How to check whether a directory is a sub directory of another directory

I like to write a template system in Python, which allows to include files.

e.g.

    This is a template
    You can safely include files with safe_include`othertemplate.rst`

As you know, including files might be dangerous. For example, if I use the template system in a web application which allows users to create their own templates, they might do something like

I want your passwords: safe_include`/etc/password`

So therefore, I have to restrict the inclusion of files to files which are for example in a certain subdirectory (e.g. /home/user/templates)

The question is now: How can I check, whether /home/user/templates/includes/inc1.rst is in a subdirectory of /home/user/templates?

Would the following code work and be secure?

import os.path

def in_directory(file, directory, allow_symlink = False):
    #make both absolute    
    directory = os.path.abspath(directory)
    file = os.path.abspath(file)

    #check whether file is a symbolic link, if yes, return false if they are not allowed
    if not allow_symlink and os.path.islink(file):
        return False

    #return true, if the common prefix of both is equal to directory
    #e.g. /a/b/c/d.rst and directory is /a/b, the common prefix is /a/b
    return os.path.commonprefix([file, directory]) == directory

As long, as allow_symlink is False, it should be secure, I think. Allowing symlinks of course would make it insecure if the user is able to create such links.

UPDATE - Solution The code above does not work, if intermediate directories are symbolic links. To prevent this, you have to use realpath instead of abspath.

UPDATE: adding a trailing / to directory to solve the problem with commonprefix() Reorx pointed out.

This also makes allow_symlink unnecessary as symlinks are expanded to their real destination

import os.path

def in_directory(file, directory):
    #make both absolute    
    directory = os.path.join(os.path.realpath(directory), '')
    file = os.path.realpath(file)

    #return true, if the common prefix of both is equal to directory
    #e.g. /a/b/c/d.rst and directory is /a/b, the common prefix is /a/b
    return os.path.commonprefix([file, directory]) == directory