%形成輸入矢量P和目標(biāo)矢量t的matlab程序
clear?all;
filename?=?dir（‘*.bmp‘）;?????%圖像文件格式
for?k=1:100
????p1=ones（1616）;??????????%縮放后用于存儲(chǔ)的像素空間
????x=imread（filename（k）.name）;??%讀取圖像文件
????bw=im2bw（x0.5）;?????????????%二值化
??[ij]=?find（bw==0）;????????????%尋找數(shù)字所在的像素索引
???imin=min（i）;??????????????????%求取數(shù)字像素占據(jù)空間的最小行索引
???imax=max（i）;??????????????????%求取數(shù)字像素占據(jù)空間的最大行的索引
???jmin=min（j）;??????????????????%求取數(shù)字像素占據(jù)空間的最小列的索引
???jmax=max（j）;??????????????????%求取數(shù)字像素占據(jù)空間的最大列的索引?
???bwl=bw（imin:imaxjmin:jmax）;??%把圖像由39×39縮放為實(shí)際數(shù)字像素所需的空間
???rate=16/max（size（bwl））;???????%求取放大比率
???bwl=imresize（bwlrate）;???????%按比率放大圖像
???[ij]=size（bwl）;??????????????%求取行列數(shù)
??i1=round（（16-i）/2）;????????????%取整
???j1=round（（16-j）/2）;
???p1（i1+1:i1+ij1+1:j1+j）=bwl;??%圖像從右向暫存
???p1=-1.*p1+ones（1616）;????????%將圖像反色
???for?m=0:15????????????????????%樣本特征存于輸入矢量
???????p（m*16+1:（m+1）*16k+1）=p1（1:16m+1）;
???end
switch?k?????????????????????????%對應(yīng)各個(gè)輸入樣本求取對應(yīng)的目標(biāo)矢量
????case{?12345678910}
??????t（k+1）=0;
????case{?11121314151617181920}
??????t（k+1）=1;
????case{?21222324252627282930}
????????t（k+1）=2;?
????case{?31323334353637383940}
?????????t（k+1）=3;
?????case{?41424344454647484950}
?????????t（k+1）=4;
????case{?51525354555657585960}
?????????t（k+1）=5;
???case{?61626364656667686970}
????????t（k+1）=6;
???case{?71727374757677787980}
???????t（k+1）=7;
????case{?81828384858687888990}
???????t（k+1）=8;
?????case{?919293949596979899100}
????????t（k+1）=9;
????end
end
save?swjPT?p?t;???????????????????????????????????%保存輸入矢量和目標(biāo)矢量

?屬性????????????大小?????日期????時(shí)間???名稱
-----------?---------??----------?-----??----

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\0?（10）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\0?（2）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\0?（3）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\0?（4）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\0?（5）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\0?（6）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\0?（7）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\0?（8）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\0?（9）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\0（1）?.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\1?（1）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\1?（10）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\1?（2）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\1?（3）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\1?（4）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\1?（5）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\1?（6）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\1?（7）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\1?（8）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\1?（9）.bmp

?????文件????????374??2009-01-07?19:38??數(shù)字識別代碼\1.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\2?（1）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\2?（10）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\2?（2）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\2?（3）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\2?（4）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\2?（5）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\2?（6）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\2?（7）.bmp

?????文件???????4734??2008-12-22?16:10??數(shù)字識別代碼\2?（8）.bmp

............此處省略84個(gè)文件信息