使用张量流解卷积/Transpose_Convolutions [英] Deconvolutions/Transpose_Convolutions with tensorflow

问题描述

我正在尝试使用tf.nn.conv3d_transpose，但是，我收到一条错误消息，表明我的过滤器和输出形状不兼容.

I am attempting to use tf.nn.conv3d_transpose, however, I am getting an error indicating that my filter and output shape is not compatible.

• 我的张量为[1,16,16,4,192]
• 我正在尝试使用[1,1,1,192,192]过滤器
• 我相信输出形状将为[1,16,16,4,192]
• 我使用的是相同"填充，跨度为1.

最终，我希望输出的形状为[1,32,32,7，无关紧要"]，但是我试图首先得到一个简单的案例.

Eventually, I want to have an output shape of [1,32,32,7,"does not matter"], but I am attempting to get a simple case to work first.

由于这些张量在常规卷积中是兼容的，因此我相信相反的解卷积也是可能的.

Since these tensors are compatible in a regular convolution, I believed that the opposite, a deconvolution, would also be possible.

为什么无法在这些张量上执行反卷积.我可以得到一个有效的滤波器大小和输出形状的示例，用于对形状为[1,16,16,4,192]的张量进行去卷积

谢谢.

推荐答案

• 我的张量为[1,16,16,4,192]
• 我正在尝试使用[1,1,1,192,192]过滤器
• 我相信输出形状将为[1,16,16,4,192]
• 我使用的是相同"填充，跨度为1.

• I have a tensor of size [1,16,16,4,192]
• I am attempting to use a filter of [1,1,1,192,192]
• I believe that the output shape would be [1,16,16,4,192]
• I am using "same" padding and a stride of 1.

是的，输出形状将为[1,16,16,4,192]

Yes the output shape will be [1,16,16,4,192]

这是一个简单的示例，显示尺寸是兼容的:

Here is a simple example showing that the dimensions are compatible:

import tensorflow as tf

i = tf.Variable(tf.constant(1., shape=[1, 16, 16, 4, 192]))

w = tf.Variable(tf.constant(1., shape=[1, 1, 1, 192, 192]))

o = tf.nn.conv3d_transpose(i, w, [1, 16, 16, 4, 192], strides=[1, 1, 1, 1, 1])

print(o.get_shape())

除了尺寸之外，您的实现中还必须存在其他一些问题.

There must be some other problem in your implementation than the dimensions.

这篇关于使用张量流解卷积/Transpose_Convolutions的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！