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資源簡介

LBP(局部二值模式)特征提取,用于人臉識別

資源截圖

代碼片段和文件信息

function?xunlian=extract()
xp=zeros(52059);
n=1;
%提取40個人臉樣本數據(男性20人,女性20人,每人13張圖片)共520幅圖片‘
%對每一幅圖片提取LBP特征(分塊的韋伯中心化局部特征)
for?i=1:9
????for?j=1:9
????????f=imread(strcat(‘訓練樣本\‘‘m-00‘num2str(i)‘-0‘num2str(j)‘.pgm‘));
????????subplot(121);
????????imshow(f);title(‘訓練圖像‘);?drawnow;
????????title(‘訓練圖像‘)
????????[tzxll]=getLBPFea(f);
????????xp(n:)=tzxl;
????????n=n+1;
????end
????for?j=10:13
????????f=imread(strcat(‘訓練樣本\‘‘m-00‘num2str(i)‘-‘num2str(j)‘.pgm‘));
????????subplot(121);
????????imshow(f);title(‘訓練圖像‘);?drawnow;
????????title(‘訓練圖像‘)
????????[tzxll]=getLBPFea(f);
????????xp(n:)=tzxl;
????????n=n+1;?
????end
end
for?i=10:20
????for?j=1:9
??????f=imread(strcat(‘訓練樣本\‘‘m-0‘num2str(i)‘-0‘num2str(j)‘.pgm‘));
???????subplot(121);
???????imshow(f);title(‘訓練圖像‘);?drawnow;
???????title(‘訓練圖像‘)
???????[tzxll]=getLBPFea(f);
???????xp(n:)=tzxl;
???????n=n+1;??
????end
????for?j=10:13
??????f=imread(strcat(‘訓練樣本\‘‘m-0‘num2str(i)‘-‘num2str(j)‘.pgm‘));
????????subplot(121);
????????imshow(f);title(‘訓練圖像‘);?drawnow;
????????title(‘訓練圖像‘)
????????[tzxll]=getLBPFea(f);
????????xp(n:)=tzxl;
????????n=n+1;?
?????end
end

for?i=1:9
????for?j=1:9
????????f=imread(strcat(‘訓練樣本\‘‘w-00‘num2str(i)‘-0‘num2str(j)‘.pgm‘));
????????subplot(121);
????????imshow(f);title(‘訓練圖像‘);?drawnow;
????????title(‘訓練圖像‘)
????????[tzxll]=getLBPFea(f);
????????xp(n:)=tzxl;
????????n=n+1;
????end
????for?j=10:13
????????f=imread(strcat(‘訓練樣本\‘‘w-00‘num2str(i)‘-‘num2str(j)‘.pgm‘));
????????subplot(121);
????????imshow(f);title(‘訓練圖像‘);?drawnow;
????????title(‘訓練圖像‘)
????????[tzxll]=getLBPFea(f);
????????xp(n:)=tzxl;
????????n=n+1;?
????end
end
for?i=10:20
????for?j=1:9
??????f=imread(strcat(‘訓練樣本\‘‘w-0‘num2str(i)‘-0‘num2str(j)‘.pgm‘));
???????subplot(121);
???????imshow(f);title(‘訓練圖像‘);?drawnow;
???????title(‘訓練圖像‘)
???????[tzxll]=getLBPFea(f);
???????xp(n:)=tzxl;
???????n=n+1;??
????end
????for?j=10:13
??????f=imread(strcat(‘訓練樣本\‘‘w-0‘num2str(i)‘-‘num2str(j)‘.pgm‘));
????????subplot(121);
????????imshow(f);title(‘訓練圖像‘);?drawnow;
????????title(‘訓練圖像‘)
????????[tzxll]=getLBPFea(f);
????????xp(n:)=tzxl;
????????n=n+1;?
?????end
end
xunlian=xp;

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????目錄???????????0??2013-08-03?22:43??LBP\
?????文件????????2408??2013-08-03?22:43??LBP\extract.asv
?????文件????????2435??2013-08-03?20:34??LBP\extract.m
?????文件????????2303??2013-08-03?19:11??LBP\getLBPFea.m
?????文件?????????416??2013-06-03?12:58??LBP\IsUniform.m
?????文件?????????171??2013-08-03?20:48??LBP\jingdu.asv
?????文件?????????172??2013-08-03?21:08??LBP\jingdu.m
?????文件?????????194??2013-08-03?13:01??LBP\kfun_rbf.m
?????文件?????????697??2013-08-03?21:37??LBP\makeLBPMap.m
?????文件?????????290??2013-08-03?21:37??LBP\MatLBPMap.mat
?????文件????????1330??2008-08-05?21:10??LBP\multiSVMClassify.m
?????文件????????1767??2013-08-03?13:03??LBP\multiSVMTrain.m
?????文件????????2153??2013-08-03?20:43??LBP\recognition.asv
?????文件????????1983??2013-08-03?20:48??LBP\recognition.m
?????文件??????????42??2013-08-03?22:44??LBP\使用說明.txt
?????目錄???????????0??2013-08-03?20:49??LBP\測試樣本\
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-14.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-15.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-16.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-17.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-18.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-19.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-20.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-21.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-22.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-23.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-24.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-25.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-001-26.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-002-14.pgm
?????文件???????19815??2007-02-23?19:32??LBP\測試樣本\m-002-15.pgm
............此處省略1026個文件信息

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