Question

我正在开展一个项目，我需要使用Google的Vision API分析图像并将响应发布到Dynamodb表格。

我已经成功实现了Vision API，但无法将其响应转换为Python Dictionary。

以下是我的尝试：

       if form.is_valid():
            obj = form
            obj.imageFile = form.cleaned_data['imageFile']
            obj.textFile = form.cleaned_data['textFile']
            obj.save()
            print(obj.imageFile)
            # Process the image using Google's vision API
            image_path = os.path.join(settings.MEDIA_ROOT, 'images/', obj.imageFile.name)
            print(image_path)
            image = vision_image_manager(image_path)
            text_path = os.path.join(settings.MEDIA_ROOT, 'texts/', obj.textFile.name)
            text = nlp_text_manager(text_path)
            # print(image)
            # print(text)
            results = {
                'imageResponse': image,
                'textResult': text
            }
            print(results.values())
            print(type(results))
            post_to_dynamo_db(image, text)

这是Vision api的实施：

def vision_image_manager(image_file):
    # Instantiates a client
    client = vision.ImageAnnotatorClient()
    file_name = str(image_file)
    with open(file_name, 'rb') as img_file:
        content = img_file.read()
    image = types.Image(content=content)
    response = client.label_detection(image=image)
    labels = response.label_annotations
    print('Labels:')
    for label in labels:
        print(label.description)
    return labels

这里是post_to_dynamo_db功能：

def post_to_dynamo_db(image, text):
session = boto3.Session(
    aws_access_key_id=settings.AWS_SERVER_PUBLIC_KEY,
    aws_secret_access_key=settings.AWS_SERVER_SECRET_KEY
)
client = session.resource('dynamodb')
table = client.Table('basetbl')
result_dict = {
    'image': image,
    'text': text
}
json_dict = dict_to_item(result_dict)
# item = dict_to_item(result_dict)
table.put_item(
    Item={
        'id': int(generate_pid()),
        'response_obj': json_dict
    }
)

现在，它没有返回任何错误，但response_obj没有在数据库表中发布，因为它不是对象的正确形式，这里的问题是{{1}来自Google的API的响应类型。

Answer 1

从Vision API获得正确的python友好响应的最佳方法是通过Google的Discovery服务使用此API。

以下是这对你有用的方法：

def vision_image_manager(image_file):
    # Instantiates a client
    service = discovery.build('vision', 'v1', credentials=credentials)
    # text.png is the image file.
    file_name = str(image_file)
    with open(file_name, 'rb') as image:
        image_content = base64.b64encode(image.read())
        service_request = service.images().annotate(body={
            'requests': [{
                'image': {
                    'content': image_content.decode('UTF-8')
                },
                'features': [{
                    'type': 'LABEL_DETECTION',
                }]
            }]
        })
    response = service_request.execute()
    print(response['responses'])
    res_dict = dict(response)
    return res_dict

谷歌的Vision Api protobuf响应对象Python字典

1 个答案: