2 回答

TA贡献1805条经验 获得超10个赞
您可以通过迭代对象来解压嵌套的 json 对象。尝试
import pandas as pd
a=[
{
"item": [
{
"value": 0,
"type": "a"
},
{
"value": 0,
"type": "b"
},
{
"value": 70,
"type": "c"
},
],
"timestamp": "2019-01-12T04:52:06.669Z"
},
{
"item": [
{
"value": 30,
"type": "a"
},
{
"value": 0,
"type": "b"
}
],
"timestamp": "2019-01-12T04:53:06.669z"
}
]
cols = ['value', 'type', 'timestamp']
rows = []
for data in a:
data_row = data['item']
timestamp = data['timestamp']
for row in data_row:
row['timestamp']=timestamp
rows.append(row)
df = pd.DataFrame(rows)
df =df.pivot_table(index='timestamp',columns=['type'],values=['value']).reset_index()
df.columns=['timestamp','a','b','c']
如果您正在寻找紧凑的解决方案,请使用json_normalize
from pandas.io.json import json_normalize
df =pd.DataFrame()
for i in range(len(a)):
df =pd.concat([df,json_normalize(a[i]['item'])])
df =df.pivot_table(index='timestamp',columns=['type'],values=['value']).reset_index()
df.columns=['timestamp','a','b','c']
最终输出
timestamp a b c
2019-01-12T04:52:06.669Z 0.0 0.0 70.0
2019-01-12T04:53:06.669z 30.0 0.0 NaN

TA贡献1827条经验 获得超9个赞
您可以从 json 中提取字典列表并将其提供给数据帧。代码可以是:
df = pd.DataFrame([dict([('timestamp', d['timestamp']), ('a', 0),
('b', 0), ('c', 0)]
+ [(item['type'], item['value'])
for item in d['item']])for d in data['data']],
columns=['timestamp', 'a', 'b', 'c'])
print(df)
按预期输出:
timestamp a b c
0 2019-01-12T04:52:06.669Z 0 0 70
1 2019-01-12T04:53:06.669z 30 0 0
这里的技巧是首先构建一个具有默认值的对列表,然后在从中构建字典之前用实际值扩展它。由于保留了最后看到的值,您实际上构建了一个包含所有相关值的字典。
columns 参数仅用于确保列的预期顺序。
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