[英]how many parameters for a conv2d layer
根据https://ai.stackexchange.com/questions/5769 中给出的答案,我试图了解 CNN 应用程序的参数数量是如何得出的。
但是,对于 Unet 架构(涉及上采样),我出错了。 我用于 conv2d 层的公式是 - k*k*i*o + o,其中:
k = kernel size, eg 3
i = input_depth
o = output_depth
仅对于紧接上采样后的 conv2d 层,我的计算与 keras 打印的不匹配。 我究竟做错了什么?
keras 代码如下:
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)
我的计算如下(根据上述公式):
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
举一个具体的例子,对于 conv2d_11:根据我的计算,它应该是 9*1024*512+512 = 4719104。但是,keras 将数字打印为 2097664。
反向计算,如果我使用 4 而不是 9,我似乎会得到 keras 结果。但是 conv2d_11 的内核大小是 3x3,那么为什么我应该只对这一层使用 4?
因为有些事情正在考虑 3 (*9) 的内核大小,而另一件事正在考虑 2 (*4) 的内核大小。
这是代码问题,不是参数计算问题。
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