python 对象序列化并压缩

python如果需要将对象存成文件，方便以后（或放在其他地方）使用，可以使用序列化pickle，

参考另一篇随笔：https://www.cnblogs.com/taoshiqian/p/9771786.html

但是一个较大的对象简单序列化成文件，可能会占用很大空间，而此时对象中可能有很多重复信息，完全可以压缩存储

因此，对象->文件：序列化+压缩，文件->对象：解压+反序列化，更省空间

import os, pickle, gzip


def save_object_to_zip(objects, filename):
    if not os.path.exists(filename):
        file_path = os.path.split(filename)[0]
        if file_path and not os.path.exists(file_path):  # 需要文件夹
            os.mkdir(os.path.split(filename)[0])  # 创建文件夹
        os.mknod(filename)  # 创建文件
    fil = gzip.open(filename, 'wb')
    pickle.dump(objects, fil)
    fil.close()


def load_object_from_zip(filename):
    fil = gzip.open(filename, 'rb')
    while True:
        try:
            return pickle.load(fil)
        except EOFError:
            break
    fil.close()


def test_pickle_and_zip():
    L1 = [1, 2, 3, 4, 5]
    filename = 'L'
    save_object_to_zip(L1, filename)
    L2 = load_object_from_zip(filename)
    print(L2)
    print(type(L2))

    L1 = 8
    filename = 'int/L'
    save_object_to_zip(L1, filename)
    L2 = load_object_from_zip(filename)
    print(L2)
    print(type(L2))

    L1 = {'a': 1, 'b': 2}
    filename = 'dict/L'
    save_object_to_zip(L1, filename)
    L2 = load_object_from_zip(filename)
    print(L2)
    print(type(L2))


if __name__ == '__main__':
    test_pickle_and_zip()

#输出

[1, 2, 3, 4, 5]
<class 'list'>

8
<class 'int'>

{'a': 1, 'b': 2}
<class 'dict'>