Question

我在黑色背景上有一些图像，图像没有方形边缘（见下图右下角）。我想将它们裁剪成最大的矩形图像（红色边框）。我知道我可能会失去原始图像。是否可以在OpenCV中使用Python执行此操作。我知道有一些功能可以裁剪到轮廓的边界框，但这仍然会让我在地方留下黑色背景。

enter image description here

Answer 1

好吧，我已经玩了一个想法并测试了它（它是c ++，但你可能能够将它转换为python）：

假设：背景为黑色，内部没有黑色边界部分
您可以使用findContours
使用该轮廓的最小/最大x / y点位置，直到由这些点构建的矩形不包含位于轮廓之外的点

我无法保证此方法始终找到“最佳”内部框，但我使用启发式方法来选择矩形是在顶部/底部/左侧/右侧缩小。

代码当然也可以优化;）

使用它作为testimage，我得到了结果（非红色区域是找到的内部矩形）：

enter image description here

认为右上方有一个不应该包含在矩形中的像素，也许是因为提取/绘制轮廓错误了？！

和这里的代码：

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);

具有缩小功能：

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;
}

Answer 2

受@micka答案启发的解决方案，在python中。

这不是一个聪明的解决方案，可以对其进行优化，但是就我而言，它（缓慢）起作用。

我修改了您的图像以添加一个正方形，例如您的示例：请参见there

最后，这段代码将picture

中的白色矩形裁剪掉

希望您会发现它有帮助！

import cv2

# Import your picture
input_picture = cv2.imread("LenaWithBG.png")

# Color it in gray
gray = cv2.cvtColor(input_picture, cv2.COLOR_BGR2GRAY)

# Create our mask by selecting the non-zero values of the picture
ret, mask = cv2.threshold(gray,0,255,cv2.THRESH_BINARY)

# Select the contour
mask , cont, _ = cv2.findContours(mask, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_SIMPLE)
# if your mask is incurved or if you want better results, 
# you may want to use cv2.CHAIN_APPROX_NONE instead of cv2.CHAIN_APPROX_SIMPLE, 
# but the rectangle search will be longer

cv2.drawContours(gray, cont, -1, (255,0,0), 1)
cv2.imshow("Your picture with contour", gray)
cv2.waitKey(0)

# Get all the points of the contour
contour = cont[0].reshape(len(cont[0]),2)

# we assume a rectangle with at least two points on the contour gives a 'good enough' result
# get all possible rectangles based on this hypothesis
rect = []

for i in range(len(contour)):
    x1, y1 = contour[i]
    for j in range(len(contour)):
        x2, y2 = contour[j]
        area = abs(y2-y1)*abs(x2-x1)
        rect.append(((x1,y1), (x2,y2), area))

# the first rect of all_rect has the biggest area, so it's the best solution if he fits in the picture
all_rect = sorted(rect, key = lambda x : x[2], reverse = True)

# we take the largest rectangle we've got, based on the value of the rectangle area
# only if the border of the rectangle is not in the black part

# if the list is not empty
if all_rect:

    best_rect_found = False
    index_rect = 0
    nb_rect = len(all_rect)

    # we check if the rectangle is  a good solution
    while not best_rect_found and index_rect < nb_rect:

        rect = all_rect[index_rect]
        (x1, y1) = rect[0]
        (x2, y2) = rect[1]

        valid_rect = True

        # we search a black area in the perimeter of the rectangle (vertical borders)
        x = min(x1, x2)
        while x <max(x1,x2)+1 and valid_rect:
            if mask[y1,x] == 0 or mask[y2,x] == 0:
                # if we find a black pixel, that means a part of the rectangle is black
                # so we don't keep this rectangle
                valid_rect = False
            x+=1

        y = min(y1, y2)
        while y <max(y1,y2)+1 and valid_rect:
            if mask[y,x1] == 0 or mask[y,x2] == 0:
                valid_rect = False
            y+=1

        if valid_rect:
            best_rect_found = True

        index_rect+=1

    if best_rect_found:

        cv2.rectangle(gray, (x1,y1), (x2,y2), (255,0,0), 1)
        cv2.imshow("Is that rectangle ok?",gray)
        cv2.waitKey(0)

        # Finally, we crop the picture and store it
        result = input_picture[min(y1, y2):max(y1, y2), min(x1,x2):max(x1,x2)]

        cv2.imwrite("Lena_cropped.png",result)
    else:
        print("No rectangle fitting into the area")

else:
    print("No rectangle found")

如果蒙版弯曲，或者只是想获得更好的效果，则可能要使用cv2.CHAIN_APPROX_NONE而不是cv2.CHAIN_APPROX_SIMPLE，但是矩形搜索会花费更多时间（因为在最佳情况下，这是二次解法）。

Answer 3

在ImageMagick 6.9.10-30（或7.0.8.30）或更高版本中，可以将-trim函数与新定义一起使用。

输入：

convert image.png -fuzz 5% -define trim:percent-background=0% -trim +repage result.png

或者对于下面显示的图像：

输入：

convert image2.png -bordercolor black -border 1 -define trim:percent-background=0% -trim +repage result2.png

如何裁剪到OpenCV中最大的内部边界框？

3 个答案: