从文件路径列表中创建具有不同扩展名的“空”文件

Question

I have a list called viszFiles and it has all the file paths to .visz files in a folder ['/Users/Blake/Deskop/blake.shelton.visz', '/Users/Blake/Deskop/bill.bob-1.visz'] .

The list is generated by this snippet of code:

viszFiles = [] #List of .visz file paths
for root, dirs, files in os.walk('/Users/Blake/Desktop/'):
    [viszFiles.append(os.path.join('/Users/Blake/Deskop/', _file))
        for _file in files if
            fnmatch.fnmatch(_file, '*.visz')]

I have searched long and hard and have been unsuccessful in finding out how to exactly go about this: Opening the .visz files (which are basically .tar.gz files) and then creating "empty" .ovpn versions of those files keeping just their root name, such as blake.shelton.visz would become blake.shelton.ovpn and so on for any other files with paths in the list. I don't want to copy any data, just create a .ovpn (basically a .tar.gz file I believe) "empty" copy with the same root name while maintaining the originals.

I'd really appreciate the help!

Answer 1

If the list of files is being created correctly, you can use some additional lines to create the empty files. Hope this helps.

viszFiles = [] #List of .visz file paths
    for root, dirs, files in os.walk('/Users/Blake/Desktop/'):
        [viszFiles.append(os.path.join('/Users/Blake/Deskop/', _file))
            for _file in files if
            fnmatch.fnmatch(_file, '*.visz')]

for file in viszFiles:
    filename, file_extension = os.path.splitext(file)
    filename += '.visz'
    with open(filename,'w') as f:
        pass