Is there a maximum size for python's shelve module?

Question

I'm getting this exception when trying to open shelve persisted files over a certain size which is actually pretty small (< 1MB) but I'm not sure where the exactly number is. Now, I know pickle is sort of the bastard child of python and shelve isn't thought of as a particularly robust solution, but it happens to solve my problem wonderfully (in theory) and I haven't been able to find a reason for this exception.

Traceback (most recent call last):
  File "test_shelve.py", line 27, in <module>
    print len(f.keys())
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/shelve.py", line 101, in keys
    return self.dict.keys()
SystemError: Negative size passed to PyString_FromStringAndSize

I can reproduce it consistently, but I haven't found much on google. Here's a script that will reproduce.

import shelve
import random
import string
import pprint

f = shelve.open('test')
# f = {}

def rand_list(list_size=20, str_size=40):
    return [''.join([random.choice(string.ascii_uppercase + string.digits) for j in range(str_size)]) for i in range(list_size)]

def recursive_dict(depth=3):
    if depth==0:
        return rand_list()
    else:
        d = {}
        for k in rand_list():
            d[k] = recursive_dict(depth-1)
        return d

for k,v in recursive_dict(2).iteritems():
    f[k] = v

f.close()

f = shelve.open('test')
print len(f.keys())

Answer 1

Regarding error itself:

The idea circulating on the web is the data size exceeded the largest integer possible on that machine (the largest 32 bit (signed) integer is 2 147 483 647), interpreted as a negative size by Python.

Your code is running with 2.7.3, so may be a fixed bug.