첫번째
clear all
%park=1; moon=2; etc=3;
voter=100000; constituency=300;
vote=rand(voter, constituency);
vote(vote<0.51)=1; % 51%
vote(vote<0.99)=2; % 48%
vote(vote<1.00)=3; % 1%
%
% The plan에서 설명한 k=1이 나와야 하는 이유
shunt=rand(voter, constituency);
accuracy=0.97;
pass=zeros(voter, constituency);
fail=zeros(voter, constituency);
pass(shunt<=accuracy)=vote(shunt<=accuracy);
fail(shunt>accuracy)=vote(shunt>accuracy);
for i=1:1:constituency
k(i)=(sum(fail(:,i)==1)/sum(fail(:,i)==2))/(sum(pass(:,i)==1)/sum(pass(:,i)==2));
unclassified(i)=sum(pass(:,i)==3)+sum(fail(:,i)==1)+sum(fail(:,i)==2)+sum(fail(:,i)==3);
end
mean_unclassified=mean(unclassified)/voter*100;
mean_k=mean(k);
histogram(k)
2번째
clear all
%park=1; moon=2; etc=3;
voter=100000; constituency=300;
vote=rand(voter, constituency);
vote(vote<0.51)=1; % 51%
vote(vote<0.99)=2; % 48%
vote(vote<1.00)=3; % 1%
% 분류기 정확도에 따른 상황
pass=zeros(voter, constituency);
fail=zeros(voter, constituency);
out=zeros(voter, constituency);
accuracy=0.95;
for c=1:1:constituency
for v=1:1:voter
if rand<accuracy
pass(v,c)=vote(v,c); % 잘 분류된것
else
switch vote(v,c);
case 1 % 잘못 분류된 1번표가
if rand>0.5
fail(v,c)=2;% 2번에 섞여들어가거나
else
out(v,c)=1; % 미분류로 나오거나
end
case 2
if rand>0.5
fail(v,c)=1;% 1번에 섞여들어가거나
else
out(v,c)=2; % 미분류로 나오거나
end
case 3
if rand>0.5
fail(v,c)=1;% 1번에 섞여들어가거나
else
fail(v,c)=2;% 2번에 섞여들어가거나
end
end
end
end
end
classified=pass+fail;
for i=1:1:constituency
k(i)=(sum(out(:,i)==1)/sum(out(:,i)==2))/(sum(classified(:,i)==1)/sum(classified(:,i)==2));
unclassified(i)=sum(classified(:,i)==3)+sum(out(:,i)==1)+sum(out(:,i)==2)+sum(out(:,i)==3);
end
mean_unclassified=mean(unclassified)/voter*100;
mean_k=mean(k);
histogram(k)
no1=sum(sum(classified==1)+sum(out==1))/(voter*constituency)*100;
no2=sum(sum(classified==2)+sum(out==2))/(voter*constituency)*100;
