如何基于组内首行第四列日期排序CSV分组列表？

需求说明 #

• 处理CSV文件，将地址相同的行划分为1-10行的组
• 按每个组首行第四列的日期对组进行排序，再合并所有组得到目标列表
• 每组内部已按日期正确排序

数据伪示例 #

原分组数据：

list_of_parcel_groups = [
    [ [0,1,2,"11/22/2022"], [0,1,2,"01/01/2001"] ],
    [ [0,1,2,"3/11/2022"], [0,1,2,"3/4/2016"], [0,1,2,"5/18/2011"], [0,1,2,"03/13/2009"] ],
    [ [0,1,2,"5/13/2019"], [0,1,2,"4/20/2018"], [0,1,2,"7/13/1999"] ]
]

预期合并排序后的结果：

qualified = [
    [0,1,2,"5/13/2019"], [0,1,2,"4/20/2018"], [0,1,2,"7/13/1999"],
    [0,1,2,"3/11/2022"], [0,1,2,"3/4/2016"], [0,1,2,"5/18/2011"], [0,1,2,"03/13/2009"],
    [0,1,2,"11/22/2022"], [0,1,2,"01/01/2001"]
]

排序依据是各组首行的日期：

• list_of_parcel_groups[0][0][3]（11/22/2022）
• list_of_parcel_groups[1][0][3]（3/11/2022）
• list_of_parcel_groups[2][0][3]（5/13/2019）

当前尝试的代码 #

import datetime

qualified = []
list_of_parcel_groups = [
    [ [0,1,2,"11/22/2022"], [0,1,2,"01/01/2001"] ],
    [ [0,1,2,"3/11/2022"], [0,1,2,"3/4/2016"], [0,1,2,"5/18/2011"], [0,1,2,"03/13/2009"] ],
    [ [0,1,2,"5/13/2019"], [0,1,2,"4/20/2018"], [0,1,2,"7/13/1999"] ]
]

for i in range(len(list_of_parcel_groups)):
    if i == 0:
        string_input_with_date = list_of_parcel_groups[i][0][3]
        date = datetime.strptime(string_input_with_date, "%m/%d/%Y")
        for item in list_of_parcel_groups[i]:
            qualified.append(item)
        continue
    
    string_input_with_date1 = list_of_parcel_groups[i][0][3]
    date1 = datetime.strptime(string_input_with_date1, "%m/%d/%Y")
    if (date1.date() <= date.date()):
        
        list_of_parcel_groups[i].reverse()
        for item in list_of_parcel_groups[i]:
            qualified.insert(0,item)
        list_of_parcel_groups[i].reverse()
        date = datetime.strptime(list_of_parcel_groups[i][0][3], "%m/%d/%Y")
        continue
    if (date1.date() > date.date()):
        
        for item in list_of_parcel_groups[i]:
            qualified.append(item)
        date = datetime.strptime(list_of_parcel_groups[i][0][3], "%m/%d/%Y")

问题点 #

逐个比较组并插入头部/尾部的逻辑无法实现全局排序，因为新组可能需要插入到列表中间位置，而非仅头部或尾部，导致最终结果不符合预期。（注：每组内部已按日期正确排序，插入时反转组是为了保持原组结构，后续仍能通过list_of_parcel_groups[i][0][3]获取日期）

解决方案 #

正确的做法是先对所有组按首行日期完成全局排序，再依次合并各组：

import datetime

list_of_parcel_groups = [
    [ [0,1,2,"11/22/2022"], [0,1,2,"01/01/2001"] ],
    [ [0,1,2,"3/11/2022"], [0,1,2,"3/4/2016"], [0,1,2,"5/18/2011"], [0,1,2,"03/13/2009"] ],
    [ [0,1,2,"5/13/2019"], [0,1,2,"4/20/2018"], [0,1,2,"7/13/1999"] ]
]

# 按组首行第四列的日期降序排序各组
sorted_groups = sorted(
    list_of_parcel_groups,
    key=lambda group: datetime.datetime.strptime(group[0][3], "%m/%d/%Y"),
    reverse=True
)

# 合并所有组到qualified列表
qualified = []
for group in sorted_groups:
    qualified.extend(group)

# 验证结果
print(qualified)

逻辑说明 #

1. 全局排序：使用sorted()函数，通过key参数指定按每组首行的日期转换为datetime对象进行排序，reverse=True实现降序，匹配预期结果的顺序
2. 合并组：直接用extend()将每组的所有元素追加到结果列表，因每组内部已完成正确排序，无需额外处理
3. 优势：代码简洁，避免逐个插入的逻辑错误，确保全局排序准确

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