資源簡介
圖像低頻用平均法進行融合,高頻用發差法和最大系數法融合,有源代碼,有圖像,可正常運行
代碼片段和文件信息
function?X=matfusion(ABe1e2)?
%附錄2;要先定義一個XH3嗎?或怎么打?矢量還是標量
%對矩陣AB按照邊緣檢測融合
L=size(A);
%零延拓
for?i=1:L(1)
??for?j=1:L(2)
????A1(1:)=0;
????A1(:1)=0;
????A1(L(1)+2:)=0;
????A1(:L(2)+2)=0;
????A1(i+1j+1)=A(ij);
??end
end?
for?i=1:L(1)
??for?j=1:L(2)
????B1(1:)=0;
????B1(:1)=0;
????B1(L(1)+2:)=0;
????B1(:L(2)+2)=0;
????B1(i+1j+1)=B(ij);
??end
end
%矩陣融合
for?i=2:1:(L(1)+1)
??for?j=2:1:(L(2)+1)
????if?e1(i-1j-1)==1|?e2(i-1j-1)==1
?????VA1=var([A1(i-1j-1)??A1(i-1j)??A1(i-1j+1)???A1(ij-1)??A1(ij)??A1(ij+1)??A1(i+1j-1)??A1(i+1j)??A1(i+1j+1)]);
??????VB1=var([B1(i-1j-1)??B1(i-1j)??B1(i-1j+1)???B1(ij-1)??B1(ij)??B1(ij+1)??B1(i+1j-1)??B1(i+1j)??B1(i+1j+1)]);
??????M=max([VA1??VB1]);
????????if??VA1>=M???%多了個‘;’
????????XH3(i-1j-1)=A1(ij);
????????else
????????XH3(i-1j-1)=B1(ij);
????????end
??????else
??????a1=abs(A1(ij));
??????b1=abs(B1(ij));
??????M=max([a1?b1]);
????????if?a1>=M
????????XH3(i-1j-1)=A1(ij);
???
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件??????22199??2009-02-14?21:52??111\clockA.jpg
?????文件??????19882??2009-02-14?21:52??111\clockB.jpg
?????文件???????8395??2011-06-07?21:15??111\hs_err_pid5956.log
?????文件???????1128??2011-05-20?08:38??111\matfusion.m
?????文件????????277??2011-05-25?11:29??111\nmaxabs.m
?????文件???????1322??2011-05-25?11:39??111\ronghe.m
?????文件???????1126??2011-05-25?11:37??111\ronghe2.m
?????文件???????1087??2011-05-25?11:49??111\ronghe3.m
?????文件????????772??2011-05-20?08:59??111\varmatfusion.m
?????文件????1006134??2011-05-20?09:16??111\最大系數與方差法.bmp
?????文件????1017654??2011-05-20?09:14??111\耗時的方差法.bmp
?????目錄??????????0??2011-06-07?21:15??111
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??????????????2079976????????????????????12
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