如何对字典列表按plate分组求和distance与time字段值？

需求说明 #

按车牌（plate）分组，汇总对应车辆的行驶距离（distance）与行驶时间（time），原始实现功能正常但存在代码冗余、效率可优化的问题。

原始代码与结果 #

原始实现代码：

# plate: license plate of the vehicule
# distance: the distance traveled by the vehicule in KM
# time: the time the vehicule travels in minutes
from collections import defaultdict
vehicles = [
    {"plate":"ABC123", "distance":80.3, "time":180},
    {"plate":"ABC123", "distance":80.3, "time":180},
    {"plate":"XYZ246", "distance":40.0, "time":30},
    {"plate":"XYZ246", "distance":60.7, "time":100},
    {"plate":"ABC123", "distance":70.2, "time":200},
    {"plate":"MNL357", "distance":40.3, "time":70}
]
returnList = []
dst, time = defaultdict(int), defaultdict(int)

for v in vehicles:
    dst[v["plate"]] += v["distance"]
    time[v["plate"]] += v["time"]

dst = dict(dst)
time = dict(time)

for i in dst:
    returnList.append({"plate":i,"distance":dst[i], "time":time[i]})

print(returnList)

运行结果：

[
  {'plate': 'ABC123', 'distance': 230.8, 'time': 560},
  {'plate': 'XYZ246', 'distance': 100.7, 'time': 130},
  {'plate': 'MNL357', 'distance': 40.3, 'time': 70}
]

优化方案 #

方案一：单字典存储汇总结果（高效简洁）

用单个defaultdict存储每个车牌对应的汇总数据，只需一次遍历即可完成统计，避免维护多个字典的冗余操作，最后通过列表推导式快速生成目标格式。

from collections import defaultdict

vehicles = [
    {"plate":"ABC123", "distance":80.3, "time":180},
    {"plate":"ABC123", "distance":80.3, "time":180},
    {"plate":"XYZ246", "distance":40.0, "time":30},
    {"plate":"XYZ246", "distance":60.7, "time":100},
    {"plate":"ABC123", "distance":70.2, "time":200},
    {"plate":"MNL357", "distance":40.3, "time":70}
]

# 初始化默认字典，每个键对应包含distance和time的汇总字典
summary = defaultdict(lambda: {"distance": 0.0, "time": 0})
for v in vehicles:
    plate = v["plate"]
    summary[plate]["distance"] += v["distance"]
    summary[plate]["time"] += v["time"]

# 转换为目标列表格式
return_list = [{"plate": plate, **data} for plate, data in summary.items()]
print(return_list)

优势：时间复杂度O(n)，仅需一次遍历，代码简洁易维护，无额外开销。

方案二：使用itertools.groupby（适合有序列表）

如果原始列表已按plate排序，groupby可以高效分组；若未排序，需先排序（注意排序的O(n log n)开销）。

from itertools import groupby
from operator import itemgetter

vehicles = [
    {"plate":"ABC123", "distance":80.3, "time":180},
    {"plate":"ABC123", "distance":80.3, "time":180},
    {"plate":"XYZ246", "distance":40.0, "time":30},
    {"plate":"XYZ246", "distance":60.7, "time":100},
    {"plate":"ABC123", "distance":70.2, "time":200},
    {"plate":"MNL357", "distance":40.3, "time":70}
]

# 先按plate排序（原列表无序时必须执行）
vehicles_sorted = sorted(vehicles, key=itemgetter("plate"))

return_list = []
# 按plate分组并汇总
for plate, group in groupby(vehicles_sorted, key=itemgetter("plate")):
    total_dist = 0.0
    total_time = 0
    for item in group:
        total_dist += item["distance"]
        total_time += item["time"]
    return_list.append({"plate": plate, "distance": total_dist, "time": total_time})

print(return_list)

优势：代码逻辑直观，适合已有序的数据集；劣势：无序数据集需排序，大数据量下性能不如方案一。

方案三：使用pandas（适合大数据/复杂统计）

如果处理的数据量较大，或后续需要更多统计操作（如平均值、最大值等），pandas的分组聚合功能会更高效且简洁。

import pandas as pd

vehicles = [
    {"plate":"ABC123", "distance":80.3, "time":180},
    {"plate":"ABC123", "distance":80.3, "time":180},
    {"plate":"XYZ246", "distance":40.0, "time":30},
    {"plate":"XYZ246", "distance":60.7, "time":100},
    {"plate":"ABC123", "distance":70.2, "time":200},
    {"plate":"MNL357", "distance":40.3, "time":70}
]

# 转换为DataFrame
df = pd.DataFrame(vehicles)
# 按plate分组求和，重置索引后转换为字典列表
summary_df = df.groupby("plate").sum().reset_index()
return_list = summary_df.to_dict("records")

print(return_list)

优势：pandas内部优化了聚合逻辑，大数据量下性能优于纯Python实现，支持多维度统计；劣势：需要额外安装pandas库，小数据量下没必要。

内容的提问来源于stack exchange，提问作者Juan Duque