如何裁剪OpenCV中最大的內部邊框？

Question

我在黑色背景上有圖像沒有方形邊緣（見下圖右下方）。我想將它們裁剪成最大的矩形圖像（紅色邊框）。我知道我可能會失去原來的形象。是否有可能在Python中使用OpenCV來完成此操作。我知道有一些功能可以裁剪出一個輪廓的邊框，但這仍然會讓我在黑色的背景中出現。

Answer 1

好了，我打了一個想法，測試它（它的C++但你可能就可以將其轉換成蟒蛇）：

1. 假設：背景是黑色，內部沒有黑色邊框部分
2. 你可以找到findContours
3. 使用最小值/最大值從輪廓X/Y點位置的外部輪廓，直到由這些點建立的矩形不包含點t1帽子躺在輪廓之外

我不能保證此方法總能找到「最好」的內部框，但我使用啓發式方法來選擇矩形在頂部/底部/左側/右側是否縮小。

代碼當然可以被優化，太;

以此作爲testimage），我得到的結果（非紅色區域是所找到的內部矩形）：

認爲在右上角有一個像素不應該包含在矩形中，也許這是從引出/繪製輪廓錯誤？！？

和這裏的代碼：

cv::Mat input = cv::imread("LenaWithBG.png"); 

cv::Mat gray; 
cv::cvtColor(input,gray,CV_BGR2GRAY); 

cv::imshow("gray", gray); 

// extract all the black background (and some interior parts maybe) 
cv::Mat mask = gray>0; 
cv::imshow("mask", mask); 

// now extract the outer contour 
std::vector<std::vector<cv::Point> > contours; 
std::vector<cv::Vec4i> hierarchy; 

cv::findContours(mask,contours,hierarchy, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_NONE, cv::Point(0,0)); 

std::cout << "found contours: " << contours.size() << std::endl; 


cv::Mat contourImage = cv::Mat::zeros(input.size(), CV_8UC3);; 

//find contour with max elements 
// remark: in theory there should be only one single outer contour surrounded by black regions!! 

unsigned int maxSize = 0; 
unsigned int id = 0; 
for(unsigned int i=0; i<contours.size(); ++i) 
{ 
    if(contours.at(i).size() > maxSize) 
    { 
     maxSize = contours.at(i).size(); 
     id = i; 
    } 
} 

std::cout << "chosen id: " << id << std::endl; 
std::cout << "max size: " << maxSize << std::endl; 

/// Draw filled contour to obtain a mask with interior parts 
cv::Mat contourMask = cv::Mat::zeros(input.size(), CV_8UC1); 
cv::drawContours(contourMask, contours, id, cv::Scalar(255), -1, 8, hierarchy, 0, cv::Point()); 
cv::imshow("contour mask", contourMask); 

// sort contour in x/y directions to easily find min/max and next 
std::vector<cv::Point> cSortedX = contours.at(id); 
std::sort(cSortedX.begin(), cSortedX.end(), sortX); 

std::vector<cv::Point> cSortedY = contours.at(id); 
std::sort(cSortedY.begin(), cSortedY.end(), sortY); 


unsigned int minXId = 0; 
unsigned int maxXId = cSortedX.size()-1; 

unsigned int minYId = 0; 
unsigned int maxYId = cSortedY.size()-1; 

cv::Rect interiorBB; 

while((minXId<maxXId)&&(minYId<maxYId)) 
{ 
    cv::Point min(cSortedX[minXId].x, cSortedY[minYId].y); 
    cv::Point max(cSortedX[maxXId].x, cSortedY[maxYId].y); 

    interiorBB = cv::Rect(min.x,min.y, max.x-min.x, max.y-min.y); 

// out-codes: if one of them is set, the rectangle size has to be reduced at that border 
    int ocTop = 0; 
    int ocBottom = 0; 
    int ocLeft = 0; 
    int ocRight = 0; 

    bool finished = checkInteriorExterior(contourMask, interiorBB, ocTop, ocBottom,ocLeft, ocRight); 
    if(finished) 
    { 
     break; 
    } 

// reduce rectangle at border if necessary 
    if(ocLeft)++minXId; 
    if(ocRight) --maxXId; 

    if(ocTop) ++minYId; 
    if(ocBottom)--maxYId; 


} 

std::cout << "done! : " << interiorBB << std::endl; 

cv::Mat mask2 = cv::Mat::zeros(input.rows, input.cols, CV_8UC1); 
cv::rectangle(mask2,interiorBB, cv::Scalar(255),-1); 

cv::Mat maskedImage; 
input.copyTo(maskedImage); 
for(unsigned int y=0; y<maskedImage.rows; ++y) 
    for(unsigned int x=0; x<maskedImage.cols; ++x) 
    { 
     maskedImage.at<cv::Vec3b>(y,x)[2] = 255; 
    } 
input.copyTo(maskedImage,mask2); 

cv::imshow("masked image", maskedImage); 
cv::imwrite("interiorBoundingBoxResult.png", maskedImage); 

與還原功能：

bool checkInteriorExterior(const cv::Mat&mask, const cv::Rect&interiorBB, int&top, int&bottom, int&left, int&right) 
{ 
// return true if the rectangle is fine as it is! 
bool returnVal = true; 

cv::Mat sub = mask(interiorBB); 

unsigned int x=0; 
unsigned int y=0; 

// count how many exterior pixels are at the 
unsigned int cTop=0; // top row 
unsigned int cBottom=0; // bottom row 
unsigned int cLeft=0; // left column 
unsigned int cRight=0; // right column 
// and choose that side for reduction where mose exterior pixels occured (that's the heuristic) 

for(y=0, x=0 ; x<sub.cols; ++x) 
{ 
    // if there is an exterior part in the interior we have to move the top side of the rect a bit to the bottom 
    if(sub.at<unsigned char>(y,x) == 0) 
    { 
     returnVal = false; 
     ++cTop; 
    } 
} 

for(y=sub.rows-1, x=0; x<sub.cols; ++x) 
{ 
    // if there is an exterior part in the interior we have to move the bottom side of the rect a bit to the top 
    if(sub.at<unsigned char>(y,x) == 0) 
    { 
     returnVal = false; 
     ++cBottom; 
    } 
} 

for(y=0, x=0 ; y<sub.rows; ++y) 
{ 
    // if there is an exterior part in the interior 
    if(sub.at<unsigned char>(y,x) == 0) 
    { 
     returnVal = false; 
     ++cLeft; 
    } 
} 

for(x=sub.cols-1, y=0; y<sub.rows; ++y) 
{ 
    // if there is an exterior part in the interior 
    if(sub.at<unsigned char>(y,x) == 0) 
    { 
     returnVal = false; 
     ++cRight; 
    } 
} 

// that part is ugly and maybe not correct, didn't check whether all possible combinations are handled. Check that one please. The idea is to set `top = 1` iff it's better to reduce the rect at the top than anywhere else. 
if(cTop > cBottom) 
{ 
    if(cTop > cLeft) 
     if(cTop > cRight) 
      top = 1; 
} 
else 
    if(cBottom > cLeft) 
     if(cBottom > cRight) 
      bottom = 1; 

if(cLeft >= cRight) 
{ 
    if(cLeft >= cBottom) 
     if(cLeft >= cTop) 
      left = 1; 
} 
else 
    if(cRight >= cTop) 
     if(cRight >= cBottom) 
      right = 1; 



return returnVal; 
} 

bool sortX(cv::Point a, cv::Point b) 
{ 
    bool ret = false; 
    if(a.x == a.x) 
     if(b.x==b.x) 
      ret = a.x < b.x; 

    return ret; 
} 

bool sortY(cv::Point a, cv::Point b) 
{ 
    bool ret = false; 
    if(a.y == a.y) 
     if(b.y == b.y) 
      ret = a.y < b.y; 


    return ret; 
}