可以使用哪些图像处理技术来实现检测以下图像中显示的圣诞树的应用程序?

我正在寻找解决方案,将工作在所有这些图像。因此,需要训练haar级联分类器或模板匹配的方法不是很有趣。

我正在寻找可以用任何编程语言编写的东西,只要它只使用开源技术。解决方案必须使用在此问题上共享的图像进行测试。有6个输入图像,答案应该显示处理每个图像的结果。最后,对于每个输出图像,必须在检测到的树周围绘制红线。

您将如何通过编程来检测这些图像中的树木呢?


当前回答

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。

其他回答

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。

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:

我有一个方法,我认为是有趣的,有点不同于其他。与其他一些方法相比,我的方法的主要区别在于如何执行图像分割步骤——我使用了Python的scikit-learn中的DBSCAN聚类算法;它被优化用于寻找一些无定形的形状,这些形状可能不一定有一个清晰的形心。

At the top level, my approach is fairly simple and can be broken down into about 3 steps. First I apply a threshold (or actually, the logical "or" of two separate and distinct thresholds). As with many of the other answers, I assumed that the Christmas tree would be one of the brighter objects in the scene, so the first threshold is just a simple monochrome brightness test; any pixels with values above 220 on a 0-255 scale (where black is 0 and white is 255) are saved to a binary black-and-white image. The second threshold tries to look for red and yellow lights, which are particularly prominent in the trees in the upper left and lower right of the six images, and stand out well against the blue-green background which is prevalent in most of the photos. I convert the rgb image to hsv space, and require that the hue is either less than 0.2 on a 0.0-1.0 scale (corresponding roughly to the border between yellow and green) or greater than 0.95 (corresponding to the border between purple and red) and additionally I require bright, saturated colors: saturation and value must both be above 0.7. The results of the two threshold procedures are logically "or"-ed together, and the resulting matrix of black-and-white binary images is shown below:

您可以清楚地看到,每张图像都有一个大的像素集群,大致对应于每棵树的位置,加上一些图像还有一些其他的小集群,对应于一些建筑物窗户上的灯光,或地平线上的背景场景。下一步是让计算机识别出这些是独立的集群,并用集群成员ID号正确地标记每个像素。

对于这个任务,我选择了DBSCAN。这里有一个关于DBSCAN与其他聚类算法的典型行为的很好的可视化比较。正如我前面所说,它可以很好地处理非晶态形状。DBSCAN的输出,每个集群用不同的颜色绘制,如下所示:

There are a few things to be aware of when looking at this result. First is that DBSCAN requires the user to set a "proximity" parameter in order to regulate its behavior, which effectively controls how separated a pair of points must be in order for the algorithm to declare a new separate cluster rather than agglomerating a test point onto an already pre-existing cluster. I set this value to be 0.04 times the size along the diagonal of each image. Since the images vary in size from roughly VGA up to about HD 1080, this type of scale-relative definition is critical.

Another point worth noting is that the DBSCAN algorithm as it is implemented in scikit-learn has memory limits which are fairly challenging for some of the larger images in this sample. Therefore, for a few of the larger images, I actually had to "decimate" (i.e., retain only every 3rd or 4th pixel and drop the others) each cluster in order to stay within this limit. As a result of this culling process, the remaining individual sparse pixels are difficult to see on some of the larger images. Therefore, for display purposes only, the color-coded pixels in the above images have been effectively "dilated" just slightly so that they stand out better. It's purely a cosmetic operation for the sake of the narrative; although there are comments mentioning this dilation in my code, rest assured that it has nothing to do with any calculations that actually matter.

一旦识别并标记了集群,第三步也是最后一步就很容易了:我只需要在每个图像中选择最大的集群(在这种情况下,我选择根据成员像素的总数来衡量“大小”,尽管也可以使用某种类型的度量标准来衡量物理范围),然后计算该集群的凸包。凸包就变成了树的边界。通过该方法计算的6个凸包如下图中红色部分所示:

源代码是为Python 2.7.6编写的,它依赖于numpy, scipy, matplotlib和scikit-learn。我把它分成了两部分。第一部分负责实际图像处理:

from PIL import Image
import numpy as np
import scipy as sp
import matplotlib.colors as colors
from sklearn.cluster import DBSCAN
from math import ceil, sqrt

"""
Inputs:

    rgbimg:         [M,N,3] numpy array containing (uint, 0-255) color image

    hueleftthr:     Scalar constant to select maximum allowed hue in the
                    yellow-green region

    huerightthr:    Scalar constant to select minimum allowed hue in the
                    blue-purple region

    satthr:         Scalar constant to select minimum allowed saturation

    valthr:         Scalar constant to select minimum allowed value

    monothr:        Scalar constant to select minimum allowed monochrome
                    brightness

    maxpoints:      Scalar constant maximum number of pixels to forward to
                    the DBSCAN clustering algorithm

    proxthresh:     Proximity threshold to use for DBSCAN, as a fraction of
                    the diagonal size of the image

Outputs:

    borderseg:      [K,2,2] Nested list containing K pairs of x- and y- pixel
                    values for drawing the tree border

    X:              [P,2] List of pixels that passed the threshold step

    labels:         [Q,2] List of cluster labels for points in Xslice (see
                    below)

    Xslice:         [Q,2] Reduced list of pixels to be passed to DBSCAN

"""

def findtree(rgbimg, hueleftthr=0.2, huerightthr=0.95, satthr=0.7, 
             valthr=0.7, monothr=220, maxpoints=5000, proxthresh=0.04):

    # Convert rgb image to monochrome for
    gryimg = np.asarray(Image.fromarray(rgbimg).convert('L'))
    # Convert rgb image (uint, 0-255) to hsv (float, 0.0-1.0)
    hsvimg = colors.rgb_to_hsv(rgbimg.astype(float)/255)

    # Initialize binary thresholded image
    binimg = np.zeros((rgbimg.shape[0], rgbimg.shape[1]))
    # Find pixels with hue<0.2 or hue>0.95 (red or yellow) and saturation/value
    # both greater than 0.7 (saturated and bright)--tends to coincide with
    # ornamental lights on trees in some of the images
    boolidx = np.logical_and(
                np.logical_and(
                  np.logical_or((hsvimg[:,:,0] < hueleftthr),
                                (hsvimg[:,:,0] > huerightthr)),
                                (hsvimg[:,:,1] > satthr)),
                                (hsvimg[:,:,2] > valthr))
    # Find pixels that meet hsv criterion
    binimg[np.where(boolidx)] = 255
    # Add pixels that meet grayscale brightness criterion
    binimg[np.where(gryimg > monothr)] = 255

    # Prepare thresholded points for DBSCAN clustering algorithm
    X = np.transpose(np.where(binimg == 255))
    Xslice = X
    nsample = len(Xslice)
    if nsample > maxpoints:
        # Make sure number of points does not exceed DBSCAN maximum capacity
        Xslice = X[range(0,nsample,int(ceil(float(nsample)/maxpoints)))]

    # Translate DBSCAN proximity threshold to units of pixels and run DBSCAN
    pixproxthr = proxthresh * sqrt(binimg.shape[0]**2 + binimg.shape[1]**2)
    db = DBSCAN(eps=pixproxthr, min_samples=10).fit(Xslice)
    labels = db.labels_.astype(int)

    # Find the largest cluster (i.e., with most points) and obtain convex hull   
    unique_labels = set(labels)
    maxclustpt = 0
    for k in unique_labels:
        class_members = [index[0] for index in np.argwhere(labels == k)]
        if len(class_members) > maxclustpt:
            points = Xslice[class_members]
            hull = sp.spatial.ConvexHull(points)
            maxclustpt = len(class_members)
            borderseg = [[points[simplex,0], points[simplex,1]] for simplex
                          in hull.simplices]

    return borderseg, X, labels, Xslice

第二部分是一个用户级脚本,它调用第一个文件并生成上面所有的图:

#!/usr/bin/env python

from PIL import Image
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from findtree import findtree

# Image files to process
fname = ['nmzwj.png', 'aVZhC.png', '2K9EF.png',
         'YowlH.png', '2y4o5.png', 'FWhSP.png']

# Initialize figures
fgsz = (16,7)        
figthresh = plt.figure(figsize=fgsz, facecolor='w')
figclust  = plt.figure(figsize=fgsz, facecolor='w')
figcltwo  = plt.figure(figsize=fgsz, facecolor='w')
figborder = plt.figure(figsize=fgsz, facecolor='w')
figthresh.canvas.set_window_title('Thresholded HSV and Monochrome Brightness')
figclust.canvas.set_window_title('DBSCAN Clusters (Raw Pixel Output)')
figcltwo.canvas.set_window_title('DBSCAN Clusters (Slightly Dilated for Display)')
figborder.canvas.set_window_title('Trees with Borders')

for ii, name in zip(range(len(fname)), fname):
    # Open the file and convert to rgb image
    rgbimg = np.asarray(Image.open(name))

    # Get the tree borders as well as a bunch of other intermediate values
    # that will be used to illustrate how the algorithm works
    borderseg, X, labels, Xslice = findtree(rgbimg)

    # Display thresholded images
    axthresh = figthresh.add_subplot(2,3,ii+1)
    axthresh.set_xticks([])
    axthresh.set_yticks([])
    binimg = np.zeros((rgbimg.shape[0], rgbimg.shape[1]))
    for v, h in X:
        binimg[v,h] = 255
    axthresh.imshow(binimg, interpolation='nearest', cmap='Greys')

    # Display color-coded clusters
    axclust = figclust.add_subplot(2,3,ii+1) # Raw version
    axclust.set_xticks([])
    axclust.set_yticks([])
    axcltwo = figcltwo.add_subplot(2,3,ii+1) # Dilated slightly for display only
    axcltwo.set_xticks([])
    axcltwo.set_yticks([])
    axcltwo.imshow(binimg, interpolation='nearest', cmap='Greys')
    clustimg = np.ones(rgbimg.shape)    
    unique_labels = set(labels)
    # Generate a unique color for each cluster 
    plcol = cm.rainbow_r(np.linspace(0, 1, len(unique_labels)))
    for lbl, pix in zip(labels, Xslice):
        for col, unqlbl in zip(plcol, unique_labels):
            if lbl == unqlbl:
                # Cluster label of -1 indicates no cluster membership;
                # override default color with black
                if lbl == -1:
                    col = [0.0, 0.0, 0.0, 1.0]
                # Raw version
                for ij in range(3):
                    clustimg[pix[0],pix[1],ij] = col[ij]
                # Dilated just for display
                axcltwo.plot(pix[1], pix[0], 'o', markerfacecolor=col, 
                    markersize=1, markeredgecolor=col)
    axclust.imshow(clustimg)
    axcltwo.set_xlim(0, binimg.shape[1]-1)
    axcltwo.set_ylim(binimg.shape[0], -1)

    # Plot original images with read borders around the trees
    axborder = figborder.add_subplot(2,3,ii+1)
    axborder.set_axis_off()
    axborder.imshow(rgbimg, interpolation='nearest')
    for vseg, hseg in borderseg:
        axborder.plot(hseg, vseg, 'r-', lw=3)
    axborder.set_xlim(0, binimg.shape[1]-1)
    axborder.set_ylim(binimg.shape[0], -1)

plt.show()

使用一种与我所看到的完全不同的方法,我创建了一个php脚本,通过它们的灯来检测圣诞树。结果总是一个对称的三角形,如果需要的话,还会有数值,比如树的角度(“肥度”)。

这个算法最大的威胁显然是树旁边(大量)或树前面的灯(在进一步优化之前是更大的问题)。 编辑(补充):它不能做的:找出是否有一棵圣诞树,在一张图片中找到多棵圣诞树,正确地检测出拉斯维加斯中心的圣诞树,检测出严重弯曲、颠倒或被砍倒的圣诞树……;)

不同的阶段是:

Calculate the added brightness (R+G+B) for each pixel Add up this value of all 8 neighbouring pixels on top of each pixel Rank all pixels by this value (brightest first) - I know, not really subtle... Choose N of these, starting from the top, skipping ones that are too close Calculate the median of these top N (gives us the approximate center of the tree) Start from the median position upwards in a widening search beam for the topmost light from the selected brightest ones (people tend to put at least one light at the very top) From there, imagine lines going 60 degrees left and right downwards (christmas trees shouldn't be that fat) Decrease those 60 degrees until 20% of the brightest lights are outside this triangle Find the light at the very bottom of the triangle, giving you the lower horizontal border of the tree Done

标记说明:

大红色十字在树的中心:最亮的N个灯的中间 虚线从那里往上:“搜索光束”,寻找树的顶部 小十字:树顶 小红叉:所有最亮的灯光 红三角:呃!

源代码:

<?php

ini_set('memory_limit', '1024M');

header("Content-type: image/png");

$chosenImage = 6;

switch($chosenImage){
    case 1:
        $inputImage     = imagecreatefromjpeg("nmzwj.jpg");
        break;
    case 2:
        $inputImage     = imagecreatefromjpeg("2y4o5.jpg");
        break;
    case 3:
        $inputImage     = imagecreatefromjpeg("YowlH.jpg");
        break;
    case 4:
        $inputImage     = imagecreatefromjpeg("2K9Ef.jpg");
        break;
    case 5:
        $inputImage     = imagecreatefromjpeg("aVZhC.jpg");
        break;
    case 6:
        $inputImage     = imagecreatefromjpeg("FWhSP.jpg");
        break;
    case 7:
        $inputImage     = imagecreatefromjpeg("roemerberg.jpg");
        break;
    default:
        exit();
}

// Process the loaded image

$topNspots = processImage($inputImage);

imagejpeg($inputImage);
imagedestroy($inputImage);

// Here be functions

function processImage($image) {
    $orange = imagecolorallocate($image, 220, 210, 60);
    $black = imagecolorallocate($image, 0, 0, 0);
    $red = imagecolorallocate($image, 255, 0, 0);

    $maxX = imagesx($image)-1;
    $maxY = imagesy($image)-1;

    // Parameters
    $spread = 1; // Number of pixels to each direction that will be added up
    $topPositions = 80; // Number of (brightest) lights taken into account
    $minLightDistance = round(min(array($maxX, $maxY)) / 30); // Minimum number of pixels between the brigtests lights
    $searchYperX = 5; // spread of the "search beam" from the median point to the top

    $renderStage = 3; // 1 to 3; exits the process early


    // STAGE 1
    // Calculate the brightness of each pixel (R+G+B)

    $maxBrightness = 0;
    $stage1array = array();

    for($row = 0; $row <= $maxY; $row++) {

        $stage1array[$row] = array();

        for($col = 0; $col <= $maxX; $col++) {

            $rgb = imagecolorat($image, $col, $row);
            $brightness = getBrightnessFromRgb($rgb);
            $stage1array[$row][$col] = $brightness;

            if($renderStage == 1){
                $brightnessToGrey = round($brightness / 765 * 256);
                $greyRgb = imagecolorallocate($image, $brightnessToGrey, $brightnessToGrey, $brightnessToGrey);
                imagesetpixel($image, $col, $row, $greyRgb);
            }

            if($brightness > $maxBrightness) {
                $maxBrightness = $brightness;
                if($renderStage == 1){
                    imagesetpixel($image, $col, $row, $red);
                }
            }
        }
    }
    if($renderStage == 1) {
        return;
    }


    // STAGE 2
    // Add up brightness of neighbouring pixels

    $stage2array = array();
    $maxStage2 = 0;

    for($row = 0; $row <= $maxY; $row++) {
        $stage2array[$row] = array();

        for($col = 0; $col <= $maxX; $col++) {
            if(!isset($stage2array[$row][$col])) $stage2array[$row][$col] = 0;

            // Look around the current pixel, add brightness
            for($y = $row-$spread; $y <= $row+$spread; $y++) {
                for($x = $col-$spread; $x <= $col+$spread; $x++) {

                    // Don't read values from outside the image
                    if($x >= 0 && $x <= $maxX && $y >= 0 && $y <= $maxY){
                        $stage2array[$row][$col] += $stage1array[$y][$x]+10;
                    }
                }
            }

            $stage2value = $stage2array[$row][$col];
            if($stage2value > $maxStage2) {
                $maxStage2 = $stage2value;
            }
        }
    }

    if($renderStage >= 2){
        // Paint the accumulated light, dimmed by the maximum value from stage 2
        for($row = 0; $row <= $maxY; $row++) {
            for($col = 0; $col <= $maxX; $col++) {
                $brightness = round($stage2array[$row][$col] / $maxStage2 * 255);
                $greyRgb = imagecolorallocate($image, $brightness, $brightness, $brightness);
                imagesetpixel($image, $col, $row, $greyRgb);
            }
        }
    }

    if($renderStage == 2) {
        return;
    }


    // STAGE 3

    // Create a ranking of bright spots (like "Top 20")
    $topN = array();

    for($row = 0; $row <= $maxY; $row++) {
        for($col = 0; $col <= $maxX; $col++) {

            $stage2Brightness = $stage2array[$row][$col];
            $topN[$col.":".$row] = $stage2Brightness;
        }
    }
    arsort($topN);

    $topNused = array();
    $topPositionCountdown = $topPositions;

    if($renderStage == 3){
        foreach ($topN as $key => $val) {
            if($topPositionCountdown <= 0){
                break;
            }

            $position = explode(":", $key);

            foreach($topNused as $usedPosition => $usedValue) {
                $usedPosition = explode(":", $usedPosition);
                $distance = abs($usedPosition[0] - $position[0]) + abs($usedPosition[1] - $position[1]);
                if($distance < $minLightDistance) {
                    continue 2;
                }
            }

            $topNused[$key] = $val;

            paintCrosshair($image, $position[0], $position[1], $red, 2);

            $topPositionCountdown--;

        }
    }


    // STAGE 4
    // Median of all Top N lights
    $topNxValues = array();
    $topNyValues = array();

    foreach ($topNused as $key => $val) {
        $position = explode(":", $key);
        array_push($topNxValues, $position[0]);
        array_push($topNyValues, $position[1]);
    }

    $medianXvalue = round(calculate_median($topNxValues));
    $medianYvalue = round(calculate_median($topNyValues));
    paintCrosshair($image, $medianXvalue, $medianYvalue, $red, 15);


    // STAGE 5
    // Find treetop

    $filename = 'debug.log';
    $handle = fopen($filename, "w");
    fwrite($handle, "\n\n STAGE 5");

    $treetopX = $medianXvalue;
    $treetopY = $medianYvalue;

    $searchXmin = $medianXvalue;
    $searchXmax = $medianXvalue;

    $width = 0;
    for($y = $medianYvalue; $y >= 0; $y--) {
        fwrite($handle, "\nAt y = ".$y);

        if(($y % $searchYperX) == 0) { // Modulo
            $width++;
            $searchXmin = $medianXvalue - $width;
            $searchXmax = $medianXvalue + $width;
            imagesetpixel($image, $searchXmin, $y, $red);
            imagesetpixel($image, $searchXmax, $y, $red);
        }

        foreach ($topNused as $key => $val) {
            $position = explode(":", $key); // "x:y"

            if($position[1] != $y){
                continue;
            }

            if($position[0] >= $searchXmin && $position[0] <= $searchXmax){
                $treetopX = $position[0];
                $treetopY = $y;
            }
        }

    }

    paintCrosshair($image, $treetopX, $treetopY, $red, 5);


    // STAGE 6
    // Find tree sides
    fwrite($handle, "\n\n STAGE 6");

    $treesideAngle = 60; // The extremely "fat" end of a christmas tree
    $treeBottomY = $treetopY;

    $topPositionsExcluded = 0;
    $xymultiplier = 0;
    while(($topPositionsExcluded < ($topPositions / 5)) && $treesideAngle >= 1){
        fwrite($handle, "\n\nWe're at angle ".$treesideAngle);
        $xymultiplier = sin(deg2rad($treesideAngle));
        fwrite($handle, "\nMultiplier: ".$xymultiplier);

        $topPositionsExcluded = 0;
        foreach ($topNused as $key => $val) {
            $position = explode(":", $key);
            fwrite($handle, "\nAt position ".$key);

            if($position[1] > $treeBottomY) {
                $treeBottomY = $position[1];
            }

            // Lights above the tree are outside of it, but don't matter
            if($position[1] < $treetopY){
                $topPositionsExcluded++;
                fwrite($handle, "\nTOO HIGH");
                continue;
            }

            // Top light will generate division by zero
            if($treetopY-$position[1] == 0) {
                fwrite($handle, "\nDIVISION BY ZERO");
                continue;
            }

            // Lights left end right of it are also not inside
            fwrite($handle, "\nLight position factor: ".(abs($treetopX-$position[0]) / abs($treetopY-$position[1])));
            if((abs($treetopX-$position[0]) / abs($treetopY-$position[1])) > $xymultiplier){
                $topPositionsExcluded++;
                fwrite($handle, "\n --- Outside tree ---");
            }
        }

        $treesideAngle--;
    }
    fclose($handle);

    // Paint tree's outline
    $treeHeight = abs($treetopY-$treeBottomY);
    $treeBottomLeft = 0;
    $treeBottomRight = 0;
    $previousState = false; // line has not started; assumes the tree does not "leave"^^

    for($x = 0; $x <= $maxX; $x++){
        if(abs($treetopX-$x) != 0 && abs($treetopX-$x) / $treeHeight > $xymultiplier){
            if($previousState == true){
                $treeBottomRight = $x;
                $previousState = false;
            }
            continue;
        }
        imagesetpixel($image, $x, $treeBottomY, $red);
        if($previousState == false){
            $treeBottomLeft = $x;
            $previousState = true;
        }
    }
    imageline($image, $treeBottomLeft, $treeBottomY, $treetopX, $treetopY, $red);
    imageline($image, $treeBottomRight, $treeBottomY, $treetopX, $treetopY, $red);


    // Print out some parameters

    $string = "Min dist: ".$minLightDistance." | Tree angle: ".$treesideAngle." deg | Tree bottom: ".$treeBottomY;

    $px     = (imagesx($image) - 6.5 * strlen($string)) / 2;
    imagestring($image, 2, $px, 5, $string, $orange);

    return $topN;
}

/**
 * Returns values from 0 to 765
 */
function getBrightnessFromRgb($rgb) {
    $r = ($rgb >> 16) & 0xFF;
    $g = ($rgb >> 8) & 0xFF;
    $b = $rgb & 0xFF;

    return $r+$r+$b;
}

function paintCrosshair($image, $posX, $posY, $color, $size=5) {
    for($x = $posX-$size; $x <= $posX+$size; $x++) {
        if($x>=0 && $x < imagesx($image)){
            imagesetpixel($image, $x, $posY, $color);
        }
    }
    for($y = $posY-$size; $y <= $posY+$size; $y++) {
        if($y>=0 && $y < imagesy($image)){
            imagesetpixel($image, $posX, $y, $color);
        }
    }
}

// From http://www.mdj.us/web-development/php-programming/calculating-the-median-average-values-of-an-array-with-php/
function calculate_median($arr) {
    sort($arr);
    $count = count($arr); //total numbers in array
    $middleval = floor(($count-1)/2); // find the middle value, or the lowest middle value
    if($count % 2) { // odd number, middle is the median
        $median = $arr[$middleval];
    } else { // even number, calculate avg of 2 medians
        $low = $arr[$middleval];
        $high = $arr[$middleval+1];
        $median = (($low+$high)/2);
    }
    return $median;
}


?>

图片:

额外奖励:来自维基百科的德国Weihnachtsbaum http://commons.wikimedia.org/wiki/File:Weihnachtsbaum_R%C3%B6merberg.jpg

编辑注意:我编辑这篇文章是为了(I)按照要求分别处理每个树图像,(ii)同时考虑物体亮度和形状,以提高结果的质量。


下面是一种考虑到物体亮度和形状的方法。换句话说,它寻找的是具有三角形形状和显著亮度的物体。它是用Java语言实现的,使用Marvin图像处理框架。

第一步是颜色阈值。这里的目标是集中分析具有显著亮度的物体。

输出图片:

源代码:

public class ChristmasTree {

private MarvinImagePlugin fill = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.fill.boundaryFill");
private MarvinImagePlugin threshold = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.thresholding");
private MarvinImagePlugin invert = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.invert");
private MarvinImagePlugin dilation = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.morphological.dilation");

public ChristmasTree(){
    MarvinImage tree;

    // Iterate each image
    for(int i=1; i<=6; i++){
        tree = MarvinImageIO.loadImage("./res/trees/tree"+i+".png");

        // 1. Threshold
        threshold.setAttribute("threshold", 200);
        threshold.process(tree.clone(), tree);
    }
}
public static void main(String[] args) {
    new ChristmasTree();
}
}

在第二步中,图像中最亮的点被放大以形成形状。这一过程的结果是具有显著亮度的物体的可能形状。应用洪水填充分割,断开的形状被检测。

输出图片:

源代码:

public class ChristmasTree {

private MarvinImagePlugin fill = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.fill.boundaryFill");
private MarvinImagePlugin threshold = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.thresholding");
private MarvinImagePlugin invert = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.invert");
private MarvinImagePlugin dilation = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.morphological.dilation");

public ChristmasTree(){
    MarvinImage tree;

    // Iterate each image
    for(int i=1; i<=6; i++){
        tree = MarvinImageIO.loadImage("./res/trees/tree"+i+".png");

        // 1. Threshold
        threshold.setAttribute("threshold", 200);
        threshold.process(tree.clone(), tree);

        // 2. Dilate
        invert.process(tree.clone(), tree);
        tree = MarvinColorModelConverter.rgbToBinary(tree, 127);
        MarvinImageIO.saveImage(tree, "./res/trees/new/tree_"+i+"threshold.png");
        dilation.setAttribute("matrix", MarvinMath.getTrueMatrix(50, 50));
        dilation.process(tree.clone(), tree);
        MarvinImageIO.saveImage(tree, "./res/trees/new/tree_"+1+"_dilation.png");
        tree = MarvinColorModelConverter.binaryToRgb(tree);

        // 3. Segment shapes
        MarvinImage trees2 = tree.clone();
        fill(tree, trees2);
        MarvinImageIO.saveImage(trees2, "./res/trees/new/tree_"+i+"_fill.png");
}

private void fill(MarvinImage imageIn, MarvinImage imageOut){
    boolean found;
    int color= 0xFFFF0000;

    while(true){
        found=false;

        Outerloop:
        for(int y=0; y<imageIn.getHeight(); y++){
            for(int x=0; x<imageIn.getWidth(); x++){
                if(imageOut.getIntComponent0(x, y) == 0){
                    fill.setAttribute("x", x);
                    fill.setAttribute("y", y);
                    fill.setAttribute("color", color);
                    fill.setAttribute("threshold", 120);
                    fill.process(imageIn, imageOut);
                    color = newColor(color);

                    found = true;
                    break Outerloop;
                }
            }
        }

        if(!found){
            break;
        }
    }

}

private int newColor(int color){
    int red = (color & 0x00FF0000) >> 16;
    int green = (color & 0x0000FF00) >> 8;
    int blue = (color & 0x000000FF);

    if(red <= green && red <= blue){
        red+=5;
    }
    else if(green <= red && green <= blue){
        green+=5;
    }
    else{
        blue+=5;
    }

    return 0xFF000000 + (red << 16) + (green << 8) + blue;
}

public static void main(String[] args) {
    new ChristmasTree();
}
}

如输出图像所示,检测到多个形状。在这个问题中,图像中只有几个亮点。但是,实现这种方法是为了处理更复杂的场景。

In the next step each shape is analyzed. A simple algorithm detects shapes with a pattern similar to a triangle. The algorithm analyze the object shape line by line. If the center of the mass of each shape line is almost the same (given a threshold) and mass increase as y increase, the object has a triangle-like shape. The mass of the shape line is the number of pixels in that line that belongs to the shape. Imagine you slice the object horizontally and analyze each horizontal segment. If they are centralized to each other and the length increase from the first segment to last one in a linear pattern, you probably has an object that resembles a triangle.

源代码:

private int[] detectTrees(MarvinImage image){
    HashSet<Integer> analysed = new HashSet<Integer>();
    boolean found;
    while(true){
        found = false;
        for(int y=0; y<image.getHeight(); y++){
            for(int x=0; x<image.getWidth(); x++){
                int color = image.getIntColor(x, y);

                if(!analysed.contains(color)){
                    if(isTree(image, color)){
                        return getObjectRect(image, color);
                    }

                    analysed.add(color);
                    found=true;
                }
            }
        }

        if(!found){
            break;
        }
    }
    return null;
}

private boolean isTree(MarvinImage image, int color){

    int mass[][] = new int[image.getHeight()][2];
    int yStart=-1;
    int xStart=-1;
    for(int y=0; y<image.getHeight(); y++){
        int mc = 0;
        int xs=-1;
        int xe=-1;
        for(int x=0; x<image.getWidth(); x++){
            if(image.getIntColor(x, y) == color){
                mc++;

                if(yStart == -1){
                    yStart=y;
                    xStart=x;
                }

                if(xs == -1){
                    xs = x;
                }
                if(x > xe){
                    xe = x;
                }
            }
        }
        mass[y][0] = xs;
        mass[y][3] = xe;
        mass[y][4] = mc;    
    }

    int validLines=0;
    for(int y=0; y<image.getHeight(); y++){
        if
        ( 
            mass[y][5] > 0 &&
            Math.abs(((mass[y][0]+mass[y][6])/2)-xStart) <= 50 &&
            mass[y][7] >= (mass[yStart][8] + (y-yStart)*0.3) &&
            mass[y][9] <= (mass[yStart][10] + (y-yStart)*1.5)
        )
        {
            validLines++;
        }
    }

    if(validLines > 100){
        return true;
    }
    return false;
}

最后,每个形状的位置类似于一个三角形,并具有显著的亮度,在这种情况下,圣诞树,突出显示在原图中,如下所示。

最终输出图片:

最终源代码:

public class ChristmasTree {

private MarvinImagePlugin fill = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.fill.boundaryFill");
private MarvinImagePlugin threshold = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.thresholding");
private MarvinImagePlugin invert = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.color.invert");
private MarvinImagePlugin dilation = MarvinPluginLoader.loadImagePlugin("org.marvinproject.image.morphological.dilation");

public ChristmasTree(){
    MarvinImage tree;

    // Iterate each image
    for(int i=1; i<=6; i++){
        tree = MarvinImageIO.loadImage("./res/trees/tree"+i+".png");

        // 1. Threshold
        threshold.setAttribute("threshold", 200);
        threshold.process(tree.clone(), tree);

        // 2. Dilate
        invert.process(tree.clone(), tree);
        tree = MarvinColorModelConverter.rgbToBinary(tree, 127);
        MarvinImageIO.saveImage(tree, "./res/trees/new/tree_"+i+"threshold.png");
        dilation.setAttribute("matrix", MarvinMath.getTrueMatrix(50, 50));
        dilation.process(tree.clone(), tree);
        MarvinImageIO.saveImage(tree, "./res/trees/new/tree_"+1+"_dilation.png");
        tree = MarvinColorModelConverter.binaryToRgb(tree);

        // 3. Segment shapes
        MarvinImage trees2 = tree.clone();
        fill(tree, trees2);
        MarvinImageIO.saveImage(trees2, "./res/trees/new/tree_"+i+"_fill.png");

        // 4. Detect tree-like shapes
        int[] rect = detectTrees(trees2);

        // 5. Draw the result
        MarvinImage original = MarvinImageIO.loadImage("./res/trees/tree"+i+".png");
        drawBoundary(trees2, original, rect);
        MarvinImageIO.saveImage(original, "./res/trees/new/tree_"+i+"_out_2.jpg");
    }
}

private void drawBoundary(MarvinImage shape, MarvinImage original, int[] rect){
    int yLines[] = new int[6];
    yLines[0] = rect[1];
    yLines[1] = rect[1]+(int)((rect[3]/5));
    yLines[2] = rect[1]+((rect[3]/5)*2);
    yLines[3] = rect[1]+((rect[3]/5)*3);
    yLines[4] = rect[1]+(int)((rect[3]/5)*4);
    yLines[5] = rect[1]+rect[3];

    List<Point> points = new ArrayList<Point>();
    for(int i=0; i<yLines.length; i++){
        boolean in=false;
        Point startPoint=null;
        Point endPoint=null;
        for(int x=rect[0]; x<rect[0]+rect[2]; x++){

            if(shape.getIntColor(x, yLines[i]) != 0xFFFFFFFF){
                if(!in){
                    if(startPoint == null){
                        startPoint = new Point(x, yLines[i]);
                    }
                }
                in = true;
            }
            else{
                if(in){
                    endPoint = new Point(x, yLines[i]);
                }
                in = false;
            }
        }

        if(endPoint == null){
            endPoint = new Point((rect[0]+rect[2])-1, yLines[i]);
        }

        points.add(startPoint);
        points.add(endPoint);
    }

    drawLine(points.get(0).x, points.get(0).y, points.get(1).x, points.get(1).y, 15, original);
    drawLine(points.get(1).x, points.get(1).y, points.get(3).x, points.get(3).y, 15, original);
    drawLine(points.get(3).x, points.get(3).y, points.get(5).x, points.get(5).y, 15, original);
    drawLine(points.get(5).x, points.get(5).y, points.get(7).x, points.get(7).y, 15, original);
    drawLine(points.get(7).x, points.get(7).y, points.get(9).x, points.get(9).y, 15, original);
    drawLine(points.get(9).x, points.get(9).y, points.get(11).x, points.get(11).y, 15, original);
    drawLine(points.get(11).x, points.get(11).y, points.get(10).x, points.get(10).y, 15, original);
    drawLine(points.get(10).x, points.get(10).y, points.get(8).x, points.get(8).y, 15, original);
    drawLine(points.get(8).x, points.get(8).y, points.get(6).x, points.get(6).y, 15, original);
    drawLine(points.get(6).x, points.get(6).y, points.get(4).x, points.get(4).y, 15, original);
    drawLine(points.get(4).x, points.get(4).y, points.get(2).x, points.get(2).y, 15, original);
    drawLine(points.get(2).x, points.get(2).y, points.get(0).x, points.get(0).y, 15, original);
}

private void drawLine(int x1, int y1, int x2, int y2, int length, MarvinImage image){
    int lx1, lx2, ly1, ly2;
    for(int i=0; i<length; i++){
        lx1 = (x1+i >= image.getWidth() ? (image.getWidth()-1)-i: x1);
        lx2 = (x2+i >= image.getWidth() ? (image.getWidth()-1)-i: x2);
        ly1 = (y1+i >= image.getHeight() ? (image.getHeight()-1)-i: y1);
        ly2 = (y2+i >= image.getHeight() ? (image.getHeight()-1)-i: y2);

        image.drawLine(lx1+i, ly1, lx2+i, ly2, Color.red);
        image.drawLine(lx1, ly1+i, lx2, ly2+i, Color.red);
    }
}

private void fillRect(MarvinImage image, int[] rect, int length){
    for(int i=0; i<length; i++){
        image.drawRect(rect[0]+i, rect[1]+i, rect[2]-(i*2), rect[3]-(i*2), Color.red);
    }
}

private void fill(MarvinImage imageIn, MarvinImage imageOut){
    boolean found;
    int color= 0xFFFF0000;

    while(true){
        found=false;

        Outerloop:
        for(int y=0; y<imageIn.getHeight(); y++){
            for(int x=0; x<imageIn.getWidth(); x++){
                if(imageOut.getIntComponent0(x, y) == 0){
                    fill.setAttribute("x", x);
                    fill.setAttribute("y", y);
                    fill.setAttribute("color", color);
                    fill.setAttribute("threshold", 120);
                    fill.process(imageIn, imageOut);
                    color = newColor(color);

                    found = true;
                    break Outerloop;
                }
            }
        }

        if(!found){
            break;
        }
    }

}

private int[] detectTrees(MarvinImage image){
    HashSet<Integer> analysed = new HashSet<Integer>();
    boolean found;
    while(true){
        found = false;
        for(int y=0; y<image.getHeight(); y++){
            for(int x=0; x<image.getWidth(); x++){
                int color = image.getIntColor(x, y);

                if(!analysed.contains(color)){
                    if(isTree(image, color)){
                        return getObjectRect(image, color);
                    }

                    analysed.add(color);
                    found=true;
                }
            }
        }

        if(!found){
            break;
        }
    }
    return null;
}

private boolean isTree(MarvinImage image, int color){

    int mass[][] = new int[image.getHeight()][11];
    int yStart=-1;
    int xStart=-1;
    for(int y=0; y<image.getHeight(); y++){
        int mc = 0;
        int xs=-1;
        int xe=-1;
        for(int x=0; x<image.getWidth(); x++){
            if(image.getIntColor(x, y) == color){
                mc++;

                if(yStart == -1){
                    yStart=y;
                    xStart=x;
                }

                if(xs == -1){
                    xs = x;
                }
                if(x > xe){
                    xe = x;
                }
            }
        }
        mass[y][0] = xs;
        mass[y][12] = xe;
        mass[y][13] = mc;   
    }

    int validLines=0;
    for(int y=0; y<image.getHeight(); y++){
        if
        ( 
            mass[y][14] > 0 &&
            Math.abs(((mass[y][0]+mass[y][15])/2)-xStart) <= 50 &&
            mass[y][16] >= (mass[yStart][17] + (y-yStart)*0.3) &&
            mass[y][18] <= (mass[yStart][19] + (y-yStart)*1.5)
        )
        {
            validLines++;
        }
    }

    if(validLines > 100){
        return true;
    }
    return false;
}

private int[] getObjectRect(MarvinImage image, int color){
    int x1=-1;
    int x2=-1;
    int y1=-1;
    int y2=-1;

    for(int y=0; y<image.getHeight(); y++){
        for(int x=0; x<image.getWidth(); x++){
            if(image.getIntColor(x, y) == color){

                if(x1 == -1 || x < x1){
                    x1 = x;
                }
                if(x2 == -1 || x > x2){
                    x2 = x;
                }
                if(y1 == -1 || y < y1){
                    y1 = y;
                }
                if(y2 == -1 || y > y2){
                    y2 = y;
                }
            }
        }
    }

    return new int[]{x1, y1, (x2-x1), (y2-y1)};
}

private int newColor(int color){
    int red = (color & 0x00FF0000) >> 16;
    int green = (color & 0x0000FF00) >> 8;
    int blue = (color & 0x000000FF);

    if(red <= green && red <= blue){
        red+=5;
    }
    else if(green <= red && green <= blue){
        green+=30;
    }
    else{
        blue+=30;
    }

    return 0xFF000000 + (red << 16) + (green << 8) + blue;
}

public static void main(String[] args) {
    new ChristmasTree();
}
}

这种方法的优点是它可能适用于包含其他发光物体的图像,因为它分析物体的形状。

圣诞快乐!


编辑说明2

There is a discussion about the similarity of the output images of this solution and some other ones. In fact, they are very similar. But this approach does not just segment objects. It also analyzes the object shapes in some sense. It can handle multiple luminous objects in the same scene. In fact, the Christmas tree does not need to be the brightest one. I'm just abording it to enrich the discussion. There is a bias in the samples that just looking for the brightest object, you will find the trees. But, does we really want to stop the discussion at this point? At this point, how far the computer is really recognizing an object that resembles a Christmas tree? Let's try to close this gap.

下面给出一个结果来说明这一点:

输入图像

输出