求助：如何将Python字典列表转换为键对应值列表的字典

问题描述 #

输入字典列表：

a = [{'Name': 'dhaya', 'Place': 'pune', 'Designation': 'fleetEngineer'},{'Name': 'rishi', 'Place': 'maharastra', 'Designation': 'Sr.Manager'}]

期望输出：

{"Name":["dhaya","rishi"],"Place":["pune","maharastra"],"Designation":["fleetEngineer","Sr.Manager"]}

解法1：手动遍历构建 #

适用于所有字典键一致的场景，逻辑直观：

a = [{'Name': 'dhaya', 'Place': 'pune', 'Designation': 'fleetEngineer'},{'Name': 'rishi', 'Place': 'maharastra', 'Designation': 'Sr.Manager'}]

result = {}
# 提取第一个字典的所有键（假设所有字典键相同）
for key in a[0].keys():
    # 遍历每个字典，收集当前键对应的值
    result[key] = [item[key] for item in a]

print(result)

解法2：用zip()和字典推导式 #

利用Python内置函数简化代码，效率较高：

a = [{'Name': 'dhaya', 'Place': 'pune', 'Designation': 'fleetEngineer'},{'Name': 'rishi', 'Place': 'maharastra', 'Designation': 'Sr.Manager'}]

# 打包所有字典的对应值，再与键配对
keys = a[0].keys()
result = {k: list(v) for k, v in zip(keys, zip(*(d.values() for d in a)))}

print(result)

解法3：collections.defaultdict处理键不一致场景 #

如果列表中字典的键可能不统一，这个方法会自动为每个键创建列表并收集值：

from collections import defaultdict

a = [{'Name': 'dhaya', 'Place': 'pune', 'Designation': 'fleetEngineer'},{'Name': 'rishi', 'Place': 'maharastra', 'Designation': 'Sr.Manager'}]

result = defaultdict(list)
for item in a:
    for key, val in item.items():
        result[key].append(val)

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

内容的提问来源于stack exchange，提问作者anand