我有一个json对象,例如
{
"id": 3590403096656,
"title": "Romania Special Zip Hoodie Blue - Version 02 A5",
"tags": [
"1ST THE WORLD FOR YOU <3",
"apparel",
],
"props": [
{
"id": 28310659235920,
"title": "S / romainia All Over Print Full Zip Hoodie for Men (Model H14)",
"position": 1,
"product_id": 3590403096656,
"created_at": "2019-05-22T00:46:19+07:00",
"updated_at": "2019-05-22T01:03:29+07:00"
},
{
"id": 444444444444,
"title": "number 2",
"position": 1,
"product_id": 3590403096656,
"created_at": "2019-05-22T00:46:19+07:00",
"updated_at": "2019-05-22T01:03:29+07:00"
}
]
}
我想将其展平,以便所需的输出看起来像
{"id": 3590403096656,"title": "Romania Special Zip Hoodie Blue - Version 02 A5","tags": ["1ST THE WORLD FOR YOU <3","apparel"],"props.id": 28310659235920,"props.title": "S / romainia All Over Print Full Zip Hoodie for Men (Model H14)","props.position": 1,"props.product_id": 3590403096656,"props.created_at": "2019-05-22T00:46:19+07:00", "props.updated_at": "2019-05-22T01:03:29+07:00"}
{"id": 3590403096656,"title": "Romania Special Zip Hoodie Blue - Version 02 A5","tags": ["1ST THE WORLD FOR YOU <3","apparel"],"props.id": 444444444444,"props.title": "number 2","props.position": 1,"props.product_id": 3590403096656,"props.created_at": "2019-05-22T00:46:19+07:00","props.updated_at": "2019-05-22T01:03:29+07:00"}
到目前为止,我已经尝试过:
from pandas.io.json import json_normalize
json_normalize(sample_object)
其中sample_object
包含json
对象,我正在循环浏览这些对象的大文件,并希望将它们以所需的格式展平。
json_normalize
不能给我想要的输出,我想保持标签不变,但要展平props
并重复父对象信息。
在这方面的任何帮助都将受到赞赏。
答案 0 :(得分:2)
您想要一些json_normalize
的行为,但是要有一个自定义的方式。因此,在部分数据上使用json_normalize
或类似数据,然后将其与其余数据组合。
下面的代码更喜欢“或相似的”途径,深入into the pandas codebase以获得nested_to_record
辅助函数,该函数使字典扁平化。它用于创建单独的行,这些行将基本数据(所有属性中共有的键/值)与特定于每个props条目的扁平化数据相结合。有一条注释掉的行在没有nested_to_record
的情况下执行了等效的操作,但是它有点过分地变平为DataFrame
,然后导出为dict
。
from collections import OrderedDict
import json
import pandas as pd
from pandas.io.json.normalize import nested_to_record
data = json.loads(rawjson)
props = data.pop('props')
rows = []
for prop in props:
rowdict = OrderedDict(data)
flattened_prop = nested_to_record({'props': prop})
# flatteded_prop = json_normalize({'props': prop}).to_dict(orient='records')[0]
rowdict.update(flattened_prop)
rows.append(rowdict)
df = pd.DataFrame(rows)
结果:
答案 1 :(得分:1)
请尝试以下操作:
import copy
obj = {
"id": 3590403096656,
"title": "Romania Special Zip Hoodie Blue - Version 02 A5",
"tags": [
"1ST THE WORLD FOR YOU <3",
"apparel",
],
"props": [
{
"id": 28310659235920,
"title": "S / romainia All Over Print Full Zip Hoodie for Men (Model H14)",
"position": 1,
"product_id": 3590403096656,
"created_at": "2019-05-22T00:46:19+07:00",
"updated_at": "2019-05-22T01:03:29+07:00"
},
{
"id": 444444444444,
"title": "number 2",
"position": 1,
"product_id": 3590403096656,
"created_at": "2019-05-22T00:46:19+07:00",
"updated_at": "2019-05-22T01:03:29+07:00"
}
]
}
props = obj.pop("props")
for p in props:
res = copy.deepcopy(obj)
for k in p:
res["props."+k] = p[k]
print(res)
基本上,它使用pop("props")
来获取不带"props"
的obj(这是在所有结果对象中使用的通用部分),
然后我们遍历道具,并创建包含基础对象的新对象,然后为每个道具中的每个键填充“ props.key”。