我有一个看起来像这样的数据结构:
my_structure = [{'description': 'description',
'network_element': 'network-elem1',
'data_json': {'2018-01-31 00:00:00': 10860,
'2018-02-28 00:00:00': 11530,
'2018-03-31 00:00:00': 11530,
'2018-04-30 00:00:00': 8100,
'2018-05-31 00:00:00': 5060,
'2018-06-30 00:00:00': 4470,
'2018-07-31 00:00:00': 4390,
'2018-08-31 00:00:00': 6620,
'2018-09-30 00:00:00': 3070,
'2018-10-31 00:00:00': 18670,
'2018-11-30 00:00:00': 19880,
'2018-12-31 00:00:00': 4700}},
{'description': 'description',
'network_element': 'network-elem-2',
'data_json': {'2015-01-01 00:00:00': 92, '2016-01-01 00:00:00': 109}},
{'description': 'description',
'network_element': 'network-elem3',
'data_json': {'2018-01-31 00:00:00': 0,
'2018-02-28 00:00:00': 0,
'2018-03-31 00:00:00': 0,
'2018-04-30 00:00:00': 0,
'2018-05-31 00:00:00': 0,
'2018-06-30 00:00:00': 0,
'2018-07-31 00:00:00': 0,
'2018-08-31 00:00:00': 1000,
'2018-09-30 00:00:00': 0,
'2018-10-31 00:00:00': 0,
'2018-11-30 00:00:00': 7230,
'2018-12-31 00:00:00': 28630}},
{'description': 'description',
'network_element': 'network-elem...',
'data_json': {'2015-01-01 00:00:00': 264, '2016-01-01 00:00:00': 37}},
{'description': 'description',
'network_element': 'network-elem5',
'data_json': {'2018-01-31 00:00:00': 69220,
'2018-02-28 00:00:00': 80120,
'2018-03-31 00:00:00': 80800,
'2018-04-30 00:00:00': 60560,
'2018-05-31 00:00:00': 35250,
'2018-06-30 00:00:00': 0,
'2018-07-31 00:00:00': 290,
'2018-08-31 00:00:00': 0,
'2018-09-30 00:00:00': 540,
'2018-10-31 00:00:00': 69350,
'2018-11-30 00:00:00': 59410,
'2018-12-31 00:00:00': 70670}},
{'description': 'descr',
'network_element': 'network-elem',
'data_json': {'2015-01-01 00:00:00': 498, '2016-01-01 00:00:00': 526}},
.....
所以基本上是一个包含其他字典的字典列表。
我要从中创建一个DataFrame,其中network_element
的值是我DataFrame的列。嵌套字典的键应为我的索引,嵌套字典的值应为我的值。
我实际上是使用两个列表推导来完成此工作的,然后像这样将df进行转置:
columns = [elem["network_element"] for elem in my_structure]
df_data = [elem["data_json"] for elem in my_structure]
result = pd.DataFrame(df_data, index=columns).T.sort_index()
但是我认为这并不是一个好的解决方案,因为我将数据分为两个列表。我正在寻找可以在单个循环中完成此操作的pandas
解决方案。
像这样做loc
df = pd.DataFrame()
for elem in my_structure:
result.loc[elem["data_json"].keys(), elem["network_element"]] = elem["data_json"].values()
向我抛出了一个关键错误:
KeyError: "None of [Index .... ] are in the [index]"
有没有简单的解决方案来实现这一目标?一个帮助helping,将不胜感激:)预先感谢!
根据建议输出pd.DataFrame.from_dict(....)
ne1 ne2 ne3 ne4 ne5 ne6 ne7 ne8 \
2015-01-01 00:00:00 92 264 498 1086 1022 116 713 40
2016-01-01 00:00:00 109 37 526 1177 1168 123 733 40
ne9 ne10 ne11 ne12 ne13 ne14 ne15 \
2015-01-01 00:00:00 123 61 21 159 14 37 756
2016-01-01 00:00:00 117 115 23 160 8 22 777
ne16
2015-01-01 00:00:00 132
2016-01-01 00:00:00 124
答案 0 :(得分:1)
会进行以下工作吗?
pd.DataFrame.from_dict({elem['network element']: elem['data_json'] for elem in my_structure})
我无法测试,因为您的my_structure不够大。
编辑:如果要将数据作为行,则可以传递orient='index'