如何在Python中完全保存/读取类

Question

som = SOM_CLASS() # includes many big difficult data structures
som.hard_work()
som.save_to_disk(filename)
#then later or another program
som = SOM_CLASS()
som.read_from_file(filename)
som.do_anythink_else()

要么

som = SOM_CLASS()
save(som)
#...
load(som)
som.work()

最简单的方法是什么？

Answer 1

你可以(de)序列化pickle.它是向后兼容的,即它将支持未来版本中的所有旧协议.

import pickle

som = SOM_CLASS()
fileObject = <any file-like object>
pickle.dump(som, fileObject)
#...
som = pickle.load(fileObject)
som.work()

但请注意,如果您将腌制对象转移到另一台计算机,请确保连接不会被篡改,因为pickle可能不安全(这是每个pickle用户应该知道的文章).

另一种选择是较旧的模块编组.

Answer 2

我使用这个代码：

import cPickle
import traceback

class someClass():
    def __init__(self):
        #set name from variable name. http://stackoverflow.com/questions/1690400/getting-an-instance-name-inside-class-init
        (filename,line_number,function_name,text)=traceback.extract_stack()[-2]
        def_name = text[:text.find('=')].strip()
        self.name = def_name

        try:
            self.load()
        except:
            ##############
            #to demonstrate
            self.someAttribute = 'bla'
            self.someAttribute2 = ['more']
            ##############

            self.save()

    def save(self):
        """save class as self.name.txt"""
        file = open(self.name+'.txt','w')
        file.write(cPickle.dumps(self.__dict__))
        file.close()

    def load(self):
        """try load self.name.txt"""
        file = open(self.name+'.txt','r')
        dataPickle = file.read()
        file.close()

        self.__dict__ = cPickle.loads(dataPickle)

此代码从其实际类实例名称保存和加载类。代码来自我的博客http://www.schurpf.com/python-save-a-class/。