这是我从其他地方获得的结构,即深度嵌套字典的列表:
{
"foo_code": 404,
"foo_rbody": {
"query": {
"info": {
"acme_no": "444444",
"road_runner": "123"
},
"error": "no_lunch",
"message": "runner problem."
}
},
"acme_no": "444444",
"road_runner": "123",
"xyzzy_code": 200,
"xyzzy_rbody": {
"api": {
"items": [
{
"desc": "OK",
"id": 198,
"acme_no": "789",
"road_runner": "123",
"params": {
"bicycle": "2wheel",
"willie": "hungry",
"height": "1",
"coyote_id": "1511111"
},
"activity": "TRAP",
"state": "active",
"status": 200,
"type": "chase"
}
]
}
}
}
{
"foo_code": 200,
"foo_rbody": {
"query": {
"result": {
"acme_no": "260060730303258",
"road_runner": "123",
"abyss": "26843545600"
}
}
},
"acme_no": "260060730303258",
"road_runner": "123",
"xyzzy_code": 200,
"xyzzy_rbody": {
"api": {
"items": [
{
"desc": "OK",
"id": 198,
"acme_no": "789",
"road_runner": "123",
"params": {
"bicycle": "2wheel",
"willie": "hungry",
"height": "1",
"coyote_id": "1511111"
},
"activity": "TRAP",
"state": "active",
"status": 200,
"type": "chase"
}
]
}
}
}
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询问不同的结构是不可能的(传统的apis等).
所以我想知道是否有一些聪明的方法从这样的结构中提取选定的值.
我想到的候选人:
压扁特定字典,构建复合键,如:
{"foo_rbody.query.info.acme_no":"444444","foo_rbody.query.info.road_runner":"123",...}
Pro:通过一次访问获取每个值,如果没有可预测的键,则表示结构不存在(您可能已经注意到,字典可能具有不同的结构,具体取决于它是否成功运行,错误发生等).
骗局:如何处理清单?
有更好的候选人?
您可以尝试使用这个相当简单的函数来访问嵌套属性:
import re
def get_path(dct, path):
for i, p in re.findall(r'(\d+)|(\w+)', path):
dct = dct[p or int(i)]
return dct
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用法:
value = get_path(data, "xyzzy_rbody.api.items[0].params.bicycle")
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