Python pickle实例变量

Question

I am doing some calculations on an instance variable, and after that is done I want to pickle the class instance, such that I don't have to do the calculations again. Here an example:

import cPickle as pickle

class Test(object):
    def __init__(self, a, b):
        self.a = a
        self.b = b
        self.c = None
    def compute(self, x):
        print 'calculating c...'
        self.c = x * 2

test = Test(10, 'hello')
test.compute(6)

# I have computed c and I want to store it, so I don't have to recompute it again:

pickle.dump(test, open('test_file.pkl', 'wb'))

After test.compute(6) I can check to see what test.__dict__ is:

>>> test.__dict__
{'a': 10, 'c': 12, 'b': 'hello'}

I thought that is what going to get pickled; however,

When I go to load the class instance:

import cPickle as pickle

from pickle_class_object import Test

t2 = pickle.load(open('test_file.pkl', 'rb'))

I see this in the shell:

calculating c...

Which means that I did not pickle c and I am computing it over again.

Is there a way to pickle test how I want to? So I don't have to compute c over again. I see that I could just pickle test.__dict__ , but I am wondering if there is a better solutions. Also, my understanding about what is going on here is weak, so any comment about what is going would be great. I've read about __getstate__ and __setstate__ , but I don't see how to apply them here.

Answer 1

You are importing the pickle_class_object module again, and Python runs all code in that module.

Your top-level module code includes a call to .compute() , that is what is being called.

You may want to move the code that creates the pickle out of the module, or move it to a if __name__ == '__main__': guarded section:

if __name__ == '__main__':
    test = Test(10, 'hello')
    test.compute(6)

    pickle.dump(test, open('test_file.pkl', 'wb'))

Only when running a python file as the main script is __name__ set to __main__ ; when imported as a module __name__ is set to the module name instead and the if branch will not run.

Answer 2

Pickling works as you expect it to work. The problem here is when you run the new script, you import the module that contains the class Test . That entire module is run including the bit where you create test .

The typical way to handle this sort of thing would be to protect the stuff in a if __name__ == "__main__: block.

class Test(object):
    def __init__(self, a, b):
        self.a = a
        self.b = b
        self.c = None
    def compute(self, x):
        print 'calculating c...'
        self.c = x * 2

if __name__ == "__main__":
    import cPickle as pickle

    test = Test(10, 'hello')
    test.compute(6)

    # I have computed c and I want to store it, so I don't have to recompute it again:

    pickle.dump(test, open('test_file.pkl', 'wb'))

Answer 3

That isn't what's happening. You import a python module that has code in it at the top level, which executes when you import the module. You can see that your code works as you intended:

import cPickle as pickle

class Test(object):
    def __init__(self, a, b):
        self.a = a
        self.b = b
        self.c = None
    def compute(self, x):
        print 'calculating c...'
        self.c = x * 2

test = Test(10, 'hello')
test.compute(6)

pickle.dump(test, open('test_file.pkl', 'wb'))

t2 = pickle.load(open('test_file.pkl', 'rb'))
print t2.c


--output:--
calculating c...
12

If your code worked as you describe, then you would see "calculating c..." twice.