我在opencv中使用python。

我的算法是这样的:

首先，它从图像中取出红色通道 对红色通道应用阈值(最小值200) 然后应用形态梯度，然后做一个“关闭”(扩张，然后侵蚀) 然后它找到平面上的轮廓然后选择最长的轮廓。

代码:

import numpy as np
import cv2
import copy


def findTree(image,num):
    im = cv2.imread(image)
    im = cv2.resize(im, (400,250))
    gray = cv2.cvtColor(im, cv2.COLOR_RGB2GRAY)
    imf = copy.deepcopy(im)

    b,g,r = cv2.split(im)
    minR = 200
    _,thresh = cv2.threshold(r,minR,255,0)
    kernel = np.ones((25,5))
    dst = cv2.morphologyEx(thresh, cv2.MORPH_GRADIENT, kernel)
    dst = cv2.morphologyEx(dst, cv2.MORPH_CLOSE, kernel)

    contours = cv2.findContours(dst,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)[0]
    cv2.drawContours(im, contours,-1, (0,255,0), 1)

    maxI = 0
    for i in range(len(contours)):
        if len(contours[maxI]) < len(contours[i]):
            maxI = i

    img = copy.deepcopy(r)
    cv2.polylines(img,[contours[maxI]],True,(255,255,255),3)
    imf[:,:,2] = img

    cv2.imshow(str(num), imf)

def main():
    findTree('tree.jpg',1)
    findTree('tree2.jpg',2)
    findTree('tree3.jpg',3)
    findTree('tree4.jpg',4)
    findTree('tree5.jpg',5)
    findTree('tree6.jpg',6)

    cv2.waitKey(0)
    cv2.destroyAllWindows()

if __name__ == "__main__":
    main()

如果我把核函数从(25,5)改成(10,5) 我在所有树上都得到了更好的结果，除了左下角，

我的算法假设树上有灯，而且 在左下角的树中，顶部的光线比其他树要少。

这是我简单而愚蠢的解决方案。 它是基于这样一个假设:树将是图片中最明亮、最大的东西。

//g++ -Wall -pedantic -ansi -O2 -pipe -s -o christmas_tree christmas_tree.cpp `pkg-config --cflags --libs opencv`
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <iostream>

using namespace cv;
using namespace std;

int main(int argc,char *argv[])
{
    Mat original,tmp,tmp1;
    vector <vector<Point> > contours;
    Moments m;
    Rect boundrect;
    Point2f center;
    double radius, max_area=0,tmp_area=0;
    unsigned int j, k;
    int i;

    for(i = 1; i < argc; ++i)
    {
        original = imread(argv[i]);
        if(original.empty())
        {
            cerr << "Error"<<endl;
            return -1;
        }

        GaussianBlur(original, tmp, Size(3, 3), 0, 0, BORDER_DEFAULT);
        erode(tmp, tmp, Mat(), Point(-1, -1), 10);
        cvtColor(tmp, tmp, CV_BGR2HSV);
        inRange(tmp, Scalar(0, 0, 0), Scalar(180, 255, 200), tmp);

        dilate(original, tmp1, Mat(), Point(-1, -1), 15);
        cvtColor(tmp1, tmp1, CV_BGR2HLS);
        inRange(tmp1, Scalar(0, 185, 0), Scalar(180, 255, 255), tmp1);
        dilate(tmp1, tmp1, Mat(), Point(-1, -1), 10);

        bitwise_and(tmp, tmp1, tmp1);

        findContours(tmp1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
        max_area = 0;
        j = 0;
        for(k = 0; k < contours.size(); k++)
        {
            tmp_area = contourArea(contours[k]);
            if(tmp_area > max_area)
            {
                max_area = tmp_area;
                j = k;
            }
        }
        tmp1 = Mat::zeros(original.size(),CV_8U);
        approxPolyDP(contours[j], contours[j], 30, true);
        drawContours(tmp1, contours, j, Scalar(255,255,255), CV_FILLED);

        m = moments(contours[j]);
        boundrect = boundingRect(contours[j]);
        center = Point2f(m.m10/m.m00, m.m01/m.m00);
        radius = (center.y - (boundrect.tl().y))/4.0*3.0;
        Rect heightrect(center.x-original.cols/5, boundrect.tl().y, original.cols/5*2, boundrect.size().height);

        tmp = Mat::zeros(original.size(), CV_8U);
        rectangle(tmp, heightrect, Scalar(255, 255, 255), -1);
        circle(tmp, center, radius, Scalar(255, 255, 255), -1);

        bitwise_and(tmp, tmp1, tmp1);

        findContours(tmp1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
        max_area = 0;
        j = 0;
        for(k = 0; k < contours.size(); k++)
        {
            tmp_area = contourArea(contours[k]);
            if(tmp_area > max_area)
            {
                max_area = tmp_area;
                j = k;
            }
        }

        approxPolyDP(contours[j], contours[j], 30, true);
        convexHull(contours[j], contours[j]);

        drawContours(original, contours, j, Scalar(0, 0, 255), 3);

        namedWindow(argv[i], CV_WINDOW_NORMAL|CV_WINDOW_KEEPRATIO|CV_GUI_EXPANDED);
        imshow(argv[i], original);

        waitKey(0);
        destroyWindow(argv[i]);
    }

    return 0;
}

第一步是检测图片中最亮的像素，但我们必须在树本身和反射其光的雪之间做区分。在这里，我们试图排除雪应用一个非常简单的滤镜的颜色代码:

GaussianBlur(original, tmp, Size(3, 3), 0, 0, BORDER_DEFAULT);
erode(tmp, tmp, Mat(), Point(-1, -1), 10);
cvtColor(tmp, tmp, CV_BGR2HSV);
inRange(tmp, Scalar(0, 0, 0), Scalar(180, 255, 200), tmp);

然后我们找到每个“亮”像素:

dilate(original, tmp1, Mat(), Point(-1, -1), 15);
cvtColor(tmp1, tmp1, CV_BGR2HLS);
inRange(tmp1, Scalar(0, 185, 0), Scalar(180, 255, 255), tmp1);
dilate(tmp1, tmp1, Mat(), Point(-1, -1), 10);

最后我们将两个结果结合起来:

bitwise_and(tmp, tmp1, tmp1);

现在我们寻找最大的明亮物体:

findContours(tmp1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
max_area = 0;
j = 0;
for(k = 0; k < contours.size(); k++)
{
    tmp_area = contourArea(contours[k]);
    if(tmp_area > max_area)
    {
        max_area = tmp_area;
        j = k;
    }
}
tmp1 = Mat::zeros(original.size(),CV_8U);
approxPolyDP(contours[j], contours[j], 30, true);
drawContours(tmp1, contours, j, Scalar(255,255,255), CV_FILLED);

现在我们已经基本完成了，但是由于下雪，我们仍然有一些不完善的地方。 为了切断它们，我们将使用一个圆形和一个矩形来近似树的形状来创建一个蒙版，以删除不需要的部分:

m = moments(contours[j]);
boundrect = boundingRect(contours[j]);
center = Point2f(m.m10/m.m00, m.m01/m.m00);
radius = (center.y - (boundrect.tl().y))/4.0*3.0;
Rect heightrect(center.x-original.cols/5, boundrect.tl().y, original.cols/5*2, boundrect.size().height);

tmp = Mat::zeros(original.size(), CV_8U);
rectangle(tmp, heightrect, Scalar(255, 255, 255), -1);
circle(tmp, center, radius, Scalar(255, 255, 255), -1);

bitwise_and(tmp, tmp1, tmp1);

最后一步是找到我们的树的轮廓，并把它画在原图上。

findContours(tmp1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
max_area = 0;
j = 0;
for(k = 0; k < contours.size(); k++)
{
    tmp_area = contourArea(contours[k]);
    if(tmp_area > max_area)
    {
        max_area = tmp_area;
        j = k;
    }
}

approxPolyDP(contours[j], contours[j], 30, true);
convexHull(contours[j], contours[j]);

drawContours(original, contours, j, Scalar(0, 0, 255), 3);

很抱歉，目前我的网络连接不好，无法上传图片。我以后再试着做。

圣诞快乐。

编辑:

以下是最终输出的一些图片:

代码用Matlab R2007a编写。我用k-means粗略地提取了圣诞树。我 将只显示一张图像的中间结果，并显示所有六张图像的最终结果。

首先，我将RGB空间映射到Lab空间上，这样可以增强红色b通道的对比度:

colorTransform = makecform('srgb2lab');
I = applycform(I, colorTransform);
L = double(I(:,:,1));
a = double(I(:,:,2));
b = double(I(:,:,3));

除了颜色空间的特征，我还使用了与纹理相关的特征 而不是每个像素本身。这里我线性组合了强度 3个原始频道(R、G、B)。我之所以这样格式化是因为圣诞节 图中的树都有红灯，有时是绿色，有时是蓝色 还有照明。

R=double(Irgb(:,:,1));
G=double(Irgb(:,:,2));
B=double(Irgb(:,:,3));
I0 = (3*R + max(G,B)-min(G,B))/2;

我在I0上应用了一个3X3局部二进制模式，以中心像素作为阈值，并且 通过计算平均像素强度值的差值得到对比度 高于阈值，平均值低于阈值。

I0_copy = zeros(size(I0));
for i = 2 : size(I0,1) - 1
    for j = 2 : size(I0,2) - 1
        tmp = I0(i-1:i+1,j-1:j+1) >= I0(i,j);
        I0_copy(i,j) = mean(mean(tmp.*I0(i-1:i+1,j-1:j+1))) - ...
            mean(mean(~tmp.*I0(i-1:i+1,j-1:j+1))); % Contrast
    end
end

因为我总共有4个特征，所以在我的聚类方法中我会选择K=5。的代码 k-均值如下所示(它来自Andrew Ng博士的机器学习课程。我拿了 在他的编程作业中，我自己写了代码)。

[centroids, idx] = runkMeans(X, initial_centroids, max_iters);
mask=reshape(idx,img_size(1),img_size(2));

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [centroids, idx] = runkMeans(X, initial_centroids, ...
                                  max_iters, plot_progress)
   [m n] = size(X);
   K = size(initial_centroids, 1);
   centroids = initial_centroids;
   previous_centroids = centroids;
   idx = zeros(m, 1);

   for i=1:max_iters    
      % For each example in X, assign it to the closest centroid
      idx = findClosestCentroids(X, centroids);

      % Given the memberships, compute new centroids
      centroids = computeCentroids(X, idx, K);

   end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function idx = findClosestCentroids(X, centroids)
   K = size(centroids, 1);
   idx = zeros(size(X,1), 1);
   for xi = 1:size(X,1)
      x = X(xi, :);
      % Find closest centroid for x.
      best = Inf;
      for mui = 1:K
        mu = centroids(mui, :);
        d = dot(x - mu, x - mu);
        if d < best
           best = d;
           idx(xi) = mui;
        end
      end
   end 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function centroids = computeCentroids(X, idx, K)
   [m n] = size(X);
   centroids = zeros(K, n);
   for mui = 1:K
      centroids(mui, :) = sum(X(idx == mui, :)) / sum(idx == mui);
   end

Since the program runs very slow in my computer, I just ran 3 iterations. Normally the stop criteria is (i) iteration time at least 10, or (ii) no change on the centroids any more. To my test, increasing the iteration may differentiate the background (sky and tree, sky and building,...) more accurately, but did not show a drastic changes in christmas tree extraction. Also note k-means is not immune to the random centroid initialization, so running the program several times to make a comparison is recommended.

After the k-means, the labelled region with the maximum intensity of I0 was chosen. And boundary tracing was used to extracted the boundaries. To me, the last christmas tree is the most difficult one to extract since the contrast in that picture is not high enough as they are in the first five. Another issue in my method is that I used bwboundaries function in Matlab to trace the boundary, but sometimes the inner boundaries are also included as you can observe in 3rd, 5th, 6th results. The dark side within the christmas trees are not only failed to be clustered with the illuminated side, but they also lead to so many tiny inner boundaries tracing (imfill doesn't improve very much). In all my algorithm still has a lot improvement space.

Some publications indicates that mean-shift may be more robust than k-means, and many graph-cut based algorithms are also very competitive on complicated boundaries segmentation. I wrote a mean-shift algorithm myself, it seems to better extract the regions without enough light. But mean-shift is a little bit over-segmented, and some strategy of merging is needed. It ran even much slower than k-means in my computer, I am afraid I have to give it up. I eagerly look forward to see others would submit excellent results here with those modern algorithms mentioned above.

但我始终认为特征选择是图像分割的关键部分。与 一个适当的特征选择，可以最大化的边缘之间的对象和背景，许多 分割算法肯定会起作用。不同的算法可以改善结果 从1到10，但特征选择可以将其从0提高到1。

圣诞快乐!

我的解决步骤:

Get R channel (from RGB) - all operations we make on this channel: Create Region of Interest (ROI) Threshold R channel with min value 149 (top right image) Dilate result region (middle left image) Detect eges in computed roi. Tree has a lot of edges (middle right image) Dilate result Erode with bigger radius ( bottom left image) Select the biggest (by area) object - it's the result region ConvexHull ( tree is convex polygon ) ( bottom right image ) Bounding box (bottom right image - grren box )

循序渐进:

第一个结果——最简单但不是开源软件——“自适应视觉工作室+自适应视觉库”: 这不是开源的，但是很快就能原型化:

整个圣诞树检测算法(11块):

下一个步骤。我们需要开源解决方案。将AVL过滤器更改为OpenCV过滤器: 这里我做了一些小改动，例如边缘检测使用cvCanny过滤器，为了尊重roi，我将区域图像与边缘图像相乘，为了选择最大的元素，我使用findContours + contourArea，但想法是一样的。

https://www.youtube.com/watch?v=sfjB3MigLH0&index=1&list=UUpSRrkMHNHiLDXgylwhWNQQ

我现在不能显示中间步骤的图像，因为我只能放2个链接。

好吧，现在我们使用开源过滤器，但它仍然不是完全开源的。 最后一步-移植到c++代码。我在2.4.4版本中使用了OpenCV

最终的c++代码的结果是:

c++代码也很短:

#include "opencv2/highgui/highgui.hpp"
#include "opencv2/opencv.hpp"
#include <algorithm>
using namespace cv;

int main()
{

    string images[6] = {"..\\1.png","..\\2.png","..\\3.png","..\\4.png","..\\5.png","..\\6.png"};

    for(int i = 0; i < 6; ++i)
    {
        Mat img, thresholded, tdilated, tmp, tmp1;
        vector<Mat> channels(3);

        img = imread(images[i]);
        split(img, channels);
        threshold( channels[2], thresholded, 149, 255, THRESH_BINARY);                      //prepare ROI - threshold
        dilate( thresholded, tdilated,  getStructuringElement( MORPH_RECT, Size(22,22) ) ); //prepare ROI - dilate
        Canny( channels[2], tmp, 75, 125, 3, true );    //Canny edge detection
        multiply( tmp, tdilated, tmp1 );    // set ROI

        dilate( tmp1, tmp, getStructuringElement( MORPH_RECT, Size(20,16) ) ); // dilate
        erode( tmp, tmp1, getStructuringElement( MORPH_RECT, Size(36,36) ) ); // erode

        vector<vector<Point> > contours, contours1(1);
        vector<Point> convex;
        vector<Vec4i> hierarchy;
        findContours( tmp1, contours, hierarchy, CV_RETR_TREE, CV_CHAIN_APPROX_SIMPLE, Point(0, 0) );

        //get element of maximum area
        //int bestID = std::max_element( contours.begin(), contours.end(), 
        //  []( const vector<Point>& A, const vector<Point>& B ) { return contourArea(A) < contourArea(B); } ) - contours.begin();

            int bestID = 0;
        int bestArea = contourArea( contours[0] );
        for( int i = 1; i < contours.size(); ++i )
        {
            int area = contourArea( contours[i] );
            if( area > bestArea )
            {
                bestArea  = area;
                bestID = i;
            }
        }

        convexHull( contours[bestID], contours1[0] ); 
        drawContours( img, contours1, 0, Scalar( 100, 100, 255 ), img.rows / 100, 8, hierarchy, 0, Point() );

        imshow("image", img );
        waitKey(0);
    }


    return 0;
}

我在opencv中使用python。

我的算法是这样的:

首先，它从图像中取出红色通道 对红色通道应用阈值(最小值200) 然后应用形态梯度，然后做一个“关闭”(扩张，然后侵蚀) 然后它找到平面上的轮廓然后选择最长的轮廓。

代码:

import numpy as np
import cv2
import copy


def findTree(image,num):
    im = cv2.imread(image)
    im = cv2.resize(im, (400,250))
    gray = cv2.cvtColor(im, cv2.COLOR_RGB2GRAY)
    imf = copy.deepcopy(im)

    b,g,r = cv2.split(im)
    minR = 200
    _,thresh = cv2.threshold(r,minR,255,0)
    kernel = np.ones((25,5))
    dst = cv2.morphologyEx(thresh, cv2.MORPH_GRADIENT, kernel)
    dst = cv2.morphologyEx(dst, cv2.MORPH_CLOSE, kernel)

    contours = cv2.findContours(dst,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)[0]
    cv2.drawContours(im, contours,-1, (0,255,0), 1)

    maxI = 0
    for i in range(len(contours)):
        if len(contours[maxI]) < len(contours[i]):
            maxI = i

    img = copy.deepcopy(r)
    cv2.polylines(img,[contours[maxI]],True,(255,255,255),3)
    imf[:,:,2] = img

    cv2.imshow(str(num), imf)

def main():
    findTree('tree.jpg',1)
    findTree('tree2.jpg',2)
    findTree('tree3.jpg',3)
    findTree('tree4.jpg',4)
    findTree('tree5.jpg',5)
    findTree('tree6.jpg',6)

    cv2.waitKey(0)
    cv2.destroyAllWindows()

if __name__ == "__main__":
    main()

如果我把核函数从(25,5)改成(10,5) 我在所有树上都得到了更好的结果，除了左下角，

我的算法假设树上有灯，而且 在左下角的树中，顶部的光线比其他树要少。

这是我使用传统图像处理方法的最后一篇文章…

在这里，我以某种方式结合了我的另外两个建议，取得了更好的结果。事实上，我看不出这些结果还能有什么更好的(特别是当你看到该方法产生的掩码图像时)。

该方法的核心是结合三个关键假设:

图像在树形区域应该有很高的波动 图像在树形区域应该有更高的强度 背景区域应该是低强度的，大部分是蓝色的

考虑到这些假设，方法如下:

将图像转换为HSV 用LoG滤波器过滤V通道 应用硬阈值对LoG过滤图像得到'活动'掩码A 对V通道进行硬阈值处理得到强度掩码B 采用H通道阈值法将低强度淡蓝色区域捕获到背景掩模C中 使用AND组合蒙版得到最终的蒙版 放大蒙版以扩大区域并连接分散的像素 消除小区域，得到最终的蒙版，最终只代表树

下面是MATLAB中的代码(同样，脚本加载当前文件夹中的所有jpg图像，同样，这远非一段优化的代码):

% clear everything
clear;
pack;
close all;
close all hidden;
drawnow;
clc;

% initialization
ims=dir('./*.jpg');
imgs={};
images={}; 
blur_images={}; 
log_image={}; 
dilated_image={};
int_image={};
back_image={};
bin_image={};
measurements={};
box={};
num=length(ims);
thres_div = 3;

for i=1:num, 
    % load original image
    imgs{end+1}=imread(ims(i).name);

    % convert to HSV colorspace
    images{end+1}=rgb2hsv(imgs{i});

    % apply laplacian filtering and heuristic hard thresholding
    val_thres = (max(max(images{i}(:,:,3)))/thres_div);
    log_image{end+1} = imfilter( images{i}(:,:,3),fspecial('log')) > val_thres;

    % get the most bright regions of the image
    int_thres = 0.26*max(max( images{i}(:,:,3)));
    int_image{end+1} = images{i}(:,:,3) > int_thres;

    % get the most probable background regions of the image
    back_image{end+1} = images{i}(:,:,1)>(150/360) & images{i}(:,:,1)<(320/360) & images{i}(:,:,3)<0.5;

    % compute the final binary image by combining 
    % high 'activity' with high intensity
    bin_image{end+1} = logical( log_image{i}) & logical( int_image{i}) & ~logical( back_image{i});

    % apply morphological dilation to connect distonnected components
    strel_size = round(0.01*max(size(imgs{i})));        % structuring element for morphological dilation
    dilated_image{end+1} = imdilate( bin_image{i}, strel('disk',strel_size));

    % do some measurements to eliminate small objects
    measurements{i} = regionprops( logical( dilated_image{i}),'Area','BoundingBox');

    % iterative enlargement of the structuring element for better connectivity
    while length(measurements{i})>14 && strel_size<(min(size(imgs{i}(:,:,1)))/2),
        strel_size = round( 1.5 * strel_size);
        dilated_image{i} = imdilate( bin_image{i}, strel('disk',strel_size));
        measurements{i} = regionprops( logical( dilated_image{i}),'Area','BoundingBox');
    end

    for m=1:length(measurements{i})
        if measurements{i}(m).Area < 0.05*numel( dilated_image{i})
            dilated_image{i}( round(measurements{i}(m).BoundingBox(2):measurements{i}(m).BoundingBox(4)+measurements{i}(m).BoundingBox(2)),...
                round(measurements{i}(m).BoundingBox(1):measurements{i}(m).BoundingBox(3)+measurements{i}(m).BoundingBox(1))) = 0;
        end
    end
    % make sure the dilated image is the same size with the original
    dilated_image{i} = dilated_image{i}(1:size(imgs{i},1),1:size(imgs{i},2));
    % compute the bounding box
    [y,x] = find( dilated_image{i});
    if isempty( y)
        box{end+1}=[];
    else
        box{end+1} = [ min(x) min(y) max(x)-min(x)+1 max(y)-min(y)+1];
    end
end 

%%% additional code to display things
for i=1:num,
    figure;
    subplot(121);
    colormap gray;
    imshow( imgs{i});
    if ~isempty(box{i})
        hold on;
        rr = rectangle( 'position', box{i});
        set( rr, 'EdgeColor', 'r');
        hold off;
    end
    subplot(122);
    imshow( imgs{i}.*uint8(repmat(dilated_image{i},[1 1 3])));
end

结果

高分辨率的结果仍然可以在这里! 更多的实验图片可以在这里找到。