Reputation: 429
I am running code that creates large objects, containing multiple user-defined classes, which I must then serialize for later use. From what I can tell, only pickling is versatile enough for my requirements. I've been using cPickle to store them but the objects it generates are approximately 40G in size, from code that runs in 500 mb of memory. Speed of serialization isn't an issue, but size of the object is. Are there any tips or alternate processes I can use to make the pickles smaller?
Upvotes: 37
Views: 30505
Reputation: 2382
You might want to use a more efficient pickling protocol.
As of now, there are three pickle protocols:
- Protocol version 0 is the original ASCII protocol and is backwards compatible with earlier versions of Python.
- Protocol version 1 is the old binary format which is also compatible with earlier versions of Python.
- Protocol version 2 was introduced in Python 2.3. It provides much more efficient pickling of new-style classes.
and furthermore, the default is protocol 0, the least efficient one:
If a protocol is not specified, protocol 0 is used. If protocol is specified as a negative value or HIGHEST_PROTOCOL, the highest protocol version available will be used.
Let's check the difference in size between using the latest protocol, which is currently protocol 2 (the most efficient one) and using protocol 0 (the default) for an arbitrary example. Note that I use protocol=-1 here, to make sure we are always using the latest protocol, and that I import cPickle to make sure we are using the faster C implementation:
import numpy
from sys import getsizeof
import cPickle as pickle
# Create list of 10 arrays of size 100x100
a = [numpy.random.random((100, 100)) for _ in xrange(10)]
# Pickle to a string in two ways
str_old = pickle.dumps(a, protocol=0)
str_new = pickle.dumps(a, protocol=-1)
# Measure size of strings
size_old = getsizeof(str_old)
size_new = getsizeof(str_new)
# Print size (in kilobytes) using old, using new, and the ratio
print size_old / 1024.0, size_new / 1024.0, size_old / float(size_new)
The print out I get is:
2172.27246094 781.703125 2.77889698975
Indicating that pickling using the old protocol used up 2172KB, pickling using the new protocol used up 782KB and the difference is a factor of x2.8. Note that this factor is specific to this example - your results might vary, depending on the object you are pickling.
Upvotes: 4
Reputation: 11420
You can combine your cPickle dump
call with a zipfile:
import cPickle
import gzip
def save_zipped_pickle(obj, filename, protocol=-1):
with gzip.open(filename, 'wb') as f:
cPickle.dump(obj, f, protocol)
And to re-load a zipped pickled object:
def load_zipped_pickle(filename):
with gzip.open(filename, 'rb') as f:
loaded_object = cPickle.load(f)
return loaded_object
Upvotes: 60
Reputation: 46596
If you must use pickle and no other method of serialization works for you, you can always pipe the pickle through bzip2
. The only problem is that bzip2
is a little bit slowish... gzip
should be faster, but the file size is almost 2x bigger:
In [1]: class Test(object):
def __init__(self):
self.x = 3841984789317471348934788731984731749374
self.y = 'kdjsaflkjda;sjfkdjsf;klsdjakfjdafjdskfl;adsjfl;dasjf;ljfdlf'
l = [Test() for i in range(1000000)]
In [2]: import cPickle as pickle
with open('test.pickle', 'wb') as f:
pickle.dump(l, f)
!ls -lh test.pickle
-rw-r--r-- 1 viktor staff 88M Aug 27 22:45 test.pickle
In [3]: import bz2
import cPickle as pickle
with bz2.BZ2File('test.pbz2', 'w') as f:
pickle.dump(l, f)
!ls -lh test.pbz2
-rw-r--r-- 1 viktor staff 2.3M Aug 27 22:47 test.pbz2
In [4]: import gzip
import cPickle as pickle
with gzip.GzipFile('test.pgz', 'w') as f:
pickle.dump(l, f)
!ls -lh test.pgz
-rw-r--r-- 1 viktor staff 4.8M Aug 27 22:51 test.pgz
So we see that the file size of the bzip2
is almost 40x smaller, gzip
is 20x smaller. And gzip is pretty close in performance to the raw cPickle, as you can see:
cPickle : best of 3: 18.9 s per loop
bzip2 : best of 3: 54.6 s per loop
gzip : best of 3: 24.4 s per loop
Upvotes: 49