修正Tensorflow中的去卷積層

Question

我試圖用Tensorflow構建一個變分自動編碼器。我從最簡單的模型開始。我有以下方法：

def conv_layer(x, w_shape, b_shape, padding='SAME'): 
    W = weight_variable(w_shape) 
    tf.summary.histogram(W.name, W) 

    b = bias_variable(b_shape) 
    tf.summary.histogram(b.name, b) 

    # Note that I used a stride of 2 on purpose in order not to use max pool layer. 
    activations = tf.nn.relu(tf.nn.conv2d(x, W, strides=[1, 1, 1, 1], padding=padding) + b) 
    tf.summary.histogram(activations.name, activations) 
    return activations 

def deconv_layer(x, w_shape, b_shape, padding="SAME"): 
    W = weight_variable(w_shape) 
    tf.summary.histogram(W.name, W) 

    b = bias_variable(b_shape) 
    tf.summary.histogram('bias', b) 

    x_shape = tf.shape(x) 

    out_shape = tf.stack([x_shape[0], x_shape[1], x_shape[2], w_shape[2]]) 
    # Note that I have used a stride of 2 since I used a stride of 2 in conv layer. 
    transposed_activations = tf.nn.conv2d_transpose(x, W, out_shape, [1, 1, 1, 1], padding=padding) + b 
    tf.summary.histogram(transposed_activations.name, transposed_activations) 
    return transposed_activations 

而整個網絡的模型如下：

with tf.name_scope('conv1'): 
    conv1 = conv_layer(image, [3, 3, 3, 32], [32]) 
with tf.name_scope('conv2'): 
    conv2 = conv_layer(conv1, [3, 3, 32, 64], [64]) 
with tf.name_scope('conv3'): 
    conv3 = conv_layer(conv2, [3, 3, 64, 128], [128]) 
with tf.name_scope('conv4'): 
    conv4 = conv_layer(conv3, [3, 3, 128, 256], [256]) 

with tf.name_scope('z'): 
    z = conv_layer(conv4, [3, 3, 256, 256], [256]) 

with tf.name_scope('deconv4'): 
    deconv4 = deconv_layer(z, [3, 3, 128, 256], [128]) 
with tf.name_scope('deconv3'): 
    deconv3 = deconv_layer(deconv4, [3, 3, 64, 128], [64]) 
with tf.name_scope('deconv2'): 
    deconv2 = deconv_layer(deconv3, [3, 3, 32, 64], [32]) 
with tf.name_scope('deconv1'): 
    deconv_image = deconv_layer(deconv2, [3, 3, 3, 32], [3]) 

我從FIFOQueue讓我的圖像，並將它們送入這一模式。我的圖像尺寸是112, 112, 3。我的問題是在這兩個CONV和deconv層我有以下錯誤從 [1, 1, 1, 1] to [1, 2, 2, 1]改變步幅時：

InvalidArgumentError (see above for traceback): Conv2DSlowBackpropInput: Size of out_backprop doesn't match computed: actual = 4, computed = 2 
    [[Node: deconv4/conv2d_transpose = Conv2DBackpropInput[T=DT_FLOAT, data_format="NHWC", padding="SAME", strides=[1, 2, 2, 1], use_cudnn_on_gpu=true, _device="/job:localhost/replica:0/task:0/gpu:0"](deconv4/stack, deconv4/Variable/read, z/Relu)]] 
    [[Node: deconv1/add/_17 = _Recv[client_terminated=false, recv_device="/job:localhost/replica:0/task:0/cpu:0", send_device="/job:localhost/replica:0/task:0/gpu:0", send_device_incarnation=1, tensor_name="edge_85_deconv1/add", tensor_type=DT_FLOAT, _device="/job:localhost/replica:0/task:0/cpu:0"]()]] 

PS：我知道，我的思念在deconv層的激活功能，但我猜這與我得到的錯誤沒有關係。 任何幫助非常感謝！

Answer 1

這樣做的原因錯誤是如下：

如果我們假定我們有stride = 2在每個CONV層，然後，在我的情況下，當輸入圖像的大小爲112, 112, 3，每個conv layer大小後frames或feature maps在應用卷積後減半。也就是說，在conv1之後，圖像的大小（高度，寬度）變爲[56, 56]。在conv2之後，尺寸變成[28, 28]。 conv3之後：[14, 14]，之後conv4：[7, 7]。因此，應用一個名爲z的額外conv layer已將尺寸減小到[3, 3]。這是問題：7不能被2整除。所以我們得到了不同的維度。從[3, 3]到[112, 112]從申請deconv layer後是不可能的。此外：

[3, 3] -> [6, 6] -> [12, 12] -> [24, 24] -> [48, 48] 

第二錯誤：在deconv layer輸出形狀應該如下：

# we should multiply x_shape[1] and x_shape[2] by 2. 
out_shape = tf.stack([x_shape[0], x_shape[1] * 2, x_shape[2] * 2, w_shape[2]]) 

因此，最終deconv layer變爲如下：

def deconv_layer(x, w_shape, b_shape, is_training, padding="SAME", activation='selu'): 
    W = weight_variable(w_shape) 
    tf.summary.histogram("weights", W) 

    b = bias_variable(b_shape) 
    tf.summary.histogram('biases', b) 

    x_shape = tf.shape(x) 
    # output shape: [batch_size, h * 2, w * 2, input_shape from w]. 
    out_shape = tf.stack([x_shape[0], x_shape[1] * 2, x_shape[2] * 2, w_shape[2]]) 
    # Note that I have used a stride of 2 since I used a stride of 2 in conv layer. 
    if activation == 'selu': 
     conv_trans = tf.nn.conv2d_transpose(x, W, out_shape, [1, 2, 2, 1], padding=padding) + b 

     transposed_activations = tf.nn.elu(conv_trans) 

    else: 
     conv_trans = tf.nn.conv2d_transpose(x, W, out_shape, [1, 2, 2, 1], padding=padding) + b 
     transposed_activations = tf.nn.sigmoid(conv_trans) 

    tf.summary.histogram("transpose_activation", transposed_activations) 
    return transposed_activations 

因此，大小的輸出與輸入的大小不同，這就是我得到錯誤的原因。而爲了做back propagation我們需要一個成本函數。此成本函數將取決於output和input。因此，如果他們有不同的大小，那會導致錯誤。

可以解決的方法是讓conv layer, z,擁有stride of 1。

Answer 2

當上述代碼中的步幅爲[1, 2, 2, 1]時，input和deconv_image的形狀不匹配。這種情況下的decov_image大小將爲[64x64]。你在代碼中的其他地方照顧這個嗎？