Python从包含字典的嵌套列表创建新列表

Question

给定的是输入列表，其中包含字典

Input=[
       {"rbd":"x","cat":"xxx","origin":"us"},
       {"rbd":"y","cat":"xxx","origin":"us"},
       {"rbd":"z","cat":"xxx","origin":"us"},
       {"rbd":"q","cat":"xxx","origin":"us"},
       {"rbd":"1","cat":"xxy","origin":"us"},
       {"rbd":"2","cat":"xxy","origin":"us"}
                                              ]

我试图把它作为一个输出。 这里，如果输入列表中的字典的键“cat”和键“origin”相同，那么它会将它们添加到一个“rbd”中。否则它将创建一个新的。下面是输出

Output=[{"rbd":["x","y","z","q"],"cat":"xxx","origin":"us"},
        {"rbd":["l","2"],"cat":"xxy","origin":"us"}]

我尝试了以下不完整的实现。首先将列表转换为字典。它抛出了一个错误

def Convert(lst): 
    Input_Dict = {lst[i]: lst[i + 1] for i in range(0, len(lst), 2)} 
    return Input_Dict

Answer 1

解决此问题的一种方法是通过来自collections模块的defaultdict：

from collections import defaultdict

d = defaultdict(list)

for entry in Input:
    d[(entry["cat"], entry["origin"])].append(entry["rbd"])

print(d)
defaultdict(list,
        {('xxx', 'us'): ['x', 'y', 'z', 'q'], ('xxy', 'us'): ['1', '2']})


keys = ("rbd", "cat", "origin")

#list comprehension to get final output
[dict(zip(keys, (value, *key))) for key, value in d.items()]

[{'rbd': ['x', 'y', 'z', 'q'], 'cat': 'xxx', 'origin': 'us'},
 {'rbd': ['1', '2'], 'cat': 'xxy', 'origin': 'us'}]

Answer 2

您应该循环输入，并为每个元素检查输出中是否存在cat和原点。如果已经存在，则将输入的rbd放在输出中。如果不存在，请在输出中创建一个新映射

请注意，这不是最有效的答案

Answer 3

试试itertools.groupby：

from itertools import groupby

Input = [{"rbd": "x", "cat": "xxx", "origin": "us"},
         {"rbd": "y", "cat": "xxx", "origin": "us"},
         {"rbd": "z", "cat": "xxx", "origin": "us"},
         {"rbd": "q", "cat": "xxx", "origin": "us"},
         {"rbd": "1", "cat": "xxy", "origin": "us"},
         {"rbd": "2", "cat": "xxy", "origin": "us"}]

print([{"rbd": [i["rbd"] for i in item], "cat": k[0], "origin": k[1]} for k, item in groupby(Input, key=lambda x: (x["cat"], x["origin"]))])

这给了我：

[{'rbd': ['x', 'y', 'z', 'q'], 'cat': 'xxx', 'origin': 'us'}, {'rbd': ['1', '2'], 'cat': 'xxy', 'origin': 'us'}]