python - 如何在 Tensorflow Serving 中进行批处理？

Question

部署了 Tensorflow Serving 并为 Inception-V3 运行了测试。工作正常。

现在，想为 Inception-V3 服务做批处理。例如。想要发送 10 张图像而不是一张图像进行预测。

该怎么做？要更新哪些文件(inception_saved_model.py 或 inception_client.py)？这些更新是什么样的？图像是如何传递到服务的 - 它是作为包含图像的文件夹传递还是如何传递？

感谢对这个问题的一些见解。与此相关的任何代码片段都将非常有帮助。

=================================

更新了 inception_client.py

# Copyright 2016 Google Inc. All Rights Reserved.
#
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
#     http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
# ==============================================================================

#!/usr/bin/env python2.7

"""Send JPEG image to tensorflow_model_server loaded with inception model.
"""

from __future__ import print_function

"""Send JPEG image to tensorflow_model_server loaded with inception model.
"""

from __future__ import print_function

# This is a placeholder for a Google-internal import.

from grpc.beta import implementations
import tensorflow as tf
from tensorflow.python.platform import flags
from tensorflow_serving.apis import predict_pb2
from tensorflow_serving.apis import prediction_service_pb2


tf.app.flags.DEFINE_string('server', 'localhost:9000',
                            'PredictionService host:port')
tf.app.flags.DEFINE_string('image', '', 'path to image in JPEG format')
FLAGS = tf.app.flags.FLAGS


def main(_):
   host, port = FLAGS.server.split(':')
   channel = implementations.insecure_channel(host, int(port))
   stub = prediction_service_pb2.beta_create_PredictionService_stub(channel)
   # Send request
   #with open(FLAGS.image, 'rb') as f:
     # See prediction_service.proto for gRPC request/response details.
     #data = f.read()
     #request = predict_pb2.PredictRequest()
     #request.model_spec.name = 'inception'
     #request.model_spec.signature_name = 'predict_images'


 #    request.inputs['images'].CopyFrom(
 #        tf.contrib.util.make_tensor_proto(data, shape=[1]))
 #    result = stub.Predict(request, 10.0)  # 10 secs timeout
 #    print(result)


# Build a batch of images

    request = predict_pb2.PredictRequest()
 request.model_spec.name = 'inception'
 request.model_spec.signature_name = 'predict_images'
  
  image_data = []
  for image in FLAGS.image.split(','):
   with open(image, 'rb') as f:
     image_data.append(f.read())
  
  request.inputs['images'].CopyFrom(
      tf.contrib.util.make_tensor_proto(image_data, shape=[len(image_data)]))
  
  result = stub.Predict(request, 10.0)  # 10 secs timeout
  print(result)
 if __name__ == '__main__':
   tf.app.run()

Answer 1

您应该能够通过对 inception_client.py 中的请求构造代码进行少量更改来计算一批图像的预测。 .该文件中的以下行使用包含单个图像的“批处理”创建请求(注意 shape=[1]，这意味着“长度为 1 的向量”):

with open(FLAGS.image, 'rb') as f:
  # See prediction_service.proto for gRPC request/response details.
  data = f.read()
  request = predict_pb2.PredictRequest()
  request.model_spec.name = 'inception'
  request.model_spec.signature_name = 'predict_images'
  request.inputs['images'].CopyFrom(
      tf.contrib.util.make_tensor_proto(data, shape=[1]))
  result = stub.Predict(request, 10.0)  # 10 secs timeout
  print(result)

您可以在同一向量中传递更多图像以对一批数据运行预测。例如，如果 FLAGS.image 是逗号分隔的文件名列表:

request = predict_pb2.PredictRequest()
request.model_spec.name = 'inception'
request.model_spec.signature_name = 'predict_images'

# Build a batch of images.
image_data = []
for image in FLAGS.image.split(','):
  with open(image, 'rb') as f:
    image_data.append(f.read())

request.inputs['images'].CopyFrom(
    tf.contrib.util.make_tensor_proto(image_data, shape=[len(image_data)]))

result = stub.Predict(request, 10.0)  # 10 secs timeout
print(result)

 if __name__ == '__main__':
   tf.app.run()