conv2d 層有多少個參數

Question

根據https://ai.stackexchange.com/questions/5769 中給出的答案，我試圖了解 CNN 應用程序的參數數量是如何得出的。

但是，對於 Unet 架構（涉及上采樣），我出錯了。 我用於 conv2d 層的公式是 - k*k*i*o + o，其中：

k = kernel size, eg 3
i = input_depth
o = output_depth

僅對於緊接上采樣后的 conv2d 層，我的計算與 keras 打印的不匹配。 我究竟做錯了什么？

keras 代碼如下：

inputs = Input(256,256,1)
conv1 = Conv2D(64, 3, activation = 'relu', padding = 'same')(inputs)
conv1 = Conv2D(64, 3, activation = 'relu', padding = 'same')(conv1)
pool1 = MaxPooling2D(pool_size=(2, 2))(conv1)
conv2 = Conv2D(128, 3, activation = 'relu', padding = 'same')(pool1)
conv2 = Conv2D(128, 3, activation = 'relu', padding = 'same')(conv2)
pool2 = MaxPooling2D(pool_size=(2, 2))(conv2)
conv3 = Conv2D(256, 3, activation = 'relu', padding = 'same')(pool2)
conv3 = Conv2D(256, 3, activation = 'relu', padding = 'same')(conv3)
pool3 = MaxPooling2D(pool_size=(2, 2))(conv3)
conv4 = Conv2D(512, 3, activation = 'relu', padding = 'same')(pool3)
conv4 = Conv2D(512, 3, activation = 'relu', padding = 'same')(conv4)
drop4 = Dropout(0.5)(conv4)
pool4 = MaxPooling2D(pool_size=(2, 2))(drop4)
conv5 = Conv2D(1024, 3, activation = 'relu', padding = 'same')(pool4)
conv5 = Conv2D(1024, 3, activation = 'relu', padding = 'same')(conv5)
drop5 = Dropout(0.5)(conv5)
up6 = Conv2D(512, 2, activation = 'relu', padding = 'same')(UpSampling2D(size = (2,2))(drop5))
merge6 = concatenate([drop4,up6], axis = 3)
conv6 = Conv2D(512, 3, activation = 'relu', padding = 'same')(merge6)
conv6 = Conv2D(512, 3, activation = 'relu', padding = 'same')(conv6)
up7 = Conv2D(256, 2, activation = 'relu', padding = 'same')(UpSampling2D(size = (2,2))(conv6))
merge7 = concatenate([conv3,up7], axis = 3)
conv7 = Conv2D(256, 3, activation = 'relu', padding = 'same')(merge7)
conv7 = Conv2D(256, 3, activation = 'relu', padding = 'same')(conv7)
up8 = Conv2D(128, 2, activation = 'relu', padding = 'same')(UpSampling2D(size = (2,2))(conv7))
merge8 = concatenate([conv2,up8], axis = 3)
conv8 = Conv2D(128, 3, activation = 'relu', padding = 'same')(merge8)
conv8 = Conv2D(128, 3, activation = 'relu', padding = 'same')(conv8)
up9 = Conv2D(64, 2, activation = 'relu', padding = 'same')(UpSampling2D(size = (2,2))(conv8))
merge9 = concatenate([conv1,up9], axis = 3)
conv9 = Conv2D(64, 3, activation = 'relu', padding = 'same')(merge9)
conv9 = Conv2D(64, 3, activation = 'relu', padding = 'same')(conv9)
conv9 = Conv2D(2, 3, activation = 'relu', padding = 'same')(conv9)
conv10 = Conv2D(1, 1, activation = 'sigmoid')(conv9)
model = Model(input = inputs, output = conv10)

我的計算如下（根據上述公式）：

name,     depth,  kernel, out-size, params,     should be
input,     1,     ,       256
conv2d_1,  64,    3,      256,      640
conv2d_2,  64,    3,      256,      36928
pool2d_1,  64,    ,       128,      0,
conv2d_3,  128,   3,      128,      73856,
conv2d_4,  128,   3,      128,      147584,
pool2d_2,  128,   ,       64,       0,
conv2d_5,  256,   3,      64,       295168,
conv2d_6,  256,   3,      64,       590080,
pool2d_3,  256,   ,       32,       0,
conv2d_7,  512,   3,      32,       1180160,
conv2d_8,  512,   3,      32,       2359808,
dropout_1, 512,   ,       32,       0,
pool2d_4,  512,   ,       16,       0,
conv2d_9,  1024,  3,      16,       4719616,
conv2d_10, 1024,  3,      16,       9438208,
dropout_2, 1024,  ,       16,       0
upsample_1,1024,  ,       32,       0
conv2d_11, 512,   3,      32,       4719104,    2097664
concat_1,  1024,  ,       32,       0
conv2d_12, 512,   3,      32,       4719104
conv2d_13, 512,   3,      32,       2359808
upsample_2,512,   ,       64,       0
conv2d_14, 256,   3,      64,       1179904,    524544
concat_2,  512,   ,       64,       0
conv2d_15, 256,   3,      64,       1179904
conv2d_16, 256,   3,      64,       590080
upsample_3,256,   ,       128,      0
conv2d_17, 128,   3,      128,      295040,     131200
concat_3,  256,   ,       128,      0
conv2d_18, 128,   3,      128,      295040
conv2d_19, 128,   3,      128,      147584
upsample_4,128,   ,       256,      0
conv2d_20, 64,    3,      256,      73792,      32832
concat_4,  128,   ,       256,      0
conv2d_21, 64,    3,      256,      73792
conv2d_22, 64,    3,      256,      36928
conv2d_23, 2,     3,      256,      1154
conv2d_24, 1,     1,      256,      3

舉一個具體的例子，對於 conv2d_11：根據我的計算，它應該是 9*1024*512+512 = 4719104。但是，keras 將數字打印為 2097664。

反向計算，如果我使用 4 而不是 9，我似乎會得到 keras 結果。但是 conv2d_11 的內核大小是 3x3，那么為什么我應該只對這一層使用 4？

Answer 1

因為有些事情正在考慮 3 (*9) 的內核大小，而另一件事正在考慮 2 (*4) 的內核大小。

這是代碼問題，不是參數計算問題。