调用Tstart.m，EVtype.m，SOC.m，Lrem.m脚本，生成200辆车的数据参数。随机从200辆车中抽取一辆车,定义充电方案Project，包括3个参数：1.充电类型(Ch_type)：Ch_no,Ch_slow,Ch_fast；2.充电成本(Ch_cost)：Ch_cost=E*Ch_charge/Pt_price；3.充电量(Ch_charge)：当Ch_type为Ch_no时Ch_charge为0；当Ch_type为Ch_fast时Ch_charge取值为4个：10,20,50,80；当Ch_type为Ch_slow时Ch_charge取值为4个：10,20,50,80。其中，当Ch_type为Ch_no时Pt_price为0,当Ch_type为Ch_slow时Pt_price为2.37，当Ch_type为Ch_fast时Pt_price为5.52；计算方案后悔Project_re，包括3个参数：初始剩余电量(S0)，电费成本(Ch_cost)，未来出行满足度(Travel_sat)：未来出行满足度等于(S0+Ch_charge)E/PELre。然后编写函数计算不同方案的后悔值R，计算公式1为：Ri⟶jm=ln(1+exp(βm(xjm-xim)))，其中，S0的β值为0.2，Ch_cost的β值为0.3，Travel_sat的β值为-0.1；计算公式2为：Ri=∑j≠i,j=1J∑m=1MRi⟶jm，计算公式3为Pi=exp(-Rj)/∑j=1Jexp(-Rj)。最后，循环输出这辆车的所有充电方案Project，方案后悔Project_re、对应的后悔值Ri和选择概率pi。用matlab代码描述这一过程。

以下是matlab代码描述这一过程：

% 调用脚本生成200辆车的数据参数
Tstart; EVtype; SOC; Lrem;
% 随机抽取一辆车
car_index = randi(200);
car_data = [Tstart(car_index) EVtype(car_index) SOC(car_index) Lrem(car_index)];
% 定义充电方案Project
Ch_type = ["Ch_no", "Ch_slow", "Ch_fast"];
Ch_cost = [0, 2.37, 5.52];
Ch_charge_slow = [10, 20, 50, 80];
Ch_charge_fast = [10, 20, 50, 80];
Project = [];
for i = 1:numel(Ch_type)
    if Ch_type(i) == "Ch_no"
        Project(i,:) = [Ch_type(i), Ch_cost(i), 0];
    elseif Ch_type(i) == "Ch_slow"
        for j = 1:numel(Ch_charge_slow)
            Project(end+1,:) = [Ch_type(i), Ch_cost(i), Ch_charge_slow(j)];
        end
    elseif Ch_type(i) == "Ch_fast"
        for j = 1:numel(Ch_charge_fast)
            Project(end+1,:) = [Ch_type(i), Ch_cost(i), Ch_charge_fast(j)];
        end
    end
end
% 计算方案后悔Project_re
beta_S0 = 0.2;
beta_Ch_cost = 0.3;
beta_Travel_sat = -0.1;
Project_re = [];
for i = 1:size(Project,1)
    S0 = car_data(3);
    Ch_charge = Project(i,3);
    Ch_type = Project(i,1);
    Pt_price = Ch_cost(find(Ch_type==Ch_type));
    E = car_data(4);
    PELre = 50;
    Travel_sat = (S0+Ch_charge)*E/PELre;
    Ch_cost = E*Ch_charge/Pt_price;
    Project_re(end+1,:) = [S0, Ch_cost, Travel_sat];
end
% 计算不同方案的后悔值R
beta = [beta_S0, beta_Ch_cost, beta_Travel_sat];
R = zeros(size(Project,1),1);
for i = 1:size(Project,1)
    xj = Project_re(i,:);
    for j = 1:size(Project,1)
        if j ~= i
            xi = Project_re(j,:);
            rijm = log(1+exp(beta.*(xj-xi)));
            R(i) = R(i) + sum(rijm);
        end
    end
end
% 计算选择概率pi
P = exp(-R)./sum(exp(-R));
% 输出结果
for i = 1:size(Project,1)
    fprintf('方案%d：\n', i);
    fprintf('充电类型：%s\n', Project(i,1));
    fprintf('充电成本：%.2f\n', Project(i,2));
    fprintf('充电量：%d\n', Project(i,3));
    fprintf('方案后悔Project_re：S0=%.2f，Ch_cost=%.2f，Travel_sat=%.2f\n', Project_re(i,1), Project_re(i,2), Project_re(i,3));
    fprintf('对应的后悔值Ri：%.2f\n', R(i));
    fprintf('选择概率pi：%.2f\n\n', P(i));
end