如何将 DataFrame 字典转换为单个 DataFrame（Python、Pandas）

Question

I have an original dataframe with 4 columns (for the example lets call them product_id, year_month, week, order_amount) and > 50,000 rows. There are 240 individual product_id values and each one of them behaves differently in the data, therefore I wanted to create individual dataframes from the original one based on individual product_id. I was able to do this by utilizing:

dict_of_productid = {k: v for k, v in df.groupby('product_id)}

this created a dictionary with the key being the product_id and the values being the columns: product_id, year_month, week, order_amount. Each item in the dictionary also maintained the index from the original df. for example: if product_id = dvvd56 was on row# 4035 then on the dictionary it will be on the dataframe created for product_id dvvd56 but with the index still being 4035.

What I'm stuck with now is a dictionary with df's as values but can't find a way to convert these values into individual dataframes I can use and manipulate. If there is a way to do this please let me know. I'll be very grateful. thank you

Answer 1

I found a way to go about this, but I dont know if this is the most appropriate way, but it might help for further answers in order to clarify what I want to do.

First step was to convert the unique values into a list and then sorting them in order:

product_id_list = df['product_id'].value_counts().index.to_list()
product_id_list = sorted(product_id_list)

After this was done I created a formula and then iterated over it with the individual values of the product_id_list:

def get_df(key): 
    for k in key: 
        df_productid = dict_of_productid[k]
    return df_productid

for c, i in enumerate(product_id_list):
    globals()[f'df_{c}'] = get_df([f'{i}'])

this allows me now to separate all the values of the dictionary that was created into separate dataframes that I can call without explicitly stating the product id. I can just do df_1 and get the dataframe.

(I dont know if this is the most efficient way to go about this)