The zipfile write() method supports an extra argument (arcname) which is the archive name to be stored in the zip file, so you would only need to change your code with:

from os.path import basename
...
zip.write(first_path, basename(first_path))
zip.write(second_path, basename(second_path))
zip.close()

When you have some spare time reading the documentation for zipfile will be helpful.

Your logic is wrong for identifying gdbs within the zip. Look at zip_f.namelist() to see why. You just need to concatenate the zip path with any zip member names that end with .gdb/

import os
from zipfile import ZipFile

folder = r'C:\example'

gdbfiles = []
shpfiles = []
csvfiles = []
jsonfiles = []
zipfiles = []

for root, dirs, files in os.walk(folder):
    for file in files:
        if file.endswith(".zip"):
            zipfiles.append(os.path.join(root, file))
        elif file.endswith(".csv"):
            csvfiles.append(os.path.join(root, file))
        elif file.endswith(".json"):
            jsonfiles.append(os.path.join(root, file))


#find zipped gdb files
for zip_path in zipfiles:
    zip_f = ZipFile(zip_path)
    for name in zip_f.namelist():
        if name.endswith('.gdb/'):
            gdbfiles.append(os.path.join(zip_path, name))

print(gdbfiles)

You can access it using filename attribute:

a.filename

The easiest way is to use shutil.make_archive. It supports both zip and tar formats.

import shutil
shutil.make_archive(output_filename, 'zip', dir_name)

If you need to do something more complicated than zipping the whole directory (such as skipping certain files), then you'll need to dig into the zipfile module as others have suggested.

As shown in: Python: Getting files into an archive without the directory?

The solution is:

     ''' 
    zip_file:
        @src: Iterable object containing one or more element
        @dst: filename (path/filename if needed)
        @arcname: Iterable object containing the names we want to give to the elements in the archive (has to correspond to src) 
'''
def zip_files(src, dst, arcname=None):
    zip_ = zipfile.ZipFile(dst, 'w')

    print src, dst
    for i in range(len(src)):
        if arcname is None:
            zip_.write(src[i], os.path.basename(src[i]), compress_type = zipfile.ZIP_DEFLATED)
        else:
            zip_.write(src[i], arcname[i], compress_type = zipfile.ZIP_DEFLATED)

    zip_.close()

The zipfile.write() method takes an optional arcname argument that specifies what the name of the file should be inside the zipfile

I think you need to do a modification for the destination, otherwise it will duplicate the directory. Use :arcname to avoid it. try like this:

Copyimport os
import zipfile

def zip(src, dst):
    zf = zipfile.ZipFile("%s.zip" % (dst), "w", zipfile.ZIP_DEFLATED)
    abs_src = os.path.abspath(src)
    for dirname, subdirs, files in os.walk(src):
        for filename in files:
            absname = os.path.abspath(os.path.join(dirname, filename))
            arcname = absname[len(abs_src) + 1:]
            print 'zipping %s as %s' % (os.path.join(dirname, filename),
                                        arcname)
            zf.write(absname, arcname)
    zf.close()

zip("src", "dst")