你怎么能在一个文件中腌制多个对象？

Question

在我的情况下,我希望挑选(使用pickle.dump())两个单独的列表到一个文件,然后从一个单独的文件中检索这些,但是当使用时pickle.load()我一直在努力寻找一个列表结束的地方,而下一个列表开始,因为我根本不知道如何pickle.dump()即使在查看文档之后,它们也能够轻松检索.

Answer 1

pickle 将按您倾倒的顺序读取它们.

import pickle

test1, test2 = ["One", "Two", "Three"], ["1", "2", "3"]
with open("C:/temp/test.pickle","wb") as f:
    pickle.dump(test1, f)
    pickle.dump(test2, f)
with open("C:/temp/test.pickle", "rb") as f:
    testout1 = pickle.load(f)
    testout2 = pickle.load(f)

print testout1, testout2

打印出来['One', 'Two', 'Three'] ['1', '2', '3'].要腌制任意数量的对象,或者只是让它们更容易使用,你可以将它们放在一个元组中,然后你只需要腌制一个对象.

import pickle

test1, test2 = ["One", "Two", "Three"], ["1", "2", "3"]
saveObject = (test1, test2)
with open("C:/temp/test.pickle","wb") as f:
    pickle.dump(saveObject, f)
with open("C:/temp/test.pickle", "rb") as f:
    testout = pickle.load(f)

print testout[0], testout[1]

打印出来 ['One', 'Two', 'Three'] ['1', '2', '3']