Re: Memory error while saving dictionary of size 65000X50 using pickle

[Nagu]

I didn't have the problem with dumping as a string. When I tried to
save this object to a file, memory error pops up.

I am sorry for the mention of size for a dictionary. What I meant by
65000X50 is that it has 65000 keys and each key has a list of 50
tuples.

I was able to save a dictionary object with 65000 keys and a list of
15-tuple values to a file. But I could not do the same when I have a
list of 25-tuple values for 65000 keys.

You exmple works just fine on my side.

Thank you,
Nagu

[=?ISO-8859-1?Q?=22Martin_v=2E_L=F6wis=22?=]

I didn't have the problem with dumping as a string. When I tried to

save this object to a file, memory error pops up.

That's not what the backtrace says. The backtrace says that the error
occurs inside pickle.dumps() (and it is consistent with the functions
being called, so it's plausible).

I am sorry for the mention of size for a dictionary. What I meant by
65000X50 is that it has 65000 keys and each key has a list of 50
tuples.

[...]

>
You exmple works just fine on my side.

I can get the program

import pickle

d = {}

for i in xrange(65000):
d[i]=[(x,) for x in range(50)]
print "Starting dump"
s = pickle.dumps(d)

to complete successfully, also, however, it consumes a lot
of memory. I can reduce memory usage slightly by
a) dumping directly to a file, and
b) using cPickle instead of pickle
i.e.

import cPickle as pickle

d = {}

for i in xrange(65000):
d[i]=[(x,) for x in range(50)]
print "Starting dump"
pickle.dump(d,o pen("/tmp/t.pickle","wb") )

The memory consumed originates primarily from the need to determine
shared references. If you are certain that no object sharing occurs
in your graph, you can do
import cPickle as pickle

d = {}

for i in xrange(65000):
d[i]=[(x,) for x in range(50)]
print "Starting dump"
p = pickle.Pickler( open("/tmp/t.pickle","wb") )
p.fast = True
p.dump(d)

With that, I see no additional memory usage, and pickling completes
really fast.

Regards,
Martin