循环以从两个列表创建数据框

Question

I am trying to read in multiple files in a directory into individual dataframes, but I need to make the name of each dataframe a substring from the original filename.

# master list of substrings to look for in filename
sub_list = ['ABC', 'DEF', 'GHI', 'JKL', 'MNO', 'PQR']

# set path
path = 'C:/Users/my_user/Desktop/my_folder'

# get list of files with full path
files = glob.glob(os.path.join(path, '*.xlsx'))

# empty list for extracted substrings
df_names = []

Below is how I am extracting the substrings from the filename

for filename in files:
    if any(sub in filename for sub in sub_list):
        name = [sub_str for sub_str in sub_list if(sub_str in filename)]
        helper = '' # empty string to join with list element to convert to string
        name = helper.join(name) # convert list element to a string
        df_names.append(name)

I iterate over the df_names list to create dataframes

for name in (df_names):
    exec('{} = pd.DataFrame()'.format(name))

However I'm not sure how to add the actual data to these dataframes. I assume there is another way to do this, but haven't been able to figure out how. Maybe using dictionaries?

I've tried the following, but this overwrites all previous names and leaves me with one dataframe named name .

for name, file in zip(df_names, files):
    name = pd.read_excel(file)

Answer 1

Have you considered storing your dataframes in a dictionary instead of in a list?

Instead of:

for name, file in zip(df_names, files):
    name = pd.read_excel(file)

You could use:

dfs = {}

for name, file in zip(df_names, files):
    dfs[name] = pd.read_excel(file)

You could then get the dataframe for file 'ABC' (assuming 'ABC' is a filename) like this:

dfs['ABC']