[英]Nested dictionary to pandas DataFrame
My data looks like this: 我的数据如下所示:
{ outer_key1 : [ {key1: some_value},
{key2: some_value},
{key3: some_value} ],
outer_key2 : [ {key1: some_value},
{key2: some_value},
{key3: some_value} ] }
The inner arrays are always the same lengths. 内部数组的长度始终相同。 key1, key2, key3 are also always the same.
key1,key2,key3也始终相同。
I want to convert this to a pandas DataFrame, where outer_key1, outer_key2, ... are the index and key1, key2, key3 are the columns. 我想将其转换为pandas DataFrame,其中outer_key1,outer_key2,...是索引,key1,key2,key3是列。
Edit: 编辑:
There's an issue in the data, which I believe is the reason the given solutions are not working. 数据中存在一个问题,我认为这是给定解决方案无法正常工作的原因。 In a few cases, in the inner array there are three
None
s instead of the three dictionaries. 在某些情况下,内部数组中有三个
None
而不是三个字典。 Like this: 像这样:
outer_key3: [ None, None, None ]
Here's one way: 这是一种方法:
d = { 'O1' : [ {'K1': 1},
{'K2': 2},
{'K3': 3} ],
'O2' : [ {'K1': 4},
{'K2': 5},
{'K3': 6} ] }
d = {k: { k: v for d in L for k, v in d.items() } for k, L in d.items()}
df = pd.DataFrame.from_dict(d, orient='index')
# K1 K2 K3
# O1 1 2 3
# O2 4 5 6
Alternative solution: 替代解决方案:
df = pd.DataFrame(d).T
More cumbersome method for None
data: None
数据的更麻烦的方法:
d = { 'O1' : [ {'K1': 1},
{'K2': 2},
{'K3': 3} ],
'O2' : [ {'K1': 4},
{'K2': 5},
{'K3': 6} ],
'O3' : [ {'K1': None},
{'K2': None},
{'K3': None} ] }
d = {k: v if isinstance(v[0], dict) else [{k: None} for k in ('K1', 'K2','K3')] for k, v in d.items()}
d = {k: { k: v for d in L for k, v in d.items() } for k, L in d.items()}
df = pd.DataFrame.from_dict(d, orient='index')
# K1 K2 K3
# O1 1.0 2.0 3.0
# O2 4.0 5.0 6.0
# O3 NaN NaN NaN
Data from Jpp 来自Jpp的数据
pd.Series(d).apply(lambda x : pd.Series({ k: v for y in x for k, v in y.items() }))
Out[1166]:
K1 K2 K3
O1 1 2 3
O2 4 5 6
Update 更新
pd.Series(d).apply(lambda x : pd.Series({ k: v for y in x for k, v in y.items() }))
Out[1179]:
K1 K2 K3
O1 1.0 2.0 3.0
O2 4.0 5.0 6.0
O3 NaN NaN NaN
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.