有没有办法在python中合并两个字典列表而没有重复答案

【问题标题】：Is there any way to merge two list of dictionaries in python without having duplicates有没有办法在python中合并两个字典列表而没有重复
【发布时间】：2021-08-06 09:23:51
【问题描述】：

我想在python中将这两个字典列表合并为一个-

输入

arr1 = [
{ "field": 'nickname', "direction": 'ASC' },
{ "field": 'email', "direction": 'ASC' },
{ "field": 'name', "direction": 'ASC' },
{ "field": 'first_name', "direction": 'ASC' }
]

arr2 = [
{ "field" : "nickname", "direction" : "DESC"},
{ "field" : "email", "direction" : "DESC"},
{ "field" : "last_name", "direction" : "DESC"}
]

并且需要输出 -

输出

[
    { field: "nickname",   direction: "DESC"},
    { field: "email",      direction: "DESC"},
    { field: "last_name",  direction: "DESC"},
    { field: "name",       direction: "ASC" },
    { field: "first_name", direction: "ASC" }
]

我的解决方案 -

arr1 = [
{ "field": 'nickname', "direction": 'ASC' },
{ "field": 'email', "direction": 'ASC' },
{ "field": 'name', "direction": 'ASC' },
{ "field": 'first_name', "direction": 'ASC' }
]

arr2 = [
{ "field" : "nickname", "direction" : "DESC"},
{ "field" : "email", "direction" : "DESC"},
{ "field" : "last_name", "direction" : "DESC"}
]

arr4 = arr2;
arr3 = []

for i in range(0, len(arr1)):
    for j in range(0, len(arr2)):
        flag = 0
        if(arr1[i]['field'] == arr2[j]['field']):
            arr3.append(arr2[j])
            arr4.remove(arr2[j])
            flag = 1
            break
    if flag == 0 :
        arr3.append(arr1[i])

arr3 += arr4
print(arr3)

我尝试了几种方法，但需要在 O(n) 复杂度中完成，并且不修改 arr2。有什么办法吗？

【问题讨论】：

标签： python list dictionary data-structures

【解决方案1】：

试试：

out = {}
for d in arr1 + arr2:
    out[d["field"]] = d["direction"]

out = [{"field": k, "direction": v} for k, v in out.items()]
print(out)

打印：

[
    {"field": "nickname", "direction": "DESC"},
    {"field": "email", "direction": "DESC"},
    {"field": "name", "direction": "ASC"},
    {"field": "first_name", "direction": "ASC"},
    {"field": "last_name", "direction": "DESC"},
]

【讨论】：

【解决方案2】：

单线将适用于熊猫：

df=pd.DataFrame.from_dict(arr1).append(pd.DataFrame.from_dict(arr2)).drop_duplicates('field', keep='last').reset_index(drop=True)

结果是：

        field direction
0        name       ASC
1  first_name       ASC
2    nickname      DESC
3       email      DESC
4   last_name      DESC

【讨论】：

【解决方案3】：

算法应该很简单：

为结果创建一个新地图
确定每个输入 Map（或 Map 列表）中的哪些键/值对在发生冲突时应优先考虑（重复键映射到不同的值）
在输入映射上编写迭代构造，考虑顺序和优先级。

问：地图列表中较早出现的地图的键是否应保留其值？或者地图的键是否应该出现在地图列表的后面？

【讨论】：