数学建模结课题目

 题目

这学期选修了数学建模，期末考查题挺有意思的。

 

思路

数据预处理（指标设置）

处理方式：我设置了两个指标，一个考勤率，一个作业完成度。
考勤率：

缺勤：0分
早退：1分（未上满90分钟）
完成：2分
缺勤率=(各课程考勤总分数/2*7)*100%

完成度：

未提交：0分
D：1分
C：2分
B：3分：
A：4分
A+：5分
其他：3分
得到作业的总分数
然后，判断是否迟交（在下一次上课前必须提交，即时间差是否大于14）
如果迟交：则分数-1，其他0分，得到最终作业分数。
完成度=最终作业总分/35*100%

 

综合评价

可以参考之前写的两篇博客：

https://www.cnblogs.com/Mayfly-nymph/p/13096966.html

https://www.cnblogs.com/Mayfly-nymph/p/13096812.html

 

代码与结果

预处理代码

import pandas as pd
import numpy as np

df = pd.read_excel("C:\\Users\\10245\\Desktop\\data.xlsx")
a = []
tmp = df.iloc[:,1:8]
for i in range(7):
    a.append(list(tmp.iloc[:,i]))
for i in range(len(a)):
    for j in range(len(a[0])):
        if(type(a[i][j]) != int):
            a[i][j] = 0
        if(a[i][j] < 90):
            a[i][j] = 1
        if(a[i][j] >= 90):
            a[i][j] = 2
data0 = np.sum([x for x in a], axis=0)
idata = [round((x / 14)*100,2) for x in data0]
tmp1 = df.iloc[:,8::2]
b = []
for i in range(7):
    b.append(list(tmp1.iloc[:,i]))
for i in range(len(b)):
    for j in range(len(b[0])):
        if(b[i][j] == '未提交'):
            b[i][j] = 0
        elif(b[i][j] == 'D'):
            b[i][j] = 1
        elif(b[i][j] == 'C'):
            b[i][j] = 2
        elif(b[i][j] == 'B'):
            b[i][j] = 3
        elif(b[i][j] == 'A'):
            b[i][j] = 4
        elif(b[i][j] == 'A+'):
            b[i][j] = 5
        else:
            b[i][j] = 3
tmp2 = df.iloc[:,9::2]
c = []
for i in range(7):
    c.append(list(tmp2.iloc[:,i]))
limit_time = [[3,4],[3,18],[4,1],[4,15],[4,29],[5,13],[5,20]]
month_day = [31,28,31,30,31,30,31,31,30,31,30,31,0]
d = []
for i in range(len(b)):
    for j in range(len(b[0])):
        index = limit_time[i][0] if c[i][j].month - limit_time[i][0] > 0 else -1
        days = month_day[index] + (c[i][j].day - limit_time[i][1])
        if(days <= 14):
            d.append(0)
        elif(days > 14):
            d.append(-1)
        else:
            d.append(0)
e = []
for i in range(7):
    e.append(list(np.array(b[i]) - np.array(d[i])))
data1 = np.sum([x for x in e], axis=0)
jdata = [round((x / 35)*100,2) for x in data1]
# data = [idata,jdata]
# s= list(df.iloc[:,0])
# s1 = '\n'.join(s)
# with open(r'C:\Users\10245\Desktop\results.xls','w') as f:
#    f.write(s1)
# with open(r'C:\Users\10245\Desktop\results.xls','a') as f:
#    f.write('\t')
#    for i in range(len(data)):
#        for j in range(len(data[i])):
#            f.write(str(data[i][j]))
#            f.write('\n')
#        f.write('\n')

 

熵权法和优劣解距离法代码

%% 数据读取
clear,clc
load data.mat
%% 
[n,m] = size(datas_matrix);
%% 矩阵标准化
% datas_S_matrix = datas_matrix ./ repmat(sum(datas_matrix.*datas_matrix) .^ 0.5, n, 1)
for i = 1:m
    tmp = datas_matrix(:,i)
    datas_S_matrix(:,i) = (tmp - min(tmp))/(max(tmp) - min(tmp));
end
%%
model = ["编号1",  "编号2",  "编号3",  "编号4",  "编号5",  "编号6",  "编号7",  "编号8",  "编号9",  "编号10",  "编号11",  "编号12",  "编号13",  "编号14",  "编号15",  "编号16",  "编号17",  "编号18",  "编号19",  "编号20",  "编号21",  "编号22",  "编号23",  "编号24",  "编号25",  "编号26",  "编号27",  "编号28",  "编号29",  "编号30",  "编号31",  "编号32",  "编号33",  "编号34",  "编号35",  "编号36",  "编号37",  "编号38",  "编号39",  "编号40",  "编号41",  "编号42",  "编号43",  "编号44",  "编号45",  "编号46",  "编号47",  "编号48",  "编号49",  "编号50",  "编号51",  "编号52",  "编号53",  "编号54",  "编号55",  "编号56",  "编号57",  "编号58",  "编号59",  "编号60",  "编号61",  "编号62",  "编号63",  "编号64",  "编号65",  "编号66",  "编号67",  "编号68",  "编号69",  "编号70",  "编号71",  "编号72",  "编号73",  "编号74",  "编号75",  "编号76",  "编号77",  "编号78",  "编号79",  "编号80",  "编号81",  "编号82",  "编号83",  "编号84",  "编号85",  "编号86",  "编号87",  "编号88",  "编号89",  "编号90",  "编号91",  "编号92",  "编号93",  "编号94",  "编号95",  "编号96",  "编号97",  "编号98",  "编号99",  "编号100",  "编号101",  "编号102",  "编号103",  "编号104",  "编号105",  "编号106",  "编号107",  "编号108",  "编号109",  "编号110",  "编号111",  "编号112"];
%% 熵权法
p = datas_S_matrix./sum(datas_S_matrix);
k = 1/log(n);
r = zeros(n,m);
for i = 1:n
    for j = 1:m
       if p(i,j) == 0
           r(i,j) = 0;
       else
           r(i,j) = log(p(i,j));
       end
    end
end
e = -k*sum(p.*r,1);
d = ones(1,m)-e
weight = d./sum(d)
score = sum(weight.*datas_S_matrix,2);
results1 = 0 + (100-0)/(max(score)-min(score)).*(score - min(score));
[sorted_score,index] = sort(results1 ,'descend');
a = [sorted_score,index]
rivers1 = [];
for i = 1:n
    rivers1 = [rivers1;model(index(i))];
end
s = [rivers1,sorted_score]


%% 优劣解距离法
t = ones(1,m) ./ m ; 
max_dis = sum([(datas_S_matrix - repmat(max(datas_S_matrix),n,1)) .^ 2 ] .* repmat(t,n,1) ,2) .^ 0.5;
min_dis = sum([(datas_S_matrix - repmat(min(datas_S_matrix),n,1)) .^ 2 ] .* repmat(t,n,1) ,2) .^ 0.5;
S = min_dis ./ (max_dis+min_dis); 
% results = S / sum(S);
results2 = 0 + (100-0)/(max(S)-min(S)).*(S - min(S));
[sorted_results,index] = sort(results2 ,'descend');
rivers2=[];
for i = 1:n
    rivers2 = [rivers2;model(index(i))];
end
R = [rivers2,sorted_results];
%% 绘图
subplot(2,1,1)
bar(sorted_score);
title('熵权法综合评价排行图')
set(gca,'XTick',1:112)
set(gca, 'xticklabel',{rivers1{1:112}});
xlabel('数据代码');
ylabel('综合评分')
subplot(2,1,2)
bar(sorted_results);
title('优劣解距离法')
set(gca,'XTick',1:112)
set(gca, 'xticklabel',{rivers2{1:112}});
%% 保存到文件
% xlswrite('output.xls',s,'Sheet1');
% xlswrite('output.xls',R,'Sheet2');

 

结果

 

附件

https://wwa.lanzous.com/iuYt2ejrzbg