title: How to extract the file name from a file path?
body: <p>I have the following code:</p>
  
  <pre><code>os.listdir("staging")
  
  # Seperate filename from extension
  sep = os.sep
  
  # Change the casing
  for n in os.listdir("staging"):
      print(n)
      if os.path.isfile("staging" + sep + n):
          filename_one, extension = os.path.splitext(n)
          os.rename("staging" + sep + n, "staging" + sep + filename_one.lower() + extension)
  
  # Show the new file names
  print ('\n--------------------------------\n')
  for n in os.listdir("staging"):
      print (n)
  
  # Remove the blanks, -, %, and /
  for n in os.listdir("staging"):
      print (n)
      if os.path.isfile("staging" + sep + n):
          filename_zero, extension = os.path.splitext(n)
          os.rename("staging" + sep + n , "staging" + sep + filename_zero.replace(' ','_').replace('-','_').replace('%','pct').replace('/','_') + extension)
  
  # Show the new file names
  print ('\n--------------------------------\n')
  for n in os.listdir("staging"):
      print (n)
  
  """
  In order to fix all of the column headers and to solve the encoding issues and remove nulls, 
  first read in all of the CSV's to python as dataframes, then make changes and rewrite the old files
  """
  import os
  import glob
  import pandas as pd
  
  files = glob.glob(os.path.join("staging" + "/*.csv"))
  
  print(files)
  
  # Create an empty dictionary to hold the dataframes from csvs
  dict_ = {}
  
  # Write the files into the dictionary
  for file in files:
      dict_[file] = pd.read_csv(file, header = 0, dtype = str, encoding = 'cp1252').fillna('')
  </code></pre>
  
  <p>In the dictionary, the dataframes are named as "folder/name(csv)" what I would like to do is remove the prefix "staging/" from the keys in the dictionary. </p>
  
  <p>How can I do this?</p>