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最近我正在使用Tensorflow。我正在探索如何在Tensorflow中实现多层Perceptron。如何在MLP中创建可变数量的图层
我在网上通过了很多教程。他们大多数利用一个或两个隐藏层。一个简单的例子取自here
def forwardprop(X, w_1, w_2):
"""
Forward-propagation.
IMPORTANT: yhat is not softmax since TensorFlow's
softmax_cross_entropy_with_logits() does that internally.
"""
h = tf.nn.sigmoid(tf.matmul(X, w_1)) # The \sigma function
yhat = tf.matmul(h, w_2) # The \varphi function
return yhat
X = tf.placeholder("float", shape=[None, x_size])
y = tf.placeholder("float", shape=[None, y_size])
# Weight initializations
w_1 = init_weights((x_size, h_size))
w_2 = init_weights((h_size, y_size))
# Forward propagation
out = forwardprop(X, w_1, w_2)
在这段代码中,有一个隐藏层。现在我想知道如果我想构建一个可变数量的分层全连接神经网络。
假设列表h_archi = [100 150 100 50]其中每个值表示第i层神经元的数量(在这种情况下层的总数为4)。因此,对于层实现的变量数,我编写了以下丑陋的代码,
emb_vec = tf.Variable(tf.random_normal([vocabulary_size, EMBEDDING_DIM]), name="emb_vec")
tot_layer = len(h_archi)
op = np.zeros(tot_layer+1)
hid_rep = np.zeros(tot_layer+1)
bias = np.zeros(tot_layer+1)
op[0] = tf.matmul(x, emb_vec)
for idx,tot_neu in enumerate(h_archi):
assert(tot_neu > 0)
layer_no = idx+1
r,c = op[layer_no-1].get_shape()
hid_rep[layer_no] = tf.Variable(tf.random_normal([c,tot_neu]),name="hid_{0}_{1}".format(layer_no-1,layer_no))
bias[layer_no] = tf.Variable(tf.random_normal([tot_neu]), name="bias_{0}".format(layer_no))
op[layer_no] = tf.add(tf.matmul(op[layer_no-1],hid_rep[layer_no]),bias[layer_no])
r,c = op[tot_layer].get_shape()
last_layer = tf.Variable(tf.random_normal([c,output_size]),name="hid_{0}_{1}".format(tot_layer,"last_layer"))
bias_last = tf.Variable(tf.random_normal([output_size]), name="bias_last")
output = tf.add(tf.matmul(op[tot_layer],last_layer))
prediction = tf.nn.softmax(output)
此代码是完全错误的,因为tensorflow不支持赋值操作。那么设计这种东西的正确方法是什么?
非常感谢您的帮助。 –