s_43 = ones(1,128);
s_43(1,:) = 36.3;
% 31.2 32.5 36.3
for i = 1:128
if(z(i)==2)
s_43(i) = s_43(i).*0.9;
else if(z(i)==3)
s_43(i) = s_43(i)*0.8;
else if(z(i)<=6&&z(i)>=4)
s_43(i) = s_43(i)*0.7;
else if(z(i)>6)
s_43(i) = s_43(i)*0.6;
end
end
end
end
end
StudentName = categorical(StudentName);
A2 = unique(StudentName);
z = hist(StudentName, A2);
z4 = ones(1,128);
for i = 1:128
z4(i) = s_41(i)+s_42(i)+s_43(i);
end
x = 1:128;
bar(x,z4);
i = 1;
for a = 1:length(CS3(:))/2
if(LastTime(i,1)==CS3(a,1)) % 判断学生是否为同一人
LastTime(i,4)=CS3(a,2); % 将第几题的时间赋值给LastTime
else
i = i+1; % 若不是该学生,说明是下一个学生
LastTime(i,4)=CS3(a,2); % 将第几题的时间赋值给LastTime
end
end
%创建一个128行6列的矩阵,将第一列赋值
LastTime = ones(128,4)
LastTime = string(LastTime)
LastTime(:,1) = A2
%创建一个128行6列的矩阵,将第一列赋值
ls = ones(128,4)
s = 0.161 %转换后的权重
for i=1:128
ls(i,4) = z(i)*s;
end
StudentName = categorical(CS3(:,1)); % 统一格式化
A2 = unique(StudentName); % 取出每一个单独的
[z,y] = hist(StudentName, A2); % 输出z,y
t43 = ones(length(z),3); %根据z与y构建矩阵
t43 = string(t43); %将数值矩阵转化为字符串矩阵
t43(:,1) = string(y); %将序号赋值
t43(:,2) = string(z); %将提交次数赋值
t43(:,3) = "N"; %将是否迟交钱赋值为N
for i = 1:length(CS3(:,1)) %判断迟交学生编号
if(CS3(i,2)=="Y")
[m,n] = find(t43==CS3(i,1));
t43(m,3) = "Y";
end
end