[Python] To unzip a file

I want to unzip a file of type *.sec.gz which is a zipfile. But i'm getting badfile. Can someone guide me to resolve this?

import zipfile
def unzip(path):
zfile = zipfile.ZipFile(path)
for name in zfile.namelist():
(dirname, filename) = os.path.split(name)
if filename == '':
# directory
if not os.path.exists(dirname):
os.mkdir(dirname)
else:
# file
fd = open(name, 'w')
fd.write(zfile.read(name))
fd.close()
zfile.close()

k=unzip('C://test//08October2014//DATA_INTV_NEW//Oct0814//1.sec.gz')


Output:

BadZipfile Traceback (most recent call last)
<ipython-input-7-5134b63e752e> in <module>()
27 zfile.close()
28
---> 29 k=unzip('C://test//08October2014//DATA_INTV_NEW//Oct0814//1.sec.gz')

<ipython-input-7-5134b63e752e> in unzip(path)
13
14 def unzip(path):
---> 15 zfile = zipfile.ZipFile(path)
16 for name in zfile.namelist():
17 (dirname, filename) = os.path.split(name)

C:\Python27\Lib\zipfile.pyc in __init__(self, file, mode, compression, allowZip64)
768 try:
769 if key == 'r':
--> 770 self._RealGetContents()
771 elif key == 'w':
772 # set the modified flag so central directory gets written

C:\Python27\Lib\zipfile.pyc in _RealGetContents(self)
809 raise BadZipfile("File is not a zip file")
810 if not endrec:
--> 811 raise BadZipfile, "File is not a zip file"
812 if self.debug > 1:
813 print endrec

BadZipfile: File is not a zip file

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