list of dictionaries with same key and value to data frame-CodePudding

I have the following list of dictionaries

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}]

From this one I want to get to get the following dataframe

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

below is the code I have tried but it takes all the values of 'Time' key thus I can't use pd.DataFrame(dd)

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

I also tried a=dict(ChainMap(*data)) with no luck. Thanks.

CodePudding user response：

One option is to use dict.setdefault to update a dictionary where the keys are "Time" values and the values are dictionary of Strings. Then construct the DataFrame object:

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

Output:

                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

CodePudding user response：

You can try 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"))

Prints:

               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