我会将它们解析到DataFrame的字典中：

import os
import glob
import pandas as pd

files = glob.glob('/path/to/*.xlsx')
dfs = {}

for f in files:
    dfs[os.path.splitext(os.path.basename(f))[0]] = pd.read_excel(f)

那么你可以把它们作为普通的字典元素来访问：

dfs['wild_animals']
dfs['domestic_animals']

等等。

你需要得到所有的xlsx文件，而不是使用理解dict，你可以访问任何elm

import pandas as pd
import os
import glob

path = 'Your_path'
extension = 'xlsx'
os.chdir(path)
result = [i for i in glob.glob('*.{}'.format(extension))]

{elm:pd.ExcelFile(elm) for elm in result}

为了完整起见，我想展示一下我最终使用的解决方案，非常接近Khelili的建议，只是做了一些调整以适合我的特定代码，包括在此阶段不创建DataFrame

import os
import pandas as pd
import openpyxl as excel
import glob


#setting up path

path = 'data_inputs'
extension = 'xlsx'
os.chdir(path)
files = [i for i in glob.glob('*.{}'.format(extension))]

#Grouping files - brings multiple files of same type together in a list 

wild_groups = ([s for s in files if "wild" in s])
domestic_groups = ([s for s in files if "domestic" in s])

#Sets up a dictionary associated with the file groupings to be called in another module 
file_names = {"WILD":wild_groups, "DOMESTIC":domestic_groups}
...