使用 io.BytesIO 从 Azure 存储中读取图像？

Question

我有一个包含一些图像的 Azure Blob 存储帐户。我想将它们作为图像处理的字节来读取，因此我可以将它们与 Microsoft 的 Face API 一起使用。我收到一条错误消息，告诉我我需要一个 "Bytes" 对象，但我认为我已经将图像从 blob 格式转换为字节格式。我已经提供了我使用的全部代码，但问题来自最后一段代码

%matplotlib inline
import matplotlib.pyplot as plt
import io
from io import StringIO
import numpy as np
import cv2
from PIL import Image
#from StringIO import StringIO
from PIL import Image
import os
from array import array

我的凭据：

azure_storage_account_name = 'musicsurveyphotostorage'
azure_storage_account_key = None  # dont need key... we will access public 
blob... 

if azure_storage_account_name is None:
raise Exception("You must provide a name for an Azure Storage account")   
from azure.storage.blob import BlockBlobService
blob_service = BlockBlobService(azure_storage_account_name, 
azure_storage_account_key)

select blob 存储容器

#select container (folder) name where the files resides
container_name = 'musicsurveyphotostorage'

#list files in the selected folder
generator = blob_service.list_blobs(container_name)   
blob_prefix = 'https://{0}.blob.core.windows.net/{1}/{2}'

正在加载图像..错误来自以下代码块：

# load image file to process
blob_name = 'shiba.jpg'  #name of image I have stored
blob = blob_service.get_blob_to_bytes(container_name, blob_name)
image_file_in_mem = io.BytesIO(blob)
img_bytes = Image.open(image_file_in_mem)

我收到的错误信息如下：

TypeError   Traceback (most recent call last)
<ipython-input-13-6738e5733c01> in <module>()
 36 blob_name = 'shiba.jpg'  #name of image I have stored
 37 blob = blob_service.get_blob_to_bytes(container_name, blob_name)
**---> 38 image_file_in_mem = io.BytesIO(blob)**
 39 img_bytes = Image.open(image_file_in_mem)

TypeError: a bytes-like object is required, not 'Blob'

Answer 1

我手头没有帐户可以测试，但查看文档，get_blob_to_bytes() returns a Blob 实例 - 获取调用其 content 属性 所需的实际字节数，即：

blob = blob_service.get_blob_to_bytes(container_name, blob_name)
image_file_in_mem = io.BytesIO(blob.content)
img_bytes = Image.open(image_file_in_mem)
# ...