如何将含共同键值对的字典列表转换为以共同值为键的新字典？

问题 #

给定一个包含多个字典的列表，所有字典都包含共同键id，需要将列表中的字典按id分组，生成以id为键、对应值为包含details列表的字典的新结构。

原始列表：

original_list =  [
    {"local_id":1,"id":29,"name":"Ashish"},
    {"local_id":2,"id":29,"name":"Boora"},
    {"local_id":3,"id":30,"name":"Harshdeep"},
    {"local_id":4,"id":30,"name":"Singh"},
    {"local_id":5,"id":31,"name":"Deepak"}  
]

期望结果：

result_dict = {
    29:{'details':[{"local_id":1,"id":29,"name":"Ashish"},{"local_id":2,"id":29,"name":"Boora"}]},
    30:{'details':[{"local_id":3,"id":30,"name":"Harshdeep"},{"local_id":4,"id":30,"name":"Singh"}]},
    31:{'details':[{"local_id":5,"id":31,"name":"Deepak"}]} 
}

解决方案 #

方法1：基础遍历构建 #

直接遍历原始列表，逐个将元素添加到对应分组中：

original_list =  [
    {"local_id":1,"id":29,"name":"Ashish"},
    {"local_id":2,"id":29,"name":"Boora"},
    {"local_id":3,"id":30,"name":"Harshdeep"},
    {"local_id":4,"id":30,"name":"Singh"},
    {"local_id":5,"id":31,"name":"Deepak"}  
]

result_dict = {}
for item in original_list:
    id_key = item["id"]
    if id_key not in result_dict:
        result_dict[id_key] = {"details": []}
    result_dict[id_key]["details"].append(item)

print(result_dict)

方法2：使用collections.defaultdict简化 #

借助defaultdict自动生成默认结构，减少判断逻辑：

from collections import defaultdict

original_list =  [
    {"local_id":1,"id":29,"name":"Ashish"},
    {"local_id":2,"id":29,"name":"Boora"},
    {"local_id":3,"id":30,"name":"Harshdeep"},
    {"local_id":4,"id":30,"name":"Singh"},
    {"local_id":5,"id":31,"name":"Deepak"}  
]

result_dict = defaultdict(lambda: {"details": []})
for item in original_list:
    result_dict[item["id"]]["details"].append(item)

# 可选：转换为普通字典
result_dict = dict(result_dict)

print(result_dict)

方法3：使用itertools.groupby（需先排序） #

利用groupby对排序后的列表分组，适合需要分组后做其他处理的场景：

from itertools import groupby

original_list =  [
    {"local_id":1,"id":29,"name":"Ashish"},
    {"local_id":2,"id":29,"name":"Boora"},
    {"local_id":3,"id":30,"name":"Harshdeep"},
    {"local_id":4,"id":30,"name":"Singh"},
    {"local_id":5,"id":31,"name":"Deepak"}  
]

# 先按id排序，确保相同id的元素连续
sorted_list = sorted(original_list, key=lambda x: x["id"])
result_dict = {
    key: {"details": list(group)} 
    for key, group in groupby(sorted_list, key=lambda x: x["id"])
}

print(result_dict)

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