Insert List In Pandas Dataframe Cell

I have a dictionary where each key has a list of values. Length of the list associated with each key is different. I want to convert the dictionary into a pandas dataframe with two

Solution 1:

If you pass a list, pandas considers it as several rows. However, you can trick it by placing your list as the single element of an outer list as bellow:

import pandas as pd
mapping_dict = {'A':[['a', 'b', 'c', 'd']], 'B':[['aa', 'bb', 'cc']]}
df = pd.DataFrame(mapping_dict)
df

        A                 B
0   [a, b, c, d]    [aa, bb, cc]

Solution 2:

Use pd.Series inside constructor, since dict values sizes are not equal, then set_axis to add column names i.e

mapping_dict = {'A':['a', 'b', 'c', 'd'], 'B':['aa', 'bb', 'cc']}

df = pd.DataFrame(pd.Series(mapping_dict).reset_index()).set_axis(['Key','Value'],1,inplace=False)

  Key         Value
0   A  [a, b, c, d]
1   B  [aa, bb, cc]

Option 2 , convert the dict items to list then pass it to constructor:

df = pd.DataFrame(list(mapping_dict.items()),columns=['Key','Value'])

Solution 3:

I think you might have to update your dictionary beforehand then you can use from_dict. Update to make your dictionary to make it a list of list.

import pandas as pd
mapping_dict = {'A':['a', 'b', 'c', 'd'], 'B':['aa', 'bb', 'cc']}
updated_dict = {k: [v] for k, v in mapping_dict.items()}
df = pd.DataFrame.from_dict(updated_dict,orient='index')

If you want your exact formatting

df_formatted = df.reset_index()
df_formatted.columns = ['Key', 'Value']
print(df_formatted)

  Key         Value
0   B  [aa, bb, cc]
1   A  [a, b, c, d]

UPDATE

Bharath's answer is shorter but if you still want to use from_dict then you can take part of his method to do

df2 = pd.DataFrame.from_dict(list(mapping_dict.items()))
df2.columns = ['Key', 'Value']

Solution 4:

I had to insert a list at a specific location and the following solution worked for me.

For some reason pandas doesn't complain when returning a list/array in an apply function so:

df.loc[0,'A'] = df.loc[0,A].apply(lambda x: ['a', 'b', 'c', 'd'])