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資源簡(jiǎn)介

用MatLab編寫的正態(tài)分布模式下的貝葉斯分類器。很多模式識(shí)別的課設(shè)就有相關(guān)題目。不僅有樣本分類而且會(huì)在二維坐標(biāo)系下畫出,正確分類與錯(cuò)誤分類的點(diǎn)都會(huì)有不同的標(biāo)志。

資源截圖

代碼片段和文件信息

clear;?clc;
N?=?30
C1?=[3.50;03.5];
C2?=?[3.50;03.5];
M1?=?[-22];
M2?=?[2-2];
w1=mvnrnd(M1C130);???%產(chǎn)生隨即數(shù)
w2?=mvnrnd(M2C230);
f1?=?0;
f2?=?0;
M1?=?[00];
M2?=?[00];
C1?=?[00;00];
C2?=?[00;00];
mm?=?max(abs(w1));???????%控制樣本特征值在[-55]之間
mm?=?max(mm);
w1?=?w1/mm*5
mm?=?max(abs(w2));
mm?=?max(mm);
w2?=?w2/mm*5
for?i?=?1:N
????M1?=?M1?+?w1(i:);
????M2?=?M2?+?w2(i:);
end
M1?=?M1/N
M2?=?M2/N
for?i=1:N
????C1?=?C1?+?transpose(w1(i:))*w1(i:)-transpose(M1)*M1;
????C2?=?C2?+?transpose(w2(i:))*w2(i:)-transpose(M2)*M2;
end
C1?=?C1/N
C2?=?C2/N


figure(1);??????????????????????????????????????????????????????????????????????%以樣本的第一個(gè)特征值進(jìn)行分類
axis([-55-55]);
hold?on
grid?on
f1=0;
f2=0;
for?i=1:N
???t=(w1(i1)-M1(1))*(1/C1(11))*(w1(i1)-M1(1));
???d1?=?log(0.5)-0.5*log(C1(11))-0.5*t;
???t=(w1(i1)-M2(1))*(1/C2(11))*(w1(i1)-M2(1));
???d2?=?log(0.5)-0.5*log(C2(11))-0.5*t;
???if?d1???????plot(w1(i1)w1(i2)‘rx‘)
???????f1?=?f1+1;
???else
???????plot(w1(i1)w1(i2)‘ro‘)
???end
???t=(w2(i1)-M1(1))*(1/C1(11))*(w2(i1)-M1(1));
???d1?=?log(0.5)-0.5*log(C1(11))-0.5*t;
???t=(w2(i1)-M2(1))*(1/C2(11))*(w2(i1)-M2(1));
???d2?=?log(0.5)-0.5*log(C2(11))-0.5*t;
???if?d2???????plot(w2(i1)w2(i2)‘x‘)
???????f2?=?f2+1;
???else
???????plot(w2(i1)w2(i2)‘o‘)
???end
end
p1?=?f1/N*100
p2?=?f2/N*100

figure(2);??????????????????????????????????????????????????????????????????????????????%以樣本的第二個(gè)特征值進(jìn)行分類
axis([-55-55]);
hold?on
grid?on
f1=0;
f2=0;
for?i=1:N
???t=(w1(i2)-M1(2))*(1/C1(22))*(w1(i2)-M1(2));
???d1?=?log(0.5)-0.5*log(C1(22))-0.5*t;
???t=(w1(i2)-M2(2))*(1/C2(22))*(w1(i2)-M2(2));
???d2?=?log(0.5)-0.5*log(C2(22))-0.5*t;
???if?d1???????plot(w1(i1)w1(i2)‘rx‘)
???????f1?=?f1+1;
???else
???????plot(w1(i1)w1(i2)‘ro‘)
???end
???t=(w2(i2)-M1(2))*(1/C1(22))*(w2(i2)-M1(2));
???d1?=?log(0.5)-0.5*log(C1(11))-0.5*t;
???t=(w2(i2)-M2(2))*(1/C2(22))*(w2(i2)-M2(2));
???d2?=?log(0.5)-0.5*log(C2(22))-0.5*t;
???if?d2???????plot(w2(i1)w2(i2)‘x‘)
???????f2?=?f2+1;
???else
???????plot(w2(i1)w2(i2)‘o‘)
???end
end
p1?=?f1/N*100
p2?=?f2/N*100


figure(3)???????????????????????????????????????????????????????????????????????????%以樣本的兩個(gè)特征值進(jìn)行分類
axis([-55-55]);
f1?=?0;
f2?=?0;
hold?on
grid?on
for?i=1:N
???t=(w1(i:)-M1)*inv(C1)*transpose(w1(i:)-M1);
???d1?=?log(0.5)-0.5*log(det(C1))-0.5*t;
???t=(w1(i:)-M2)*inv(C2)*transpose(w1(i:)-M2);
???d2=?log(0.5)-0.5*log(det(C2))-0.5*t;
???if?d1???????plot(w1(i1)w1(i2)‘rx‘)
???????f1?=?f1+1;
???else
???????plot(w1(i1)w1(i2)‘ro‘)
???end
???t=(w2(i:)-M1)*inv(C1)*transpose(w2(i:)-M1);
???d1?=?log(0.5)-0.5*log(det(C1))-0.5*t;
???t=(w2(i:)-M2)*inv(C2)*transpose(w2(i:)-M2);
???d2=?log(0.5)-0.5*log(det(C2))-0.5*t;
???if?d2???????plot(w2(i1)w2(i2)‘x‘)
???????f2?=?f2+1;
???else
???????plot(w2(i1)w2(i2)‘o‘)
???end
end
p1?=?f1/N*10

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

?????文件???????3316??2012-12-20?18:43??BYS.m

-----------?---------??----------?-----??----

?????????????????3316????????????????????1


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