• minboundrect

2017-02-21 19:58:06
• 对图像先进行灰度处理，接着进行二值化处理。。。。。。 matlab二值化
• minboundrect.m

2021-04-23 13:40:34
function [rectx,recty,area,perimeter] = minboundrect(x,y,metric)%%ֵͼСӾif (nargin<3) || isempty(metric)metric = 'a';elseif ~ischar(metric)error 'metric must be a character flag if it is supplied...

function [rectx,recty,area,perimeter] = minboundrect(x,y,metric)

%%ֵͼСӾ

if (nargin<3) || isempty(metric)

metric = 'a';

elseif ~ischar(metric)

error 'metric must be a character flag if it is supplied.'

else

% check for 'a' or 'p'

metric = lower(metric(:)');

ind = strmatch(metric,{'area','perimeter'});

if isempty(ind)

error 'metric does not match either ''area'' or ''perimeter'''

end

% just keep the first letter.

metric = metric(1);

end

% preprocess data

x=x(:);

y=y(:);

% not many error checks to worry about

n = length(x);

if n~=length(y)

error 'x and y must be the same sizes'

end

% start out with the convex hull of the points to

% reduce the problem dramatically. Note that any

% points in the interior of the convex hull are

% never needed, so we drop them.

if n>3

% if(x(1,1)==x(2,1)&&x(2,1)==x(3,1)&&x(3,1)== x(4,1))

if(x(1,1)==155&&x(2,1)==155&&x(3,1)== 155)

nedges = 0;

% elseif(y(1,1)==y(2,1)&&y(2,1)==y(3,1)&&y(3,1)== y(4,1))

% nedges = 0;

else

edges = convhull(x,y);

%edges = convhull(x,y,{'Qt'}); % 'Pp' will silence the warnings

% exclude those points inside the hull as not relevant

% also sorts the points into their convex hull as a

% closed polygon

x = x(edges);

y = y(edges);

% probably fewer points now, unless the points are fully convex

nedges = length(x) - 1;

end

elseif n>1

% n must be 2 or 3

nedges = n;

x(end+1) = x(1);

y(end+1) = y(1);

else

% n must be 0 or 1

nedges = n;

end

% now we must find the bounding rectangle of those

% that remain.

% special case small numbers of points. If we trip any

% of these cases, then we are done, so return.

switch nedges

case 0

% empty begets empty

rectx = [];

recty = [];

area = [];

perimeter = [];

return

case 1

% with one point, the rect is simple.

rectx = repmat(x,1,5);

recty = repmat(y,1,5);

area = 0;

perimeter = 0;

return

case 2

% only two points. also simple.

rectx = x([1 2 2 1 1]);

recty = y([1 2 2 1 1]);

area = 0;

perimeter = 2*sqrt(diff(x).^2 + diff(y).^2);

return

end

% 3 or more points.

% will need a 2x2 rotation matrix through an angle theta

Rmat = @(theta) [cos(theta) sin(theta);-sin(theta) cos(theta)];

% get the angle of each edge of the hull polygon.

ind = 1:(length(x)-1);

edgeangles = atan2(y(ind+1) - y(ind),x(ind+1) - x(ind));

% move the angle into the first quadrant.

edgeangles = unique(mod(edgeangles,pi/2));

% now just check each edge of the hull

nang = length(edgeangles);

area = inf;

perimeter = inf;

met = inf;

xy = [x,y];

for i = 1:nang

% rotate the data through -theta

rot = Rmat(-edgeangles(i));

xyr = xy*rot;

xymin = min(xyr,[],1);

xymax = max(xyr,[],1);

% The area is simple, as is the perimeter

A_i = prod(xymax - xymin);

P_i = 2*sum(xymax-xymin);

if metric=='a'

M_i = A_i;

else

M_i = P_i;

end

% new metric value for the current interval. Is it better?

if M_i

% keep this one

met = M_i;

area = A_i;

perimeter = P_i;

rect = [xymin;[xymax(1),xymin(2)];xymax;[xymin(1),xymax(2)];xymin];

rect = rect*rot';

rectx = rect(:,1);

recty = rect(:,2);

end

end

% get the final rect

% all done

end % mainline end

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展开全文 • minboundrect 最小外接矩形 ：可以求二值图像最小外接矩形（a:面积最小；p：周长最小） 下面是minboundrect.m 求白色部分的最小外接。 function [rectx,recty,area,perimeter] = minboundrect(x,y,metric) %% 矩形是...

笔记

minboundrect 最小外接矩形 ：可以求二值图像最小外接矩形（a:面积最小；p：周长最小） 斜矩形

下面是minboundrect.m 求白色部分的最小外接

function [rectx,recty,area,perimeter] = minboundrect(x,y,metric)
%% 矩形是歪的：斜外接矩形  %%这是最小外接矩形
% minboundrect: Compute the minimal bounding rectangle of points in the plane
% usage: [rectx,recty,area,perimeter] = minboundrect(x,y,metric)
% 最小外接矩形
% arguments: (input)
%  x,y - vectors of points, describing points in the plane as
%        (x,y) pairs. x and y must be the same lengths.
%
%  metric - (OPTIONAL) - single letter character flag which
%        denotes the use of minimal area or perimeter as the
%        metric to be minimized. metric may be either 'a' or 'p',
%        capitalization is ignored. Any other contraction of 'area'
%        or 'perimeter' is also accepted.
%
%        DEFAULT: 'a'    ('area')
%
% arguments: (output)
%  rectx,recty - 5x1 vectors of points that define the minimal
%        bounding rectangle.
%  rectx,recty里面5个数，1-4表示从右上角开始顺时针的4个点的横纵坐标
%  area - (scalar) area of the minimal rect itself.
%
%  perimeter - (scalar) perimeter of the minimal rect as found
%
%
% Note: For those individuals who would prefer the rect with minimum
% perimeter or area, careful testing convinces me that the minimum area
% rect was generally also the minimum perimeter rect on most problems
% (with one class of exceptions). This same testing appeared to verify my
% assumption that the minimum area rect must always contain at least
% one edge of the convex hull. The exception I refer to above is for
% problems when the convex hull is composed of only a few points,
% most likely exactly 3. Here one may see differences between the
% two metrics. My thanks to Roger Stafford for pointing out this
% class of counter-examples.
%
% Thanks are also due to Roger for pointing out a proof that the
% bounding rect must always contain an edge of the convex hull, in
% both the minimal perimeter and area cases.
%
%
%
%
% default for metric
if (nargin<3) || isempty(metric)
metric = 'a';
elseif ~ischar(metric)
error 'metric must be a character flag if it is supplied.'
else
% check for 'a' or 'p'
metric = lower(metric(:)');
ind = strmatch(metric,{'area','perimeter'});
if isempty(ind)
error 'metric does not match either ''area'' or ''perimeter'''
end

% just keep the first letter.
metric = metric(1);
end

% preprocess data
x=x(:);
y=y(:);

% not many error checks to worry about
n = length(x);
if n~=length(y)
error 'x and y must be the same sizes'
end

% if var(x)==0

% start out with the convex hull of the points to
% reduce the problem dramatically. Note that any
% points in the interior of the convex hull are
% never needed, so we drop them.
if n>3

%%%%%%%%%%%%%%%%%%%%%%%%%
if (var(x)== 0|| var(y)==0)
if var(x)== 0
x = [x-1;x(1); x+1 ];
y = [y ;y(1);y];
flag = 1;
else
y = [y-1;y(1); y+1 ];
x = [x ;x(1);x];
flag = 1;
end

else
flag = 0;
%%%%%%%%%%%%%%%%%%%%%%
edges = convhull(x,y); % 'Pp' will silence the warnings

end

% exclude those points inside the hull as not relevant
% also sorts the points into their convex hull as a
% closed polygon

%%%%%%%%%%%%%%%%%%%%
if flag == 0
%%%%%%%%%%%%%%%%%%%%

x = x(edges);
y = y(edges);
%%%%%%%%%%%%%%%%%%
end
%%%%%%%%%%%%%
% probably fewer points now, unless the points are fully convex
nedges = length(x) - 1;
elseif n>1
% n must be 2 or 3
nedges = n;
x(end+1) = x(1);
y(end+1) = y(1);
else
% n must be 0 or 1
nedges = n;
end

% now we must find the bounding rectangle of those
% that remain.

% special case small numbers of points. If we trip any
% of these cases, then we are done, so return.
switch nedges
case 0
% empty begets empty
rectx = [];
recty = [];
area = [];
perimeter = [];
return
case 1
% with one point, the rect is simple.
rectx = repmat(x,1,5);
recty = repmat(y,1,5);
area = 0;
perimeter = 0;
return
case 2
% only two points. also simple.
rectx = x([1 2 2 1 1]);
recty = y([1 2 2 1 1]);
area = 0;
perimeter = 2*sqrt(diff(x).^2 + diff(y).^2);
return
end
% 3 or more points.

% will need a 2x2 rotation matrix through an angle theta
Rmat = @(theta) [cos(theta) sin(theta);-sin(theta) cos(theta)];

% get the angle of each edge of the hull polygon.
ind = 1:(length(x)-1);
edgeangles = atan2(y(ind+1) - y(ind),x(ind+1) - x(ind));
% move the angle into the first quadrant.
edgeangles = unique(mod(edgeangles,pi/2));

% now just check each edge of the hull
nang = length(edgeangles);
area = inf;
perimeter = inf;
met = inf;
xy = [x,y];
for i = 1:nang
% rotate the data through -theta
rot = Rmat(-edgeangles(i));
xyr = xy*rot;
xymin = min(xyr,[],1);
xymax = max(xyr,[],1);

% The area is simple, as is the perimeter
A_i = prod(xymax - xymin);
P_i = 2*sum(xymax-xymin);

if metric=='a'
M_i = A_i;
else
M_i = P_i;
end

% new metric value for the current interval. Is it better?
if M_i<met
% keep this one
met = M_i;
area = A_i;
perimeter = P_i;

rect = [xymin;[xymax(1),xymin(2)];xymax;[xymin(1),xymax(2)];xymin];
rect = rect*rot';
rectx = rect(:,1);
recty = rect(:,2);
end
end
% get the final rect

% all done

end % mainline end

这个是 minboxing.m 联合求斜矩形的长和宽

function [ wid hei ] = minboxing( d_x , d_y )
% minboxing Summary of this function goes here
% Detailed explanation goes here
% 获得对应外接矩形的长和宽(这是歪的矩形：没有平行于坐标轴的矩形)
dd = [d_x, d_y];
dd1 = dd([4 1 2 3],:);
ds = sqrt(sum((dd-dd1).^2,2));  %三角形勾股定理

% 长和宽 width height  斜矩形按大小复值
% wid = min(ds(1:2))
% hei = max(ds(1:2))
% 长和宽 width height  正矩形，平行于坐标轴的矩形，按此方法
wid = ds(2);
hei = ds(1);
end

可以求散点二值图像的最小外接main.m，可以画出单个目标的最小外接main2.m

主函数main.m minboundrect是求白色部分的最小外接，所以可能会用到二值图像取反

[h,w] = size(I);
bw=im2bw(I);%二值化
bw =imcomplement(bw);%二值图像反相
[r,c]=find(bw==1); %返回行号和列号
% 'a'是按面积算的最小矩形，如果按边长用'p'
[rectx,recty,area,perimeter] = minboundrect(c,r,'p');
[wei, hei] = minboxing(rectx(1:end-1),recty(1:end-1)) %显示外接矩形长和宽
figure(2);imshow(bw);hold on
line(rectx,recty,'color','r');%画红色框

主函数main2.m

[h,w] = size(I);
bw=im2bw(I);%二值化
bw =imcomplement(bw);%二值图像反相
[labelpic,num]=bwlabel(bw,8); %num目标个数，labelpic目标索引

[r c]=find(labelpic==1);
% 'a'是按面积算的最小矩形，如果按边长用'p'
[rectx,recty,area,perimeter] = minboundrect(c,r,'p');
% [wei, hei] = minboxing(rectx(1:end-1),recty(1:end-1)) %显示外接矩形长和宽
figure(2);imshow(bw);hold on
line(rectx,recty,'color','r');

[r c]=find(labelpic==2);
% 'a'是按面积算的最小矩形，如果按边长用'p'
[rectx,recty,area,perimeter] = minboundrect(c,r,'p');
figure(2);imshow(bw);hold on
line(rectx,recty,'color','g');

[r c]=find(labelpic==3);
% 'a'是按面积算的最小矩形，如果按边长用'p'
[rectx,recty,area,perimeter] = minboundrect(c,r,'p');
figure(2);imshow(bw);hold on
line(rectx,recty,'color','b');

rectx = 330.2336 937.4036 848.9164 241.7464 330.2336
recty = 80.7667 326.9541 545.1893 299.0019 80.7667
wei =655.1822
hei =235.4923 最后一个目标的数据
rectx = 662 902 902 662 662
recty = 274 274 508 508 274
wei= 240
hei = 234 原本的目标是找到一个函数，可以求散点图最小外接矩形（平行于坐标轴），但这个函数不知道再哪里修改角度。所以另寻办法。
下面是效果，矩形是斜的 注意 matlab 的figure显示图像坐标原点在左上角 展开全文  matlab
• [rectx,recty,area,perimeter] = minboundrect(c,r,‘a’) 其中a表示以面积最小、如果是p的话则是以边长最小 这里有个问题，用minboundrect函数求得的四个点顺序是什么？ 于是做了两张图验证了一下 放上结果： ...

最近在做课程大作业时看到opencv函数cv2.minAreaRect()，
但是想用matlab实现，于是查到了John D’Errico写的matlab实现求最小外接斜矩形函数。（代码贴在最后，仅供学习使用）

[rectx,recty,area,perimeter] = minboundrect(c,r,‘a’)
其中a表示以面积最小、如果是p的话则是以边长最小

这里有个问题，用minboundrect函数求得的四个点顺序是什么？
于是做了两张图验证了一下  放上结果：  可以得到minboundrect函数得到的结果(rectx,recty)是从最上边的点开始，按照顺时针方向索引。

minboundrect代码

function [rectx,recty,area,perimeter] = minboundrect(x,y,metric)
% minboundrect: Compute the minimal bounding rectangle of points in the plane
% usage: [rectx,recty,area,perimeter] = minboundrect(x,y,metric)
%
% arguments: (input)
%  x,y - vectors of points, describing points in the plane as
%        (x,y) pairs. x and y must be the same lengths.
%
%  metric - (OPTIONAL) - single letter character flag which
%        denotes the use of minimal area or perimeter as the
%        metric to be minimized. metric may be either 'a' or 'p',
%        capitalization is ignored. Any other contraction of 'area'
%        or 'perimeter' is also accepted.
%
%        DEFAULT: 'a'    ('area')
%
% arguments: (output)
%  rectx,recty - 5x1 vectors of points that define the minimal
%        bounding rectangle.
%
%  area - (scalar) area of the minimal rect itself.
%
%  perimeter - (scalar) perimeter of the minimal rect as found
%
%
% Note: For those individuals who would prefer the rect with minimum
% perimeter or area, careful testing convinces me that the minimum area
% rect was generally also the minimum perimeter rect on most problems
% (with one class of exceptions). This same testing appeared to verify my
% assumption that the minimum area rect must always contain at least
% one edge of the convex hull. The exception I refer to above is for
% problems when the convex hull is composed of only a few points,
% most likely exactly 3. Here one may see differences between the
% two metrics. My thanks to Roger Stafford for pointing out this
% class of counter-examples.
%
% Thanks are also due to Roger for pointing out a proof that the
% bounding rect must always contain an edge of the convex hull, in
% both the minimal perimeter and area cases.
%
%
% Example usage:
%  x = rand(50000,1);
%  y = rand(50000,1);
%  tic,[rx,ry,area] = minboundrect(x,y);toc
%
%  Elapsed time is 0.105754 seconds.
%
%  [rx,ry]
%  ans =
%      0.99994  -4.2515e-06
%      0.99998      0.99999
%   2.6441e-05            1
%  -5.1673e-06   2.7356e-05
%      0.99994  -4.2515e-06
%
%  area
%  area =
%      0.99994
%
%
%
%
% Author: John D'Errico
% E-mail: woodchips@rochester.rr.com
% Release: 3.0
% Release date: 3/7/07

% default for metric
if (nargin<3) || isempty(metric)
metric = 'a';
elseif ~ischar(metric)
error 'metric must be a character flag if it is supplied.'
else
% check for 'a' or 'p'
metric = lower(metric(:)');
ind = strmatch(metric,{'area','perimeter'});
if isempty(ind)
error 'metric does not match either ''area'' or ''perimeter'''
end

% just keep the first letter.
metric = metric(1);
end

% preprocess data
x=x(:);
y=y(:);

% not many error checks to worry about
n = length(x);
if n~=length(y)
error 'x and y must be the same sizes'
end

% start out with the convex hull of the points to
% reduce the problem dramatically. Note that any
% points in the interior of the convex hull are
% never needed, so we drop them.
if n>3
edges = convhull(x,y);
%edges = convhull(x,y,{'Qt'});  % 'Pp' will silence the warnings

% exclude those points inside the hull as not relevant
% also sorts the points into their convex hull as a
% closed polygon

x = x(edges);
y = y(edges);

% probably fewer points now, unless the points are fully convex
nedges = length(x) - 1;
elseif n>1
% n must be 2 or 3
nedges = n;
x(end+1) = x(1);
y(end+1) = y(1);
else
% n must be 0 or 1
nedges = n;
end

% now we must find the bounding rectangle of those
% that remain.

% special case small numbers of points. If we trip any
% of these cases, then we are done, so return.
switch nedges
case 0
% empty begets empty
rectx = [];
recty = [];
area = [];
perimeter = [];
return
case 1
% with one point, the rect is simple.
rectx = repmat(x,1,5);
recty = repmat(y,1,5);
area = 0;
perimeter = 0;
return
case 2
% only two points. also simple.
rectx = x([1 2 2 1 1]);
recty = y([1 2 2 1 1]);
area = 0;
perimeter = 2*sqrt(diff(x).^2 + diff(y).^2);
return
end
% 3 or more points.

% will need a 2x2 rotation matrix through an angle theta
Rmat = @(theta) [cos(theta) sin(theta);-sin(theta) cos(theta)];

% get the angle of each edge of the hull polygon.
ind = 1:(length(x)-1);
edgeangles = atan2(y(ind+1) - y(ind),x(ind+1) - x(ind));
% move the angle into the first quadrant.
edgeangles = unique(mod(edgeangles,pi/2));

% now just check each edge of the hull
nang = length(edgeangles);
area = inf;
perimeter = inf;
met = inf;
xy = [x,y];
for i = 1:nang
% rotate the data through -theta
rot = Rmat(-edgeangles(i));
xyr = xy*rot;
xymin = min(xyr,[],1);
xymax = max(xyr,[],1);

% The area is simple, as is the perimeter
A_i = prod(xymax - xymin);
P_i = 2*sum(xymax-xymin);

if metric=='a'
M_i = A_i;
else
M_i = P_i;
end

% new metric value for the current interval. Is it better?
if M_i<met
% keep this one
met = M_i;
area = A_i;
perimeter = P_i;

rect = [xymin;[xymax(1),xymin(2)];xymax;[xymin(1),xymax(2)];xymin];
rect = rect*rot';
rectx = rect(:,1);
recty = rect(:,2);
end
end
% get the final rect

% all done

end % mainline end

代码仅供学习

展开全文  图像处理 matlab
• matlab的minboundrect.m 函数

千次阅读 2017-09-07 10:36:40
本来目的是想将每个连通区域用一个斜的box框起来...minboundrect.m 函数function [rectx,recty,area,perimeter] = minboundrect(x,y,metric) % minboundrect: Compute the minimal bounding rectangle of points in the

本来目的是想将每个连通区域用一个斜的box框起来，可是函数貌似针对整幅图片的。
正在改进
minboundrect.m 函数

function [rectx,recty,area,perimeter] = minboundrect(x,y,metric)
% minboundrect: Compute the minimal bounding rectangle of points in the plane
% usage: [rectx,recty,area,perimeter] = minboundrect(x,y,metric)
%
% arguments: (input)
%  x,y - vectors of points, describing points in the plane as
%        (x,y) pairs. x and y must be the same lengths.
%
%  metric - (OPTIONAL) - single letter character flag which
%        denotes the use of minimal area or perimeter as the
%        metric to be minimized. metric may be either 'a' or 'p',
%        capitalization is ignored. Any other contraction of 'area'
%        or 'perimeter' is also accepted.
%
%        DEFAULT: 'a'    ('area')
%
% arguments: (output)
%  rectx,recty - 5x1 vectors of points that define the minimal
%        bounding rectangle.
%
%  area - (scalar) area of the minimal rect itself.
%
%  perimeter - (scalar) perimeter of the minimal rect as found
%
%
% Note: For those individuals who would prefer the rect with minimum
% perimeter or area, careful testing convinces me that the minimum area
% rect was generally also the minimum perimeter rect on most problems
% (with one class of exceptions). This same testing appeared to verify my
% assumption that the minimum area rect must always contain at least
% one edge of the convex hull. The exception I refer to above is for
% problems when the convex hull is composed of only a few points,
% most likely exactly 3. Here one may see differences between the
% two metrics. My thanks to Roger Stafford for pointing out this
% class of counter-examples.
%
% Thanks are also due to Roger for pointing out a proof that the
% bounding rect must always contain an edge of the convex hull, in
% both the minimal perimeter and area cases.
%
%
%
%
% default for metric
if (nargin<3) || isempty(metric)
metric = 'a';
elseif ~ischar(metric)
error 'metric must be a character flag if it is supplied.'
else
% check for 'a' or 'p'
metric = lower(metric(:)');
ind = strmatch(metric,{'area','perimeter'});
if isempty(ind)
error 'metric does not match either ''area'' or ''perimeter'''
end

% just keep the first letter.
metric = metric(1);
end

% preprocess data
x=x(:);
y=y(:);

% not many error checks to worry about
n = length(x);
if n~=length(y)
error 'x and y must be the same sizes'
end

% if var(x)==0

% start out with the convex hull of the points to
% reduce the problem dramatically. Note that any
% points in the interior of the convex hull are
% never needed, so we drop them.
if n>3

%%%%%%%%%%%%%%%%%%%%%%%%%
if (var(x)== 0|| var(y)==0)
if var(x)== 0
x = [x-1;x(1); x+1 ];
y = [y ;y(1);y];
flag = 1;
else
y = [y-1;y(1); y+1 ];
x = [x ;x(1);x];
flag = 1;
end

else
flag = 0;
%%%%%%%%%%%%%%%%%%%%%%
edges = convhull(x,y);  % 'Pp' will silence the warnings

end

% exclude those points inside the hull as not relevant
% also sorts the points into their convex hull as a
% closed polygon

%%%%%%%%%%%%%%%%%%%%
if flag == 0
%%%%%%%%%%%%%%%%%%%%

x = x(edges);
y = y(edges);
%%%%%%%%%%%%%%%%%%
end
%%%%%%%%%%%%%
% probably fewer points now, unless the points are fully convex
nedges = length(x) - 1;
elseif n>1
% n must be 2 or 3
nedges = n;
x(end+1) = x(1);
y(end+1) = y(1);
else
% n must be 0 or 1
nedges = n;
end

% now we must find the bounding rectangle of those
% that remain.

% special case small numbers of points. If we trip any
% of these cases, then we are done, so return.
switch nedges
case 0
% empty begets empty
rectx = [];
recty = [];
area = [];
perimeter = [];
return
case 1
% with one point, the rect is simple.
rectx = repmat(x,1,5);
recty = repmat(y,1,5);
area = 0;
perimeter = 0;
return
case 2
% only two points. also simple.
rectx = x([1 2 2 1 1]);
recty = y([1 2 2 1 1]);
area = 0;
perimeter = 2*sqrt(diff(x).^2 + diff(y).^2);
return
end
% 3 or more points.

% will need a 2x2 rotation matrix through an angle theta
Rmat = @(theta) [cos(theta) sin(theta);-sin(theta) cos(theta)];

% get the angle of each edge of the hull polygon.
ind = 1:(length(x)-1);
edgeangles = atan2(y(ind+1) - y(ind),x(ind+1) - x(ind));
% move the angle into the first quadrant.
edgeangles = unique(mod(edgeangles,pi/2));

% now just check each edge of the hull
nang = length(edgeangles);
area = inf;
perimeter = inf;
met = inf;
xy = [x,y];
for i = 1:nang
% rotate the data through -theta
rot = Rmat(-edgeangles(i));
xyr = xy*rot;
xymin = min(xyr,[],1);
xymax = max(xyr,[],1);

% The area is simple, as is the perimeter
A_i = prod(xymax - xymin);
P_i = 2*sum(xymax-xymin);

if metric=='a'
M_i = A_i;
else
M_i = P_i;
end

% new metric value for the current interval. Is it better?
if M_i<met
% keep this one
met = M_i;
area = A_i;
perimeter = P_i;

rect = [xymin;[xymax(1),xymin(2)];xymax;[xymin(1),xymax(2)];xymin];
rect = rect*rot';
rectx = rect(:,1);
recty = rect(:,2);
end
end
% get the final rect

% all done

end % mainline end

使用实例1.

bw=im2bw(I,0.90);   %将图像二值化
bw = ~bw;%取反
bw = bwareaopen(bw,200,8);  %去除200以下的连通区域
img_reg= regionprops(bw);   %取连通区域，ima_reg默认有质心和box两个参数
bbboooxxx = cat(1,img_reg.BoundingBox); 数组保存所有box

for i=11
rect = [bbboooxxx(i,1),bbboooxxx(i,2),bbboooxxx(i,3),bbboooxxx(i,4)];
cutpic =imcrop(bw,rect);%每次裁一小块
cutpic1 = bwareaopen( cutpic,600,8);%去除附近的其他东西
[r c]=find(cutpic1==1);  %找出边界
[rectx,recty,area,perimeter] = minboundrect(c,r,'a');
% 'a'是按面积算的最小矩形，调用函数
figure
imshow(cutpic);
line(rectx(:),recty(:),'color','r');
end

使用实例2.
这是matlab论坛上一个大佬的，果然好用，虽然不是自己写的感觉乖乖的

function minb(x)
h=figure;
subplot(2, 2, 1); imshow(I);
title('原图');
I1 = rgb2gray(I);
subplot(2, 2, 2); imshow(I1);
title('灰度图');
I2 = medfilt2(I1);
bw1 = im2bw(I2, graythresh(I2));
%bw1=~bw1;
bw1= bwareaopen(bw1,50,4);
subplot(2, 2, 3); imshow(bw1);
B=bwboundaries(bw1);
title('二值图');
[L, n]=bwlabel(bw1, 4);
figure;
imshow(I);
stats = regionprops(L,'all');
Cen = cat(1, stats.Centroid);
hold on;
for i=1:1:n
boundary=B{i};
[rx,ry,area]=minboundrect(boundary(:,2),boundary(:,1));
hold on;
line(rx,ry ,'Color','r','LineWidth',2);
end
str = sprintf('共%d个', n);
title(str, 'Color', 'r');
end
展开全文  matlab 函数
• matlab用于计算图片中图像的最小外接矩阵，用于图片识别 matlab 识别
• 本资源包含求图片中最小矩形的函数（minboundrect.m）本函数已经流传多年，非常经典。还包括一个测试程序以及测试图片！本人已经测试过了，可以运行，希望对您有用！
• Matlab关于二值图像的最小外接矩形问题借用minboundrect 现有函数。function [rectx,recty,area,perimeter] = minboundrect(x,y,metric)% minboundrect: Compute the minimal bounding rectangle of points in the ...
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千次阅读 2020-08-05 10:27:35
%外接矩阵中心点纵坐标 end end 上面Count_rectangle函数需要调用下面minboundrect函数 function [rectx,recty,area,perimeter] = minboundrect(x,y,metric) if (nargin) || isempty(metric) %narginp判断输入函数... matlab 算法 数据挖掘 机器学习 深度学习
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• matlab计算目标最小外接矩形,主要利用minboundrect函数。
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• % minboundrect: Compute the minimal bounding rectangle of points in the plane % usage: [rectx,recty,area,perimeter] = minboundrect(x,y,metric) % % arguments: (input) % x,y - vectors of points, ...
• 这是一份用Minboundrect求最小外接矩形的代码详解，语言是Matlab 机器学习
• 又是计算几何，我感觉最近对计算几何上瘾了。当然，工作上也会用一些，不过工作上一般直接调用boost的geometry库。上次写过最小包围圆，这次是最小包围矩形，要比最小包围圆复杂些。最小包围矩形可不一定是个直立的...
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• 函数mod(a,m)

2016-08-12 14:45:00
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