低下：每统计一个name就要遍历一次全量数据集，时间复杂度为O(n*m)（n为总数据条数，m为不同name的数量），数据量大时性能损耗明显

你提到的直接循环对比字典字段值的思路确实更简单，基于这个思路可以实现一次遍历完成所有统计，效率和可读性都更高。

最优实现方案（一次遍历完成统计） #

时间复杂度为O(n)，逻辑简洁清晰：

# 如需兼容无额外库导入的场景，可把defaultdict替换为普通字典判断key是否存在
from collections import defaultdict

# 初始化统计容器：key为name，value为[Passed数量, Failed数量]
count_map = defaultdict(lambda: [0, 0])
total_passed = 0
total_failed = 0

for item in input:
    name = item["name"]
    status = item["Status"]
    if status == "Passed":
        count_map[name][0] += 1
        total_passed += 1
    elif status == "Failed":
        count_map[name][1] += 1
        total_failed += 1

# 生成分组统计结果
output = []
for name, (passed, failed) in count_map.items():
    total = passed + failed
    pass_rate = f"{int(passed / total * 100)}%"
    output.append({
        "name": name,
        "Passed": str(passed),
        "Failed": str(failed),
        "Total": str(total),
        "%Pass": pass_rate
    })

# 追加全局合计行
total_all = total_passed + total_failed
total_pass_rate = f"{int(total_passed / total_all * 100)}%"
output.append({
    "name": "Total",
    "Passed": str(total_passed),
    "Failed": str(total_failed),
    "Total": str(total_all),
    "%Pass": total_pass_rate
})

无额外导入的普通字典实现 #

如果不想导入collections标准库，可替换统计容器的初始化逻辑，其余生成结果的代码完全一致：

count_map = {}
total_passed = 0
total_failed = 0

for item in input:
    name = item["name"]
    status = item["Status"]
    # 首次遇到该name时初始化统计值
    if name not in count_map:
        count_map[name] = [0, 0]
    if status == "Passed":
        count_map[name][0] += 1
        total_passed += 1
    elif status == "Failed":
        count_map[name][1] += 1
        total_failed += 1

内容的提问来源于stack exchange，提问作者Hugo_Ludo_38