How to create a DataFrame from a list with complex structure?

Question

I have a dataframe with multiple columns (a, b, c, d). I grouped my dataframe by columns 'c' and 'd' and integrated column 'a' with respect to column 'b'. This gave me the output in the following format (the values which go like '1500, 1400 ... 700' are the result of the performed integration):

I was wondering how can I convert that output into a dataframe, where it would show columns 'c', 'd', and also create a new column which would contain the result of integration?

Kriti Pawar · Accepted Answer · 2020-06-04 18:30:08Z

2

Please try:

df = df.reset_index()

answered Jun 4, 2020 at 18:30

Kriti Pawar

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2 Comments

Rauan Saturin Over a year ago

the output itself that I get is not in a 'DataFrame' format

Kriti Pawar Over a year ago

If that is groupby object try .first()

JB Larson · Accepted Answer · 2021-10-22 04:41:38Z

1

I recently created a DataFrame from similar data.

First, create simple lists for each c-value of the complex list.

c1_vals = [1500, 1400, 1300], etc

Create a DataFrame with a dictionary with c-values as keys and lists as values.

ex_df = pd.DataFrame({'1': c1_vals, '2': c2_vals, '3': c3_vals})

The d-values from the complex-data correspond to each lists' index position.

ex_df['1'][0] represents c:1 d:10

You could create variables for individual d-values like this.

c1d10 = ex_df['1'][0]

edited Oct 22, 2021 at 4:41

answered Jun 4, 2020 at 18:53

JB Larson

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3 Comments

JB Larson Over a year ago

My answer will be helpful if your issue is that Pandas DataFrame doesn't accept the type of data your attempting to input @Rauan_Saturin

Rauan Saturin Over a year ago

Mmm, I see the logic, it makes sense. But in my case I have more than a hundred thousand rows, I can't really think of a way to work around your method without manually typing in all of the data. Any ideas?

JB Larson Over a year ago

Depending on how your data is structured, you could potentially use for-loops to append the values to lists. i.e. for val in c1: c1_val.append(val)

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