与数据帧具有相同键和值的字典列表

Question

我有以下字典列表

data=[
{'Time': 18057610.0, 'String_8': -1.4209e-15},
{'Time': 18057610.0, 'String_9': 2.7353e-16},
{'Time': 18057610.0, 'String_10': 1.1935e-15},
{'Time': 18057610.0, 'String_11': 1.1624},
{'Time': 18057610.0, 'String_12': -6.1692e-15},
{'Time': 18057610.0, 'String_13': 3.2218e-15},
{'Time': 18057620.4, 'String_8': 2.4377e-16},
{'Time': 18057620.4, 'String_9': -6.2809e-15},
{'Time': 18057620.4, 'String_10': 1.6456e-15},
{'Time': 18057620.4, 'String_11': 1.1651},
{'Time': 18057620.4, 'String_12': 1.7147e-15},
{'Time': 18057620.4, 'String_13': 9.8872e-16},
{'Time': 18057631.1, 'String_8': 4.1124e-15},
{'Time': 18057631.1, 'String_9': 1.5598e-15},
{'Time': 18057631.1, 'String_10': -2.325e-16},
{'Time': 18057631.1, 'String_11': 1.1638},
{'Time': 18057631.1, 'String_12': -3.9983e-15},
{'Time': 18057631.1, 'String_13': 4.459e-16}]

从这个我想得到下面的数据帧

df=

String 8     String 9  ...    String 12   String 13
Time                                      ...                         
1.80576100e+07  -1.4209e-15   2.7353e-16  ...  -6.1692e-15  3.2218e-15
1.80576204e+07   2.4377e-16  -6.2809e-15  ...   1.7147e-15  9.8872e-16
1.80576311e+07   4.1124e-15   1.5598e-15  ...  -3.9983e-15  4.4590e-16 

下面是我尝试过的代码，但它需要"时间"键的所有值，因此我不能使用pd.DataFrame(dd)

dd = defaultdict(list)
for d in data:
for k, v in d.items():    
dd[k].append(v)

我也尝试了a=dict(ChainMap(*data))没有运气。谢谢。

Answer 1

一个选项是使用dict.setdefault来更新字典，其中键为"Time"values和values是字符串的字典。然后构造DataFrame对象:

tmp = {}
for d in data:
tmp.setdefault(d.pop('Time'), {}).update(d)
out = pd.DataFrame.from_dict(tmp, orient='index').rename_axis(index=['Time'])

输出:

String_8      String_9     String_10  String_11     String_12    String_13
Time                                                                            
18057610.0 -1.420900e-15  2.735300e-16  1.193500e-15     1.1624 -6.169200e-15   3.221800e-15
18057620.4  2.437700e-16 -6.280900e-15  1.645600e-15     1.1651  1.714700e-15   9.887200e-16
18057631.1  4.112400e-15  1.559800e-15 -2.325000e-16     1.1638 -3.998300e-15   4.459000e-16

Answer 2

您可以尝试itertools.groupby:

from itertools import groupby
from collections import ChainMap
df = pd.DataFrame([ChainMap(*g) for _, g in groupby(data, lambda k: k["Time"])])
print(df.set_index("Time"))

打印:

String_10  String_11     String_12     String_13      String_8      String_9
Time                                                                                       
18057610.0  1.193500e-15     1.1624 -6.169200e-15  3.221800e-15 -1.420900e-15  2.735300e-16
18057620.4  1.645600e-15     1.1651  1.714700e-15  9.887200e-16  2.437700e-16 -6.280900e-15
18057631.1 -2.325000e-16     1.1638 -3.998300e-15  4.459000e-16  4.112400e-15  1.559800e-15