how to create a dictionary of pandas dataframes, and return the dataframes into excel worksheets?
dictionary, pandas, python
Solution
I think this is what you're looking for. As I said in the comments, it's probably not the right solution and it's definitely not idomatic for pandas DataFrames.
import pandas as pd
groups = data.groupby('Cluster')
#create a dictionary of dataframes, one for each cluster
c_dict = {k: pd.DataFrame(v) for k, v in groups.groups.iteritems() }
If you want to save this to an excel file, the documentation is here: http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.to_excel.html
There is a nice example at the bottom that will do what you need. Hint: use `for k,v in myDict.iteritems()` to get keys and values.
Problem
how to create a dictionary of pandas dataframes, and return the dataframes into excel worksheets? Hi All, I am learning pandas and python, and I want to create a dictionary which contains a some dataframes, which I can then run metrics over each dataframe. With each unique cluster name (one of the columns) I would like to create a dataframe (subset of original dataframe. Then I would like to be able to select it, run metrics over it, putting the results in a new dataframe, and then place the original dataframe (each subset) into a separate worksheet using xlsxwriter python library. ``` #create dictionary object c_dict = {} #get a list of the unique names c_dict= data.groupby('Cluster').groups #create a dictionary of dataframes, one for each cluster for cluster in c_dict.items(): df = data[data['Cluster']==cluster c_dict[cluster] =df <<< im getting invalid syntax here #go through the dictionary and create a worksheet and put the dataframe in it. for k,v in c_dict.items(): dataframe = GetDF(k) <<< creating worksheets and puts the data from the dataframe > worksheet is not working because of invalid syntax when trying to create dataframe dictionary ^^ dataframe.to_excel(writer,sheet_name=k) writer.save #get the dataframe from the dictionary, GetDF(dictionary_key) return c_dict[dictionary_key] ```