The following is reproducable -> so lets say you have a list with the df/company names:
companies = ['AA', 'AAPL', 'BA', 'YHOO']
you probably also have data, presumably also a list? (or rather list of lists) like:
 content_of_lists = [
 [['a', '1'], ['b', '2']],
 [['c', '3'], ['d', '4']],
 [['e', '5'], ['f', '6']],
 [['g', '7'], ['h', '8']]
]
in this special example the df´s should probably look very much alike, so this does not need to be very complicated:
dic={}
for n,m in zip(companies, range(len(content_of_lists))):
   dic["df_{}".format(n)] = pd.DataFrame(content_of_lists[m]).rename(columns = {0: "col_1", 1:"col_2"}) 
Here you would have to use dic["df_AA"] to get to the dataframe inside the dictionary.
But Should you require more "distinct" naming of the dataframes I think you would have to use for example if-conditions, like:
dic={}
    for n,m in zip(companies, range(len(content_of_lists))):
if n == 'AA':
    special_naming_1 = pd.DataFrame(content_of_lists[m]).rename(columns = {0:     
    "col_1", 1:"col_2"})
elif n == 'AAPL':
    special_naming_2 ...
It is a little more effort but it allows you to grab the dataframe object in a more conventional way by just writing special_naming_1 instead of dic['df_AA'] and gives you more controll over the dataframes names and column names if that´s important.