Matlab 中并沒有發現最小外接矩形的代碼，為了方便

下面提供最小外接矩形的代碼：

注：這個函數是源于網上找到的代碼的改進版，原版不能檢測水平線或者垂直線

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.

%

%

% See also: minboundcircle, minboundtri, minboundsphere

%

%

% 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

% 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

當然這段代碼并沒有獲取到外接矩形的長和寬，下面我在寫一個函數，就可以獲得對應外接矩形的長和寬

function [ wid hei ] = minboxing( d_x , d_y )

%minboxing Summary of this function goes here

% Detailed explanation goes here

dd = [d_x, d_y];

dd1 = dd([4 1 2 3],:);

ds = sqrt(sum((dd-dd1).^2,2));

wid = min(ds(1:2));

hei = max(ds(1:2));

end

這里默認為較短的距離為寬，較長的距離為長。

調用代碼如下：注(dataX, dataY為需要計算最小外接矩形的數據。)

[recty,rectx,area,perimeter] = minboundrect(dataX, dataY);

[wei hei] = minboxing(rectx(1:end-1),recty(1:end-1));

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