Unpack zip `filename` to `extract_dir`

    Raises ``UnrecognizedFormat`` if `filename` is not a zipfile (as determined
    by ``zipfile.is_zipfile()``).  See ``unpack_archive()`` for an explanation
    of the `progress_filter` argument.

def unpack_zipfile(filename, extract_dir, progress_filter=default_filter):
    """Unpack zip `filename` to `extract_dir`

    Raises ``UnrecognizedFormat`` if `filename` is not a zipfile (as determined
    by ``zipfile.is_zipfile()``).  See ``unpack_archive()`` for an explanation
    of the `progress_filter` argument.
    """

    if not zipfile.is_zipfile(filename):
        raise UnrecognizedFormat("%s is not a zip file" % (filename,))

    with zipfile.ZipFile(filename) as z:
        for info in z.infolist():
            name = info.filename

            # don't extract absolute paths or ones with .. in them
            if name.startswith('/') or '..' in name.split('/'):
                continue

            target = os.path.join(extract_dir, *name.split('/'))
            target = progress_filter(name, target)
            if not target:
                continue
            if name.endswith('/'):
                # directory
                ensure_directory(target)
            else:
                # file
                ensure_directory(target)
                data = z.read(info.filename)
                with open(target, 'wb') as f:
                    f.write(data)
            unix_attributes = info.external_attr >> 16
            if unix_attributes:
                os.chmod(target, unix_attributes)