问题:如何经过convTransposed1d输出指定大小的特征?
import torch
from torch import nn
import torch.nn.functional as F
conv1 = nn.Conv1d(1, 2, 3, padding=1)
conv2 = nn.Conv1d(in_channels=2, out_channels=4, kernel_size=3, padding=1)
#转置卷积
dconv1 = nn.ConvTranspose1d(4, 1, kernel_size=3, stride=2, padding=1, output_padding=1)
x = torch.randn(16, 1, 8)
print(x.size())
x1 = conv1(x)
x2 = conv2(x1)
print(x2.size())
x3 = dconv1(x2)
print(x3.size())
'''
torch.Size([16, 1, 8])
torch.Size([16, 4, 8]) #conv2输出特征图大小
torch.Size([16, 1, 16]) #转置卷积输出特征图大小
'''
计算转置卷积输出特征大小公式
输入: ( N , C i n , L i n ) (N, C_{in}, L_{in}) (N,Cin,Lin)
输出: ( N , C o u t , L o u t ) (N, C_{out}, L_{out}) (N,Cout,Lout)
计算 L o u t L_{out} Lout大小:
L o u t = ( L i n − 1 ) × s t r i d e − 2 × p a d d i n g + d i l a t i o n × ( k e r n e l s i z e − 1 ) + o u t p u t p a d d i n g + 1 L_{out}=(L_{in}-1)×stride-2×padding+dilation×(kernelsize-1)+outputpadding+1 Lout=(Lin−1)×stride−2×padding+dilation×(kernelsize−1)+outputpadding+1
dilation默认为1, 上式简写为:
L o u t = ( L i n − 1 ) × s t r i d e − 2 × p a d d i n g + k e r n e l s i z e + o u t p u t p a d d i n g L_{out}=(L_{in}-1)×stride-2×padding+kernelsize+outputpadding Lout=(Lin−1)×stride−2×padding+kernelsize+outputpadding
kernel_size固定,由stride, padding, outputpadding共同决定输出特征大小。
问题:输出特定的 L o u t L_{out} Lout大小
假如 L i n = 8 , L o u t = 23 , k e r n e l s i z e = 3 , o u t p u t p a d d i n g = 1 L_{in}=8,L_{out}=23,kernelsize=3,outputpadding=1 Lin=8,Lout=23,kernelsize=3,outputpadding=1,根据上式可以求得: s t r i d e = 3 , p a d d i n g = 1 stride=3,padding=1 stride=3,padding=1
#转置卷积
dconv1 = nn.ConvTranspose1d(1, 1, kernel_size=3, stride=3, padding=1, output_padding=1)
x = torch.randn(16, 1, 8)
print(x.size()) #torch.Size([16, 1, 23])
x3 = dconv1(x)
print(x3.size()) #torch.Size([16, 1, 23])
下面两图为演示conv1d,在padding和不padding下的输出特征图大小