如何处理键不存在问题？解决Python字典合并的KeyError异常

尝试合并两个数据结构（list1是字典列表，list2是字典）时触发KeyError，原因是list2中没有list1里的部分兄弟名称键，同时list1的brothers字段存在"N/A"值。需要给list1中的每个字典新增info键，值为对应兄弟在list2中的兴趣去重后排序的列表，没有匹配项则为空列表。

示例数据 #

list1 = [ {"Name":"Paul", "age": 34, "brothers": ["Sam","Cris","Jack"] } ,
          {"Name":"Evan", "age": 14, "brothers": ["N/A"] } ,
          {"Name":"Loki", "age": 4, "brothers": ["Thor"] }  ]

list2 = {"Sam":["ice_cream"], "Loki": ["cars"], "Jack": ["fruits","cars"],
         "Cris": ["ice_cream"] }

期望输出 #

[
    {"Name": "Paul", "age": 34, "brothers": ["Sam","Cris","Jack"],
     "info": ["ice_cream", "fruits", "cars"]},
    {"Name": "Evan", "age": 14, "brothers": ["N/A"], "info": []},
    {"Name": "Loki", "age": 4, "brothers": ["Thor"], "info": []}
]

原错误代码 #

newList = [
    {**d, "Ids": sorted(set(i for t in d["brothers"] for i in list2[t]))}
        for d in list1
 ]

修正后的代码 #

new_list = [
    {**d, "info": sorted(set(
        item for brother in d["brothers"]
        if brother != "N/A"
        for item in list2.get(brother, [])
    ))}
    for d in list1
]

代码解释 #

• 避免KeyError：用list2.get(brother, [])替代直接索引list2[brother]，当兄弟名称不在list2中时返回空列表，不会触发异常。
• 过滤无效值：添加if brother != "N/A"条件，跳过无意义的兄弟标识。
• 去重与排序：通过set()自动去重，再用sorted()整理为有序列表，匹配输出要求。
• 键名修正：将原代码中的"Ids"改为需求中的"info"，对齐预期输出的键名。

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