(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。

  • 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

  • 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

图1 最简单的3×3卷积

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<<" ";
}
  • 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;
}
}
  • 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

图3 最简单卷积(2)

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;
}
}
  • 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;
}
}
  • 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

图6 多kernel卷积

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;
}
}
  • 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

图7 膨胀卷积

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;
}
}
  • git源码https://github.com/msnh2012/ConvTest

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