根据这个深度学习课程http://cs231n.github.io/convolutional-networks/#conv,它表示如果有一个输入 x
,形状 [W,W]
(其中 W = width = height
)经过 Convolutional Layer filter 形状 [F,F]
和 stride S
,则 Layer 将返回 output
形状 [(W-F)/S +1, (W-F)/S +1]
但是,当我试图遵循Tensorflow教程时:https://www.tensorflow.org/versions/r0.11/tutorials/mnist/pros/index.html . 似乎有功能的差异 tf.nn.conv2d(inputs, filter, stride)
无论如何更改我的滤镜大小, conv2d
将不断返回一个与输入形状相同的值 .
在我的情况下,我使用 MNIST
数据集,表明每个图像的大小 [28,28]
(忽略 channel_num = 1
)
但在我定义了第一个 conv1
图层后,我使用 conv1.get_shape()
查看其输出,它给了我 [28,28, num_of_filters]
为什么是这样?我认为返回值应该遵循上面的公式 .
附录:代码段
#reshape x from 2d to 4d
x_image = tf.reshape(x, [-1, 28, 28, 1]) #[num_samples, width, height, channel_num]
## define the shape of weights and bias
w_shape = [5, 5, 1, 32] #patch_w, patch_h, in_channel, output_num(out_channel)
b_shape = [32] #bias only need to be consistent with output_num
## init weights of conv1 layers
W_conv1 = weight_variable(w_shape)
b_conv1 = bias_variable(b_shape)
## first layer x_iamge->conv1/relu->pool1
#Our convolutions uses a stride of one
#and are zero padded
#so that the output is the same size as the input
h_conv1 = tf.nn.relu(
conv2d(x_image, W_conv1) + b_conv1
)
print 'conv1.shape=',h_conv1.get_shape()
## conv1.shape= (?, 28, 28, 32)
## I thought conv1.shape should be (?, (28-5)/1+1, 24 ,32)
h_pool1 = max_pool_2x2(h_conv1) #output 32 num
print 'pool1.shape=',h_pool1.get_shape() ## pool1.shape= (?, 14, 14, 32)
2 回答
它取决于填充参数 . 'SAME'将输出保持为WxW(假设stride = 1,)'VALID'将输出的大小缩小为(W-F 1)x(W-F 1)
Conv2d有一个名为padding的参数see here
如果您将填充设置为“有效”,它将满足您的公式 . 默认为“SAME”,填充零的图像(与添加边框相同),使得输出将保持与输入相同的形状 .