将字典嵌套到 Pandas DataFrame
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/49115118/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Nested dictionary to pandas DataFrame
提问by ba_ul
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. key1, key2, key3 are also always the same.
内部数组的长度始终相同。key1、key2、key3 也总是相同的。
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. In a few cases, in the inner array there are three Nones instead of the three dictionaries. Like this:
数据中存在问题，我认为这是给定解决方案不起作用的原因。在少数情况下，内部数组中有三个Nones 而不是三个字典。像这样：
outer_key3: [ None, None, None ]
outer_key3: [ None, None, None ]
采纳答案by YOBEN_S
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
回答by jpp
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 Nonedata:
比较繁琐的None数据方法：
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
