【发布时间】:2017-11-15 12:12:35
【问题描述】:
我在 Python 中看到并编写了用于腌制对象的代码。但它们都创建物理文件来包含数据。我希望我将数据写入内存并读取它并腌制它并传输它。
有可能吗?
from PIL import ImageGrab
import io
import codecs
import pickle
# Model class to send Process Data to the server
class ProcessData:
process_id = 0
project_id = 0
task_id = 0
start_time = 0
end_time = 0
user_id = 0
weekend_id = 0
# Model class to send image data to the server
class ProcessScreen:
process_id = 0
image_data = bytearray()
image_name = "Dharmindar_screen.jpg"
ImageGrab.grab().save(image_name,"JPEG")
image_data = None
with codecs.open(image_name,'rb') as image_file:
image_data = image_file.read()
serialized_process_data = io.BytesIO()
process_data = ProcessData()
process_data.process_id = 1
process_data.project_id = 2
process_data.task_id = 3
process_data.user_id = 4
process_data.weekend_id = 5
process_data.start_time = 676876
process_data.end_time = 787987
process_screen = ProcessScreen()
process_screen.process_id = process_data.process_id
process_screen.image_data = image_data
prepared_process_data = (process_data, process_screen)
process_data_serializer = pickle.Pickler()
process_data_serializer(serialized_process_data).dump(prepared_process_data)
print('Data serialized.')
if process_data_serializer is not None:
d = process_data_serializer.getvalue()
deserialized_data = None
with open(d, 'rb') as serialized_data_file:
process_deserializer = pickle.Unpickler(serialized_data_file)
deserialized_data = process_deserializer.load()
else:
print('Empty')
上面的代码抛出 TypeError: Required argument 'file' (pos 1) not found
【问题讨论】:
-
你的意思是写到内存吗?如果您想将对象发送到其他 Python 代码,您可以简单地将其作为参数传递,您想要完成什么?
-
只是我愿意通过 Socket 将我的数据发送到服务器。为此,我需要腌制物体。但是为此,我已经看过每个代码,他们说首先将对象以字节的形式写入文件,然后再次从该文件中检索。我想在内存中完成,而不是物理写入文件。
-
你可以使用 StringIO 对象,它提供缓冲区实例,但正如 Shayn 所说,你应该真正解释一下你正在尝试做什么。
-
使用
pickle.dumps? -
不是你的问题的答案,但为什么不考虑使用像 GRPC 这样的东西,从长远来看,这可能会更易于维护 thttps://grpc.io/docs/tutorials/basic/python .html
标签: python serialization pickle