Python中如何实现按最小location分组的多条件列表排序

我有一个结构如下的字典列表：

[{'location':1, 'mgr':'Joe', 'region':'Frank'}, {'location':2, 'mgr':'Sam', 'region':'Jess'}, {'location':3, 'mgr':'Joe', 'region':'Frank'}, {'location':4, 'mgr':'Sue', 'region':'Frank'}, {'location':5, 'mgr':'Susan', 'region':'Jess'}]

需要将其排序为：同一region和mgr的条目归为一组，组的排序依据是该组内最小的location值，组内按location升序排列。期望的排序结果为：

[{'location':1,'mgr':'Joe','region':'Frank'},{'location':3,'mgr':'Joe','region':'Frank'},{'location':4, 'mgr':'Sue', 'region':'Frank'},{'location':2,'mgr':'Sam','region':'Jess'},{'location':5, 'mgr':'Susan', 'region':'Jess'}]

最终只需保证location的顺序正确即可。

尝试过遍历元素移动到新列表，但条目顺序仍错误；也尝试多键排序，却出现按字母排序或仅按location排序的问题，现寻求简洁可行的实现方案。

附上尝试的代码：

correctlist = []
originallist = [{'location':1,'mgr':'Joe','region':'Frank'},{'location':2,'mgr':'Sam','region':'Jess'},{'location':3,'mgr':'Joe','region':'Frank'},{'location':4, 'mgr':'Sue', 'region':'Frank'},{'location':5, 'mgr':'Susan', 'region':'Jess'}]

# 期望结果：[{location:1,mgr:'Joe',region:'Frank'},{location:3,mgr:'Joe',region:'Frank'},{location:4, mgr:'Sue', region:'Frank'},{location:2,mgr:'Sam',region:'Jess'},{location:5, mgr:'Susan', region:'Jess'}]

sortedlist = sorted(originallist, key=lambda k: (k['location']))

while sortedlist:
    for location in sortedlist:
        currregion = location['region']
        currmgr = location['mgr']
        for locationin sortedlist:
            if location['region'] == currregion and location['mgr'] == currmgr:
                correctlist.append(number)
                sortedlist.remove(number)
        for location in sortedlist:
            if location['region'] == curregion:
                currmgr = location['mgr']
        for locationin sortedlist:
            if location['mgr'] == currmgr and location['region'] == curregion:
                correctlist.append(location)
                sortedlist.remove(location)

可以分两步完成需求：

1. 先统计每个(region, mgr)组合对应的最小location值，作为组排序的依据
2. 对原列表进行多键排序，优先级依次为：region、组最小location、当前元素的location

具体代码如下：

originallist = [
    {'location':1, 'mgr':'Joe', 'region':'Frank'},
    {'location':2, 'mgr':'Sam', 'region':'Jess'},
    {'location':3, 'mgr':'Joe', 'region':'Frank'},
    {'location':4, 'mgr':'Sue', 'region':'Frank'},
    {'location':5, 'mgr':'Susan', 'region':'Jess'}
]

# 统计每个(region, mgr)组的最小location
group_min_loc = {}
for item in originallist:
    key = (item['region'], item['mgr'])
    if key not in group_min_loc or item['location'] < group_min_loc[key]:
        group_min_loc[key] = item['location']

# 执行排序
result = sorted(
    originallist,
    key=lambda x: (x['region'], group_min_loc[(x['region'], x['mgr'])], x['location'])
)

print(result)

代码说明 #

• 第一步用字典group_min_loc记录每个组的最小location，确保组的排序符合要求
• 排序时，先按region聚合同区域条目，再按组的最小location确定组的先后顺序，最后按当前元素的location完成组内升序排列
• 代码时间复杂度为O(n log n)，属于高效排序实现，运行结果完全匹配期望

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