I have a multi-level dictionary called mydict that looks like the one bellow and I am trying to do a concatenation that gives me a different dataframe for each keys of the dict.
mydict -> {agent_1: {step1: {...}, step2: {...}, step3: {...}}, ...}
I can do that with the following code but it means doing it by hand and I am working with large data.
boo = pd.DataFrame.from_dict(mydict[agent_1]['step_1'])
boo1 = pd.DataFrame.from_dict(mydict[agent_1]['step_2'])
boo2 = pd.DataFrame.from_dict(mydict[agent_1]['step_3'])
boo3 = pd.DataFrame.from_dict(mydict[agent_1]['step_4'])
boo_tot = pd.concat([boo, boo1, boo2, boo3])
Also I have troubles automatizing this because the number of step parameter is not the same for each key of mydict.
As an example, let's say agent_1 has 4 steps while agent_2 only has 3.
I tried following what was given in this stackoverflow accepted answer but it does not apply to my situation.
As you can see bellow, this method gives me one dataframe and all the values for each step (= etape in the df bellow) are in one row each time whereas I would like a different dataframe for each agent with pcc, time, route_type and mode as columns and their values printed on row under the other for each step
This is what I was hoping to get instead:
For agent_1:
mode pcc time route_type etape id
0 TC data1 time1 data1 etape_0 NaN
1 TC data2 time2 data2 etape_0 NaN
2 TC data3 time3 data3 etape_0 NaN
3 TC data4 time4 data4 etape_0 NaN
... ... ... ... ... ... ...
16 MAP data16 time16 data16 etape_1 NaN
17 MAP data17 time17 data17 etape_1 NaN
18 MAP data18 time18 data18 etape_1 NaN
19 MAP data19 time19 data19 etape_1 NaN
... ... ... ... ... ... ...
And then I'd get a second dataframe for agent_2, a third for agent_3 etc.
I couldn't find a way to get what I want and unfortunately my vocabulary about this subject is very limited since I don't use multi-level dicts often. Any tips on the matter would be a huge help.
