(图解)一步一步使用CPP实现深度学习中的卷积
【GiantPandaCV导语】 卷积操作在深度学习中的重要性,想必大家都很清楚了。接下来将通过图解的方式,使用cpp一步一步从简单到复杂来实现卷积操作。
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符号约定 F
为输入; width
为输入的宽; height
为输入的高; channel
为输入的通道; K
为kernel; kSizeX
为kernel的宽; kSizeY
为kernel的高; filters
为kernel的个数; O
为输出; outWidth
为输出的宽; outHeight
为输出的高; outChannel
为输出的通道; -
卷积输出尺寸计算公式
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1. 最简单的3×3卷积首先, 我们不考虑任何padding, stride, 多维度等情况,来一个最简单的3×3卷积操作.计算思路很简单, 对应元素相乘最后相加即可.此处:
-
width=3
-
height=3
-
channel=1
-
paddingX=0
-
paddingY=0
-
strideX=1
-
strideY=1
-
dilationX=1
-
dilationY=1
-
kSizeX=3
-
kSizeY=3
-
filters=1
可根据卷积输出尺寸计算公式,得到:
-
outWidth=1
-
outHeight=1
-
outChannel=1

cpp代码:
void demo0()
{
float F[] = {1,2,3,4,5,6,7,8,9};
float K[] = {1,2,3,4,5,6,7,8,9};
float O = 0;
int width = 3;
int height = 3;
int kSizeX = 3;
int kSizeY = 3;
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
O+=K[m*kSizeX+n]*F[m*width+n];
}
}
std::cout<<O<<" ";
}
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2. 最简单卷积(1)接下来考虑能适用于任何尺寸的简单卷积, 如输入为4x4x1, kernel为3x3x1. 这里考虑卷积步长为1, 则此处的参数为:
cpp代码: -
width=4
-
height=4
-
channel=1
-
paddingX=0
-
paddingY=0
-
strideX=1
-
strideY=1
-
dilationX=1
-
dilationY=1
-
kSizeX=3
-
kSizeY=3
-
filters=1
可根据卷积输出尺寸计算公式,得到:
-
outWidth=2
-
outHeight=2
-
outChannel=1
图2 最简单卷积(1)
void demo1()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4,5,6,7,8,9};
float O[] = {0,0,0,0};
int padX = 0;
int padY = 0;
int dilationX = 1;
int dilationY = 1;
int strideX = 1;
int strideY = 1;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
O[i*outW+j]+=K[m*kSizeX+n]*F[(m+i)*width+(n+j)];
}
}
}
}
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[i*outW+j]<<" ";
}
std::cout<<std::endl;
}
}
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3. 最简单卷积(2)接下来考虑在步长上为任意步长的卷积, 如输入为4x4x1, kernel为2x2x1. 这里考虑卷积步长为2, 则此处的参数为:
-
width=4
-
height=4
-
channel=1
-
paddingX=0
-
paddingY=0
-
strideX=2
-
strideY=2
-
dilationX=1
-
dilationY=1
-
kSizeX=2
-
kSizeY=2
-
filters=1
可根据卷积输出尺寸计算公式,得到:
-
outWidth=2
-
outHeight=2
-
outChannel=1

cpp代码:
void demo2()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4};
float O[] = {0,0,0,0};
int padX = 0;
int padY = 0;
int dilationX = 1;
int dilationY = 1;
int strideX = 2;
int strideY = 2;
int width = 4;
int height = 4;
int kSizeX = 2;
int kSizeY = 2;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
O[i*outW+j]+=K[m*kSizeX+n]*F[(m+i*strideY)*width+(n+j*strideX)];
}
}
}
}
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[i*outW+j]<<" ";
}
std::cout<<std::endl;
}
}
-
4. 带padding的卷积接下来考虑带padding的卷积, 如输入为4x4x1, kernel为3x3x1. 卷积步长为1, padding为1 则此处的参数为:
cpp代码: -
width=4
-
height=4
-
channel=1
-
paddingX=1
-
paddingY=1
-
strideX=1
-
strideY=1
-
dilationX=1
-
dilationY=1
-
kSizeX=3
-
kSizeY=3
-
filters=1
可根据卷积输出尺寸计算公式,得到:
-
outWidth=2
-
outHeight=2
-
outChannel=1
图4 考虑padding卷积
void demo3()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4,5,6,7,8,9};
float O[] = {0,0,0,0};
int padX = 1;
int padY = 1;
int dilationX = 1;
int dilationY = 1;
int strideX = 2;
int strideY = 2;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
float fVal = 0;
//考虑边界强情况
if((n+j*strideX-padX)>-1&&(m+i*strideY-padY>-1)&&(n+j*strideX-padX)-1)<=height)
{
fVal = F[(m+i*strideY-padX)*width+(n+j*strideX-padY)];
}
O[i*outW+j]+=K[m*kSizeX+n]*fVal;
}
}
}
}
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[i*outW+j]<<" ";
}
std::cout<<std::endl;
}
}
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5. 多通道卷积接下来考虑多通道卷积, 如输入为4x4x1, kernel为3x3x1. 卷积步长为1, padding为1, 输入通道为2, 则此处的参数为:
cpp代码: -
width=4
-
height=4
-
channel=2
-
paddingX=1
-
paddingY=1
-
strideX=1
-
strideY=1
-
dilationX=1
-
dilationY=1
-
kSizeX=3
-
kSizeY=3
-
filters=1
可根据卷积输出尺寸计算公式,得到:
-
outWidth=2
-
outHeight=2
-
outChannel=1
图5 多通道卷积
void demo4()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9};
float O[] = {0,0,0,0};
int padX = 1;
int padY = 1;
int dilationX = 1;
int dilationY = 1;
int strideX = 2;
int strideY = 2;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int channel = 2;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
for (int c = 0; c < channel; ++c)
{
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
float fVal = 0;
if((n+j*strideX-padX)>-1&&(m+i*strideY-padY>-1)&&(n+j*strideX-padX)-1)<=height)
{
fVal = F[c*width*height + (m+i*strideY-padX)*width+(n+j*strideX-padY)];
}
O[i*outW+j]+=K[c*kSizeX*kSizeY+m*kSizeX+n]*fVal;
}
}
}
}
}
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[i*outW+j]<<" ";
}
std::cout<<std::endl;
}
}
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6. 多kernel卷积接下来考虑多kernel卷积, 如输入为4x4x1, kernel为3x3x1. 卷积步长为1, padding为1, 输入通道为2, filters为2, 则此处的参数为:
-
width=4
-
height=4
-
channel=2
-
paddingX=1
-
paddingY=1
-
strideX=1
-
strideY=1
-
dilationX=1
-
dilationY=1
-
kSizeX=3
-
kSizeY=3
-
filters=2
可根据卷积输出尺寸计算公式,得到:
-
outWidth=2
-
outHeight=2
-
outChannel=2

cpp代码:
void demo5()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,
1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9
};
float O[] = {0,0,0,0,0,0,0,0};
int padX = 1;
int padY = 1;
int dilationX = 1;
int dilationY = 1;
int strideX = 2;
int strideY = 2;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int channel = 2;
int filters = 2;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
int outC = filters;
for (int oc = 0; oc < outC; ++oc)
{
for (int c = 0; c < channel; ++c)
{
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
float fVal = 0;
if((n+j*strideX-padX)>-1&&(m+i*strideY-padY>-1)&&(n+j*strideX-padX)-1)<=height)
{
fVal = F[c*width*height + (m+i*strideY-padX)*width+(n+j*strideX-padY)];
}
O[oc*outH*outW+i*outW+j]+=K[oc*outC*kSizeX*kSizeY+c*kSizeX*kSizeY+m*kSizeX+n]*fVal;
}
}
}
}
}
}
for (int oc = 0; oc < outC; ++oc)
{
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[oc*outH*outW+i*outW+j]<<" ";
}
std::cout<<std::endl;
}
std::cout<<std::endl<<std::endl;
}
}
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7. 膨胀卷积接下来考虑多膨胀卷积, 如输入为4x4x1, kernel为3x3x1. 卷积步长为1, padding为1, 输入通道为2, filters为2, dilate为2则此处的参数为:
-
width=4
-
height=4
-
channel=2
-
paddingX=1
-
paddingY=1
-
strideX=1
-
strideY=1
-
dilationX=2
-
dilationY=2
-
kSizeX=3
-
kSizeY=3
-
filters=2
可根据卷积输出尺寸计算公式,得到:
-
outWidth=2
-
outHeight=2
-
outChannel=2

cpp代码:
void demo6()
{
float F[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
float K[] = {1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,
1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9
};
float O[] = {0,0,0,0,0,0,0,0};
int padX = 1;
int padY = 1;
int dilationX = 2;
int dilationY = 2;
int strideX = 1;
int strideY = 1;
int width = 4;
int height = 4;
int kSizeX = 3;
int kSizeY = 3;
int channel = 2;
int filters = 2;
int outH = (height+2*padY-(dilationY*(kSizeY-1)+1)) / strideY + 1;
int outW = (width+2*padX-(dilationX*(kSizeX-1)+1)) / strideX + 1;
int outC = filters;
for (int oc = 0; oc < outC; ++oc)
{
for (int c = 0; c < channel; ++c)
{
for(int i=0;i<outH;i++)
{
for(int j=0;j<outW;j++)
{
for(int m=0;m<kSizeY;m++)
{
for(int n=0;n<kSizeX;n++)
{
float fVal = 0;
if( ((n+j*strideX)*dilationX-padX)>-1 && ((m+i*strideY)*dilationY-padY)>-1&&
((n+j*strideX)*dilationX-padX)-1)<=height)
{
fVal = F[c*width*height + ((m+i*strideY)*dilationX-padX)*width+((n+j*strideX)*dilationY-padY)];
}
O[oc*outH*outW+i*outW+j]+=K[oc*outC*kSizeX*kSizeY+c*kSizeX*kSizeY+m*kSizeX+n]*fVal;
}
}
}
}
}
}
for (int oc = 0; oc < outC; ++oc)
{
for (int i = 0; i < outH; ++i)
{
for (int j = 0; j < outW; ++j)
{
std::cout<<O[oc*outH*outW+i*outW+j]<<" ";
}
std::cout<<std::endl;
}
std::cout<<std::endl;
}
}
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git源码https://github.com/msnh2012/ConvTest
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