如何按给定示例合并字典列表并向list1追加元素输出结果？

需求说明 #

已有两个字典列表data和data2，需要将两个列表中对应索引位置的字典进行合并，同时移除data字典中的id字段，最终将合并后的结果存入list1。

实现方法 #

方法一：循环遍历+字典合并

通过索引遍历两个列表，先复制data中的字典并删除id字段（避免修改原数据），再和data2对应位置的字典合并，最后添加到list1中：

data = [{"id":1,"name":"vikash","roll":39},{"id":2,"name":"kumar","roll":3}] 
data2 = [{"hobby":"football","food":"any"},{"hobby":"basketball","food":"any"}]
list1 = []

for d1, d2 in zip(data, data2):
    temp = d1.copy()
    del temp["id"]
    merged_dict = {**temp, **d2}
    list1.append(merged_dict)

方法二：列表推导式简化写法

用列表推导式压缩代码逻辑，写法更简洁：

data = [{"id":1,"name":"vikash","roll":39},{"id":2,"name":"kumar","roll":3}] 
data2 = [{"hobby":"football","food":"any"},{"hobby":"basketball","food":"any"}]

list1 = [
    {k: v for k, v in d1.items() if k != "id"} | d2
    for d1, d2 in zip(data, data2)
]

注：|字典合并运算符仅支持Python 3.9及以上版本，低版本可替换为{**{k:v for k,v in d1.items() if k!='id'}, **d2}

方法三：直接修改原数据（慎用）

如果不需要保留原data列表的内容，可以直接用pop()移除id字段，减少内存开销：

data = [{"id":1,"name":"vikash","roll":39},{"id":2,"name":"kumar","roll":3}] 
data2 = [{"hobby":"football","food":"any"},{"hobby":"basketball","food":"any"}]
list1 = []

for d1, d2 in zip(data, data2):
    d1.pop("id")
    list1.append({**d1, **d2})

注意：此方法会改变原data列表的内容，若需保留原始数据，不建议使用。

验证结果 #

执行上述任意一种方法后，list1的内容都会符合预期：

[
    {"name":"vikash","roll":39,"hobby":"football","food":"any"},
    {"name":"kumar","roll":3,"hobby":"basketball","food":"any"}
]

内容的提问来源于stack exchange，提问作者Vikash