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Opencv问题, 怎样去除二值图像中面积较小的连通域?
//=======函数实现=====================================================================
void RemoveSmallRegion(Mat Src, Mat Dst, int AreaLimit, int CheckMode, int NeihborMode)
{
int RemoveCount = 0;
//新建一幅标签图像初始化为0像素点,为了记录每个像素点检验状态的标签,0代表未检查,1代表正在检查,2代表检查不合格(需要反转颜色),3代表检查合格或不需检查
//初始化的图像全部为0,未检查
Mat PointLabel = Mat::zeros(Src.size(), CV_8UC1);
if (CheckMode == 1)//去除小连通区域的白色点
{
//cout "去除小连通域.";
for (int i = 0; i Src.rows; i++)
{
for (int j = 0; j Src.cols; j++)
{
if (Src.atuchar(i, j) 10)
{
PointLabel.atuchar(i, j) = 3;//将背景黑色点标记为合格,像素为3
}
}
}
}
else//去除孔洞,黑色点像素
{
//cout "去除孔洞";
for (int i = 0; i Src.rows; i++)
{
for (int j = 0; j Src.cols; j++)
{
if (Src.atuchar(i, j) 10)
{
PointLabel.atuchar(i, j) = 3;//如果原图是白色区域,标记为合格,像素为3
}
}
}
}
vectorPoint2iNeihborPos;//将邻域压进容器
NeihborPos.push_back(Point2i(-1, 0));
NeihborPos.push_back(Point2i(1, 0));
NeihborPos.push_back(Point2i(0, -1));
NeihborPos.push_back(Point2i(0, 1));
if (NeihborMode == 1)
{
//cout "Neighbor mode: 8邻域." endl;
NeihborPos.push_back(Point2i(-1, -1));
NeihborPos.push_back(Point2i(-1, 1));
NeihborPos.push_back(Point2i(1, -1));
NeihborPos.push_back(Point2i(1, 1));
}
else int a = 0;//cout "Neighbor mode: 4邻域." endl;
int NeihborCount = 4 + 4 * NeihborMode;
int CurrX = 0, CurrY = 0;
//开始检测
for (int i = 0; i Src.rows; i++)
{
for (int j = 0; j Src.cols; j++)
{
if (PointLabel.atuchar(i, j) == 0)//标签图像像素点为0,表示还未检查的不合格点
{ //开始检查
vectorPoint2iGrowBuffer;//记录检查像素点的个数
GrowBuffer.push_back(Point2i(j, i));
PointLabel.atuchar(i, j) = 1;//标记为正在检查
int CheckResult = 0;
for (int z = 0; z GrowBuffer.size(); z++)
{
for (int q = 0; q NeihborCount; q++)
{
CurrX = GrowBuffer.at(z).x + NeihborPos.at(q).x;
CurrY = GrowBuffer.at(z).y + NeihborPos.at(q).y;
if (CurrX = 0 CurrXSrc.colsCurrY = 0 CurrYSrc.rows) //防止越界
{
if (PointLabel.atuchar(CurrY, CurrX) == 0)
{
GrowBuffer.push_back(Point2i(CurrX, CurrY)); //邻域点加入buffer
PointLabel.atuchar(CurrY, CurrX) = 1; //更新邻域点的检查标签,避免重复检查
}
}
}
}
if (GrowBuffer.size()AreaLimit) //判断结果(是否超出限定的大小),1为未超出,2为超出
CheckResult = 2;
else
{
CheckResult = 1;
RemoveCount++;//记录有多少区域被去除
}
for (int z = 0; z GrowBuffer.size(); z++)
{
CurrX = GrowBuffer.at(z).x;
CurrY = GrowBuffer.at(z).y;
PointLabel.atuchar(CurrY, CurrX) += CheckResult;//标记不合格的像素点,像素值为2
}
//********结束该点处的检查**********
}
}
}
CheckMode = 255 * (1 - CheckMode);
//开始反转面积过小的区域
for (int i = 0; i Src.rows; ++i)
{
for (int j = 0; j Src.cols; ++j)
{
if (PointLabel.atuchar(i, j) == 2)
{
Dst.atuchar(i, j) = CheckMode;
}
else if (PointLabel.atuchar(i, j) == 3)
{
Dst.atuchar(i, j) = Src.atuchar(i, j);
}
}
}
//cout RemoveCount " objects removed." endl;
}
//=======函数实现=====================================================================
//=======调用函数=====================================================================
Mat img;
img = imread("D:\\1_1.jpg", 0);//读取图片
threshold(img, img, 128, 255, CV_THRESH_BINARY_INV);
imshow("去除前", img);
Mat img1;
RemoveSmallRegion(img, img, 200, 0, 1);
imshow("去除后", img);
waitKey(0);
//=======调用函数=====================================================================
python matplotlib subplot 上面面积大下面小怎么办
在matplotlib下,一个Figure对象可以包含多个子图(Axes),可以使用subplot()快速绘制,其调用形式如下:
subplot(numRows, numCols, plotNum)
图表的整个绘图区域被分成numRows行和numCols列,plotNum参数指定创建的Axes对象所在的区域,如何理解呢?
如果numRows = 3,numCols = 2,那整个绘制图表样式为3X2的图片区域,用坐标表示为(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)。这时,当plotNum = 1时,表示的坐标为(1,3),即第一行第一列的子图;
import numpy as np
python如何删除二值化图片中小块白色区域
如果确定是纯白的话你就把rgb都小于某个极小常数的像素点的alpha设成0就好了你说的nodata应该就是透明的意思!