padding=zeros 如何在 functional.conv1d 中的 pytorch 中工作

Question

This following code below giving a output of shape (1,1,3) for the shape of xodd is (1,1,2) . The given kernel shape is (112, 1, 1) .

from torch.nn import functional as F
output = F.conv1d(xodd, kernel, padding=zeros)

How the padding=zeros works?
And also, How can I write an equivalent code in tensorflow so that the output is as same as the above output ?

Answer 1

What is padding=zeros ? If we set paddin=zeros , we don't need to add numbers at the right and the left of the tensor.

Padding=0 :

from torch.nn import functional as F
import torch
inputs = torch.randn(33, 16, 6) # (minibatch,in_channels,features)
filters = torch.randn(20, 16, 5) # (out_channels, in_channels, kernel_size)
out_tns = F.conv1d(inputs, filters, stride=1, padding=0)
print(out_tns.shape)
# torch.Size([33, 20, 2]) # (minibatch,out_channels,(features-kernel_size+1))

Padding=2: (We want to add two numbers at the right and the left of the tensor)

inputs = torch.randn(33, 16, 6) # (minibatch,in_channels,features)
filters = torch.randn(20, 16, 5) # (out_channels, in_channels, kernel_size)
out_tns = F.conv1d(inputs, filters, stride=1, padding=2)
print(out_tns.shape)
# torch.Size([33, 20, 6]) # (minibatch,out_channels,(features-kernel_size+1+2+2))

How can I write an equivalent code in tensorflow:

import tensorflow as tf
input_shape = (33, 6, 16)
x = tf.random.normal(input_shape)
out_tf = tf.keras.layers.Conv1D(filters = 20, 
                                kernel_size = 5,
                                strides = 1, 
                                input_shape=input_shape[1:])(x)
print(out_tf.shape)
# TensorShape([33, 2, 20])

# If you want that tensor have shape exactly like pytorch you can transpose
tf.transpose(out_tf, [0, 2, 1]).shape
# TensorShape([33, 20, 2])