tic
clear
str=‘F:\matlab\an\bin\tupian\目標圖片\正樣本\‘;
str1=‘F:\matlab\an\bin\tupian\目標圖片\負樣本\‘;
SAVE_train=[72?9];%設定目標矩陣?以作訓練樣本；62個樣本?每個樣本7個特征
i_SAVE_train=1;
SAVE_train_label=[1?0?1?0?1?0?1?0?1?0?1?0?1?0?1?0?1?0?...
????1?0?1?0?1?0?1?0?1?0?1?0?1?0?1?0?1?0?1?0?1?0?1?0?1?...
????0?1?0?1?0?1?0?1?0?1?0?1?0?1?0?1?0?1?0?...
????1?0?1?0?1?0?1?0?1?0?];
%SAVE_train_label=double（SAVE_train_label）;
SAVE_train_label=SAVE_train_label‘;%將SAVE_train_label轉置為列向量?以符合SVM數據訓練的格式要求
%[mm?nn]=size（SAVE_train_label）;disp（mm）;
test_train_label=1;%假設所有的測試樣本均為1類并根據判斷的正確率來選擇最終屬于哪一類

%%?求目標的特征
for?i0=1:36
????I=imread（[strnum2str（i0）‘.jpg‘]）;?%依次讀取每一幅圖像
????I1=imread（[str1num2str（i0）‘.jpg‘]）;?%依次讀取每一幅圖像
????I=imresize（I[80?160]）;%將樣本化為統一大小
????I1=imresize（I1[80?160]）;%將樣本化為統一大小
????%figure;imshow（I）;
????%I=double（rgb2gray（I））;
????%I1=double（rgb2gray（I1））;
????%I=adapthisteq（I）;
????%I1=adapthisteq（I1）;
????
????%%?求不變矩，作為目標的特征
????%hu=M_GetImageHuJu（I）;
????%hu=Hu_MV_Nicolas（I）;
????I=rgb2gray（I）;
????I=imadjust（I[0?1][1?0]）;%取反操作
????I=imadjust（I）;%增強對比度
????%I=edge（I‘sobel‘）;
????%figure;imshow（I）;
?????
%?????se90=strel（‘line‘390）;
%?????se0=strel（‘line‘30）;
%?????se?=?strel（‘disk‘10）;
%?????I=imclose（Ise）;
????%figure;imshow（I）;
????
????[m0?n0]=size（I）;
????for?i1=1:m0
???????for?j1=1:n0
????????if?I（i1j1）>=150
????????????I（i1j1）=255;
????????else?I（i1j1）=1;
????????end
???????end
????end
????%figure;imshow（I）;
%?????I=edge（I‘canny‘）;%figure;imshow（I）;
%?????I=imfill（I‘holes‘）;
????
???%%?腐蝕處理
????se90=strel（‘line‘390）;
????se0=strel（‘line‘30）;
????I=imerode（I[se90?se0]）;
????I=imerode（I[se90?se0]）;%處理兩次
????%figure;imshow（I）;
????
????image0=double（I）;
????m00=sum（sum（image0））;
????m10=0;
????m01=0;
????[rowcol]=size（image0）;
????for?i=1:row
????????for?j=1:col
????????m10=m10+i*image0（ij）;
????????m01=m01+j*image0（ij）;
????????end
????end
????u10=m10/m00;
????u01=m01/m00;
????
????n20?=?0;n02?=?0;n11?=?0;n30?=?0;n12?=?0;n21?=?0;n03?=?0;
????for?i=1:row
????????for?j=1:col
????????n20=n20+i^2*image0（ij）;
????????n02=n02+j^2*image0（ij）;
????????n11=n11+i*j*image0（ij）;
????????n30=n30+i^3*image0（ij）;
????????n03=n03+j^3*image0（ij）;
????????n12=n12+i*j^2*image0（ij）;
????????n21=n21+i^2*j*image0（ij）;
????????end
????end
????n20=n20/m00^2;
????n02=n02/m00^2;
????n11=n11/m00^2;
????n30=n30/m00^2.5;
????n03=n03/m00^2.5;
????n12=n12/m00^2.5;
????n21=n21/m00^2.5;
????????

????h1?=?n20?+?n02;??????????????????????h2?=?（n20-n02）^2?+?4*（n11）^2;
????h3?=?（n30-3*n12）^2?+?（3*n21-n03）^2;??h4?=?（n30+n12）^2?+?（n21+n03）^2;
????h5?=?（n30-3*n12）*（n30+n12）*（（n30+n12）^2-3*（n21+n03）^2）+（3*n21-n03）*（n21+n03）*（3*（n30+n12）^2-（n21+n03）^2）;
????h6?=?（n20-n02）*（（n30+n12）^2-（n21+n03）^2）+4*n11*（n30+n12）*（n21+n03）;
????h7?=?（3*n21-n03）*（n30+n12）*（（n30+n12）^2-3*（n21+n03）^2）+（3*n12-n30）*（n21+n03）*（3*（n30+n12）^2-（n21+n03）^2）;
????h8=mean（image0（:））;%求均值
????h9=st

?屬性????????????大小?????日期????時間???名稱
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?????文件??????12004??2013-08-13?23:18??first.m

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????????????????12004????????????????????1