Should I Use Regex to Get a File Name?

Solution:

import osfile = open('/some/path/to/a/test.csv')fname = os.path.splitext(str(file))[0].split('/')[-1]print(fname)# test

If you get file path and name as string, then:

import osfile = "User/Folder1/test/filename.csv"fname = os.path.splitext(file)[0].split('/')[-1]print(fname)# filename

Explanation on how it works:

Pay attention that command is os.path.splitEXT, not os.path.splitTEXT - very common mistake.

The command takes argument of type string, so if we use file = open(...), then we need to pass os.path.splitext argument of type string. Therefore in our first scenario we use:

str(file)

Now, this command splits complete file path + name string into two parts:

os.path.splitext(str(file))# result:['/some/path/to/a/test','csv']

In our case we only need first part, so we take it by specifying list index:

os.path.splitext(str(file))[0]# result: '/some/path/to/a/test'

Now, since we only need file name and not the whole path, we split it by /:

os.path.splitext(str(file))[0].split('/')# result:['some','path','to','a','test']

And out of this we only need one last element, or in other words, first from the end:

os.path.splitext(str(file)[0].split('/')[-1]

Hope this helps.

Check for more here: Extract file name from path, no matter what the os/path format