张量流中的conv1d和conv2d

Question

For conv2d, assuming an input 2D matrix with shape (W,H) and the conv kernel size is (Wk,H), which means the height of the kernel is the same with the height of input matrix. In this case, can we think that conv1 with kernel size Wk carries out the same computation as conv2d?

For example:

tf.layers.conv2d(
    kernel_size=tf.Variable(tf.truncated_normal([Wk, H, 1, out_dim], stddev=0.1),
    input=...
) 

equals to:

tf.layers.conv1d(kernel_size=Wk, input=...)

Answer 1

They're not the same; the conv2d kernel has many more weights and is going to train differently because of that. Also, depending on what padding is set to, the output size of the conv2d operation may not be 1D either.

Answer 2

tf.nn.conv1d just call the tf.nn.conv2d

This is the description of tf.nn.conv1d :

Internally, this op reshapes the input tensors and invokes tf.nn.conv2d . For example, if data_format does not start with "NC", a tensor of shape [batch, in_width, in_channels] is reshaped to [batch, 1, in_width, in_channels] , and the filter is reshaped to [1, filter_width, in_channels, out_channels] . The result is then reshaped back to [batch, out_width, out_channels] (where out_width is a function of the stride and padding as in conv2d ) and returned to the caller.