How to extract the file name from a file path?

Question

I have the following code:

os.listdir(\"staging\")

# Seperate filename from extension
sep = os.sep

# Change the casing
for n in os.listdir(\"staging\"):
           


        

Answer 1

If all you want to do is truncate the file paths to just the filename, you can use os.path.basename:

for file in files:
    fname = os.path.basename(file)
    dict_[fname] = (pd.read_csv(file, header=0, dtype=str, encoding='cp1252')
                      .fillna(''))

Example:

os.path.basename('Desktop/test.txt')
# 'test.txt'