我有大的json数据,这些数据被读入python数据帧,并每行创建一个字典列表。我需要将其转换为其他格式的数据。
数据格式如下:
{
"data": [{
"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"
}
]
}
将数据转换为以下形式的数据框的最有效方法是什么:
时间戳----------------------------- a ------- b ------ c < / p>
2019-01-12T04:52:06.669Z ------ 0 ------- 0 ------ 70
2019-01-12T04:53:06.669Z ------ 30 ------ 0 ------ 0
到目前为止,我已经设法使用for循环来做到这一点,但是它非常低效且缓慢。到目前为止,我是这个。
with open('try.json') as f:
data = json.load(f)
df_data = pandas.DataFrame(data['data'])
df_formatted = pandas.DataFrame(columns=['a','b','c'])
for d, timestamp in zip(df_data['item'], df_data['timestamp']):
row = dict()
for entry in d:
category = entry['type']
value = entry['value']
row[category] = value
row['timestamp'] = timestamp
df_formatted = df_formatted.append(row, ignore_index=True)
df = df_formatted.fillna(0)
列表中的项目数通常为数千个。是否有任何有关如何高效执行此操作的指针或示例?
答案 0 :(得分:2)
您可以通过遍历对象来解压缩嵌套的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型解决方案,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
答案 1 :(得分:0)
您可以从json提取字典列表,并将其输入数据框。代码可能是:
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)
预期的输出:
timestamp a b c
0 2019-01-12T04:52:06.669Z 0 0 70
1 2019-01-12T04:53:06.669z 30 0 0
这里的窍门是先建立一个具有默认值的对的列表,然后将其扩展为实际值,然后再从中构建字典。保留最后看到的值后,您实际上将构建一个包含所有相关值的字典。
columns参数仅用于确保预期的列顺序。