我想获取列表中嵌套字典的所有键的路径。例如，如果我的字典如下所示
{
"persons": [{
"id": "f4d322fa8f552",
"address": {
"building": "710",
"coord": "[123, 465]",
"street": "Avenue Road",
"zipcode": "12345"
},
"cuisine": "Chinese",
"grades": [{
"date": "2013-03-03T00:00:00.000Z",
"grade": "B",
"score": {
"x": 3,
"y": 2
}
}, {
"date": "2012-11-23T00:00:00.000Z",
"grade": "C",
"score": {
"x": 1,
"y": 22
}
}],
"name": "Shash"
}]
}
我想得到这样的路径 path = [['persons'], ['persons','id'],['persons','address'],['persons','address','building']...] 直到最后一个键。
我试图遍历整个字典来追加路径变量。试图从 Print complete key path for all the values of a python nested dictionary 中获得一些灵感但我无法获得列表中的路径。
还有其他可能的方法吗？
请您参考如下方法:
您可以递归地描述数据结构，这是使用队列 q 与递归的一种方法。但很难判断这是否是您要查找的内容，因为它显示了列表索引，但可以很容易地排除它们:
def get_paths(d):
q = [(d, [])]
while q:
n, p = q.pop(0)
yield p
if isinstance(n, dict):
for k, v in n.items():
q.append((v, p+[k]))
elif isinstance(n, list):
for i, v in enumerate(n):
q.append((v, p+[i])) # Change to q.append((v, p)) to remove index
In []:
list(get_paths(d))
Out[]:
[[],
['persons'],
['persons', 0],
['persons', 0, 'id'],
['persons', 0, 'address'],
['persons', 0, 'cuisine'],
['persons', 0, 'grades'],
['persons', 0, 'name'],
['persons', 0, 'address', 'building'],
['persons', 0, 'address', 'coord'],
['persons', 0, 'address', 'street'],
['persons', 0, 'address', 'zipcode'],
['persons', 0, 'grades', 0],
['persons', 0, 'grades', 1],
['persons', 0, 'grades', 0, 'date'],
['persons', 0, 'grades', 0, 'grade'],
['persons', 0, 'grades', 0, 'score'],
['persons', 0, 'grades', 1, 'date'],
['persons', 0, 'grades', 1, 'grade'],
['persons', 0, 'grades', 1, 'score'],
['persons', 0, 'grades', 0, 'score', 'x'],
['persons', 0, 'grades', 0, 'score', 'y'],
['persons', 0, 'grades', 1, 'score', 'x'],
['persons', 0, 'grades', 1, 'score', 'y'],
