ValueError: Dimension must be 4 but is 3

Question

在 tensorflow Keras 中通过一点修改实现自我关注（例如，残差（添加连接））。

我有以下输入形状：

我的输入：KerasTensor(type_spec=TensorSpec(shape=(None, 8, 6, 64), dtype=tf.float32, name=None), name='multiply/mul:0', description="created by layer 'multiply'")

我的目标是通过self attention处理TensorSpec(shape=(None, 8, 6, 64) (8个时间戳一个一个(6 * 64)) 得到self attention feature map for每个时间戳，然后再次将其连接成输出张量形状 (None, 8, 6, 64)。

实现代码：

import tensorflow as tf
from tensorflow.keras.layers import Permute


def conv1d(channels, ks=1, strides=1, padding='same'):
    conv = tf.keras.layers.Conv1D(channels, ks, strides, padding, activation='relu', use_bias=False,
                                  kernel_initializer='HeNormal')
    return conv


class my_self_attention(tf.keras.layers.Layer):
    def __init__(self, channels):
        super(my_self_attention, self).__init__()
        self.query = conv1d(channels)
        self.key = conv1d(channels)
        self.value = conv1d(channels)
        self.gamma = tf.compat.v1.get_variable("gamma", [1], initializer=tf.constant_initializer(0.0))

    def call(self, x):
        x = tf.reshape(x, shape=[-1, x.shape[2], x.shape[3]])
        f = self.query(x),
        g = self.key(x)
        h = self.value(x)
        attention_weights = tf.keras.activations.softmax(
            tf.matmul(g, Permute((2, 1))(f)))  # query multiply with key and then softmax on it
        sensor_att_fm = tf.matmul(attention_weights, h)
        o = self.gamma * sensor_att_fm + x
        # return tf.reshape(o, shape = [-1, 1, x.shape[1], x.shape[2]])
        return tf.reshape(o, shape=[-1, 1, x.shape[1], x.shape[2]])


 sa = my_self_attention(channels)
 refined_fm = tf.concat([sa(tf.expand_dims(my_input[:, t, :, :], 1)) for t in   range(my_input.shape[1])], 1)

出现跟随错误

ValueError: Dimension must be 4 but is 3 for '{{node my_self_attention/permute/transpose}} = Transpose[T=DT_FLOAT, Tperm=DT_INT32](my_self_attention/permute/transpose/a, my_self_attention/permute/transpose/perm)' with input shapes: [1,?,6,64], [3].

我该如何解决这个问题？

Answer 1

确保通道数与输入数据的最后一个维度相同。还要在此处添加一个逗号：f = self.query(x), 创建一个您可能不想要的元组：这是一个工作示例：

import tensorflow as tf
from tensorflow.keras.layers import Permute

def conv1d(channels, ks=1, strides=1, padding='same'):
  conv = tf.keras.layers.Conv1D(channels, ks, strides, padding, activation='relu', use_bias=False,
                                kernel_initializer='HeNormal')
  return conv

class my_self_attention(tf.keras.layers.Layer):
  def __init__(self, channels):
      super(my_self_attention, self).__init__()
      self.query = conv1d(channels)
      self.key = conv1d(channels)
      self.value = conv1d(channels)
      self.gamma = tf.compat.v1.get_variable("gamma", [1], initializer=tf.constant_initializer(0.0))

  def call(self, x):
      x = tf.reshape(x, shape=[-1, x.shape[2], x.shape[3]])
      f = self.query(x)
      g = self.key(x)
      h = self.value(x)

      attention_weights = tf.keras.activations.softmax(
          tf.matmul(g, Permute((2, 1))(f)))  # query multiply with key and then softmax on it
      sensor_att_fm = tf.matmul(attention_weights, h)
      o = self.gamma * sensor_att_fm + x
      return tf.reshape(o, shape=[-1, 1, x.shape[1], x.shape[2]])


my_input = tf.random.normal((50, 8, 6, 64))
sa = my_self_attention(64)
refined_fm = tf.concat([sa(tf.expand_dims(my_input[:, t, :, :], 1)) for t in   range(my_input.shape[1])], 1)

ValueError: Dimension must be 4 but is 3

ValueError: Dimension must be 4 but is 3

python

keras

tensorflow