?? harris角點提取.txt
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1.【求助】matlab中的harris角點提取的問題 Copy to clipboard
Posted by: mog
Posted on: 2005-11-16 15:23
我前幾天從網上找了一個角點提取的源碼,但是執行的時候老是出現這樣的問題:
??? Error using ==> conv2
A and B must be full double matrices.
Error in ==> D:\MATLAB\work\harris.m
On line 51 ==> Ix = conv2(im, dx, 'same'); % Image derivatives
我現在附上源程序,請大家幫忙看看:
% HARRIS - Harris corner detector
%
% Usage: [cim, r, c] = harris(im, sigma, thresh, radius, disp)
%
% Arguments:
% im - image to be processed.
% sigma - standard deviation of smoothing Gaussian. Typical
% values to use might be 1-3.
% thresh - threshold (optional). Try a value ~1000.
% radius - radius of region considered in non-maximal
% suppression (optional). Typical values to use might
% be 1-3.
% disp - optional flag (0 or 1) indicating whether you want
% to display corners overlayed on the original
% image. This can be useful for parameter tuning.
%
% Returns:
% cim - binary image marking corners.
% r - row coordinates of corner points.
% c - column coordinates of corner points.
%
% If thresh and radius are omitted from the argument list 'cim' is returned
% as a raw corner strength image and r and c are returned empty.
這部分是參數設置的說明
function [cim, r, c] = harris(im, sigma, thresh, radius, disp)
error(nargchk(2,5,nargin));
dx = [-1 0 1; -1 0 1; -1 0 1]; % Derivative masks
dy = dx';
Ix = conv2(im, dx, 'same'); % Image derivatives
Iy = conv2(im, dy, 'same');
% Generate Gaussian filter of size 6*sigma (+/- 3sigma) and of
% minimum size 1x1.
g = fspecial('gaussian',max(1,fix(6*sigma)), sigma);
Ix2 = conv2(Ix.^2, g, 'same'); % Smoothed squared image derivatives
Iy2 = conv2(Iy.^2, g, 'same');
Ixy = conv2(Ix.*Iy, g, 'same');
% Compute the Harris corner measure. Note that there are two measures
% that can be calculated. I prefer the first one below as given by
% Nobel in her thesis (reference above). The second one (commented out)
% requires setting a parameter, it is commonly suggested that k=0.04 - I
% find this a bit arbitrary and unsatisfactory.
cim = (Ix2.*Iy2 - Ixy.^2)./(Ix2 + Iy2 + eps); % My preferred measure.
% k = 0.04;
% cim = (Ix2.*Iy2 - Ixy.^2) - k*(Ix2 + Iy2).^2; % Original Harris measure.
if nargin > 2 % We should perform nonmaximal suppression and threshold
% Extract local maxima by performing a grey scale morphological
% dilation and then finding points in the corner strength image that
% match the dilated image and are also greater than the threshold.
sze = 2*radius+1; % Size of mask.
mx = ordfilt2(cim,sze^2,ones(sze)); % Grey-scale dilate.
cim = (cim==mx)&(cim>thresh); % Find maxima.
[r,c] = find(cim); % Find row,col coords.
if nargin==5 & disp % overlay corners on original image
show(im,1), hold on
plot(c,r,'r+'), title('corners detected');
end
else % leave cim as a corner strength image and make r and c empty.
r = []; c = [];
end
執行程序部分.
注意話題分類
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2.Re:【求助】matlab中的harris角點提取的問題 [Re: mog] Copy to clipboard
Posted by: talon
Posted on: 2005-11-17 09:32
要將圖像im轉換成double類型的,不知你轉換了沒有,我運行了上面的程序,沒有出錯。
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