In tensorflow, I have a two-stream neural network and I want to share the same weights in first two layers. For example,
x1_conv_w1 = tf.get_variable("x1_conv_w1", [5,5,4,32], dtype=tf.float32, initializer=tf.contrib.layers.xavier_initializer_conv2d()). I want to use this convolutional weights in the first layer of two individual networks. Could I directly use it or I need to define another variable and give it reuse setting?
Thank you very much for all of answers. A following question is that
If I only define a variable like
x_conv_w1 = tf.get_variable("x_conv_w1", [5,5,4,32], dtype=tf.float32, initializer=tf.contrib.layers.xavier_initializer_conv2d())
and then I directly use it without "reuse" setting as follows
x_conv_layer_z1 = tf.nn.conv2d(x1, parameters['x_conv_w1'], strides=[1, 1, 1, 1], padding='VALID')
x_un_conv_layer_z1 = tf.nn.conv2d(x2, parameters['x_conv_w1'], strides=[1, 1, 1, 1], padding='VALID')
I want to know whether the aforementioned one is able to effectively share the same weights (x_conv_w1)? If not, why? what is the difference from the case using reuse?
Many thanks!
While defining the first layer using these weights, use reuse = None, while defining other layers using same weights, you should use reuse=True. If you are using the variable scope with reuse, the weights will be created once and used for all the layers automatically.
Thank you very much for all of answers. A following question is that
If I only define a variable like
x_conv_w1 = tf.get_variable("x_conv_w1", [5,5,4,32], dtype=tf.float32, initializer=tf.contrib.layers.xavier_initializer_conv2d())
and then I directly use it without "reuse" setting as follows
x_conv_layer_z1 = tf.nn.conv2d(x1, parameters['x_conv_w1'], strides=[1, 1, 1, 1], padding='VALID')
x_un_conv_layer_z1 = tf.nn.conv2d(x2, parameters['x_conv_w1'], strides=[1, 1, 1, 1], padding='VALID')
I want to know whether the aforementioned one is able to effectively share the same weights (x_conv_w1)? If not, why? what is the difference from the case using reuse?
Many thanks!
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