【图像检测】基于形态学实现叶子面积周长测量matlab源码

一、二值图像的腐蚀、膨胀、开、闭操作

 

二、灰度图像的腐蚀、膨胀、开、闭操作

``` function [rectx,recty,area,perimeter,long,width] = 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(长方形). % % 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 % % Author: John D'Errico % E-mail: [email protected] % Release: 3.0 % Release date: 3/7/07

% default for metric

% global long

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 = []; long = []; width = []; return case 1 % with one point, the rect is simple. rectx = repmat(x,1,5); recty = repmat(y,1,5); area = 0; perimeter = 0; long =0; width =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); long=perimeter; width =perimeter; 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; long=inf; width =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 Ai = prod(xymax - xymin); Pi = 2*sum(xymax-xymin);

if metric=='a' Mi = Ai; else Mi = Pi; end

% new metric value for the current interval. Is it better? if Mii; area = Ai; perimeter = Pi;

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

L=sqrt(diff(rectx).^2 + diff(recty).^2); long=max(L); width=min(L);

% all done

end % mainline end ```