如何在Bluemix对​​象存储中转储joblib或pickle文件?

时间:2016-05-13 07:53:29

标签: python ibm-cloud pickle object-storage

我正在使用在Bluemix上运行Flask的Python应用程序。我知道如何使用swiftclient模块的对象存储来创建容器并在其中保存文件,但是如何转储其中包含的joblib或pickle文件?我如何在我的Python程序中加载它?

以下是存储简单文本文件的代码。

import swiftclient

app = Flask(__name__)
CORS(app)


cloudant_service = json.loads(os.environ['VCAP_SERVICES'])['Object-Storage'][0]
objectstorage_creds = cloudant_service['credentials']

if objectstorage_creds:
   auth_url = objectstorage_creds['auth_url'] + '/v3' #authorization URL
   password = objectstorage_creds['password'] #password
   project_id = objectstorage_creds['projectId'] #project id
   user_id = objectstorage_creds['userId'] #user id 
   region_name = objectstorage_creds['region'] #region name 

def predict_joblib():
  print('satart')
  conn = swiftclient.Connection(key=password,authurl=auth_url,auth_version='3',os_options={"project_id": project_id,"user_id": user_id,"region_name": region_name})
  container_name = 'new-container'

  # File name for testing
  file_name = 'requirment.txt'

  # Create a new container
  conn.put_container(container_name)
  print ("nContainer %s created successfully." % container_name)

  # List your containers
  print ("nContainer List:")
  for container in conn.get_account()[1]:
    print (container['name'])

  # Create a file for uploading
  with open(file_name, 'w') as example_file:
    conn.put_object(container_name,file_name,contents= "",content_type='text/plain')

  # List objects in a container, and prints out each object name, the file size, and last modified date
  print ("nObject List:")
  for container in conn.get_account()[1]:
    for data in conn.get_container(container['name'])[1]:
      print ('object: {0}t size: {1}t date: {2}'.format(data['name'], data['bytes'], data['last_modified']))

  # Download an object and save it to ./my_example.txt
  obj = conn.get_object(container_name, file_name)
  with open(file_name, 'w') as my_example:
    my_example.write(obj[1])
  print ("nObject %s downloaded successfully." % file_name)




@app.route('/')
def hello():
    dff = predict_joblib()
    return 'Welcome to Python Flask!'

@app.route('/signUp')
def signUp():
    return 'signUp'


port = os.getenv('PORT', '5000')
if __name__ == "__main__":
    app.debug = True
    app.run(host='0.0.0.0', port=int(port))

1 个答案:

答案 0 :(得分:1)

由于file.openpickle.dumps都返回python docs上的字节对象:

pickle.dumps(obj,protocol = None,*,fix_imports = True)     将对象的pickled表示作为bytes对象返回,而不是将其写入文件。

打开(名称[,模式[,缓冲]])     打开文件,返回文件对象部分中描述的文件类型的对象。如果无法打开文件,则引发IOError。打开文件时,最好使用open()而不是直接调用文件构造函数。

您可以直接处理要存储为obj的对象:

# Create a file for uploading
file = pickle.dumps(obj)
conn.put_object(container_name,file,contents= "",content_type='application/python-pickle')

内容类型的这种变化是由http协议中的标准引起的。这是我从另一个问题得到的,请检查。如上所述:

  

这是事实上的标准。 RFC2046规定:4.5.3。其他应用程序子类型预计未来将定义“应用程序”的许多其他子类型。 MIME实现必须至少将任何未识别的子类型视为等同于“application / octet-stream”。因此,对于非pickle-aware系统,流将看起来像任何其他八位字节流,但对于启用pickle的系统,这是至关重要的信息