|
推荐一个学习的地方，详细易懂：
https://blog.csdn.net/lemontree_____/article/details/119181395?utm_source=app&app_version=4.12.0&code=app_1562916241&uLinkId=usr1mkqgl919blen
topsis法python代码实现，推荐：
https://blog.csdn.net/weixin_41799019/article/details/97611462
基于上述两个博客，对暑假建模训练题给出了基于熵权法的topsis法的python代码实现：
import numpy as np
import xlrd
import pandas as pd
def read(file):
wb = xlrd.open_workbook(filename=file) #打开文件
sheet = wb.sheet_by_index(0) #通过索引获取表格
rows = sheet.nrows # 获取行数
all_content = [] #存放读取的数据
for j in range(1, 10): #取第1~第9列对的数据
temp = []
for i in range(1,rows) :
cell = sheet.cell_value(i, j) #获取数据
temp.append(cell)
all_content.append(temp) #按列添加到结果集中
temp = []
return np.array(all_content)
def entropy(data0):
# 返回每个样本的指数
# 样本数，指标个数
n, m = np.shape(data0)
# 一行一个样本，一列一个指标
# 下面是归一化
maxium = np.max(data0, axis=1)
minium = np.min(data0, axis=1)
data1=data0.copy()
for i in range(0,9):
data0[i]= (data0[i] - minium[i]) * 1.0 / (maxium[i] - minium[i])
##计算第j项指标，第i个样本占该指标的比重
sumzb = np.sum(data0, axis=1)
for i in range(0,9):
data0[i] = data0[i] / sumzb[i]
# 对ln0处理
a = data0* 1.0
a[np.where(data0== 0)] = 0.0001
# #计算每个指标的熵
e = (-1.0 / np.log(m)) * np.sum(data0* np.log(a), axis=1)
# #计算权重
w = (1 - e) / np.sum(1 - e)
print(w)
return w
# 极小型指标 -> 极大型指标
def dataDirection_1(datas):
return (np.max(datas) - datas) # 套公式
# 中间型指标 -> 极大型指标
def dataDirection_2(datas, x_best):
temp_datas = datas - x_best
M = np.max(abs(temp_datas))
answer_datas = 1 - abs(datas - x_best) / M # 套公式
return answer_datas
# 区间型指标 -> 极大型指标
def dataDirection_3(datas, x_min, x_max):
M = max(x_min - np.min(datas), np.max(datas) - x_max)
answer_list = []
for i in datas:
if (i < x_min):
answer_list.append(1 - (x_min - i) / M) # 套公式
elif (x_min <= i <= x_max):
answer_list.append(1)
else:
answer_list.append(1 - (i - x_max) / M)
return np.array(answer_list)
def temp2(datas):
K = np.power(np.sum(pow(datas,2),axis =1),0.5)
for i in range(0,K.size):
for j in range(0,datas[i].size):
datas[i,j] = datas[i,j] / K[i] #套用矩阵标准化的公式
return datas
def temp3(answer,w):
list_max=[]
for i in answer:
list_max.append(np.max(i[:]))#获取每一列的最大值
list_max=np.array(list_max)
list_min =[]
for i in answer:
list_min.append(np.min(i[:]))#获取每一列的最小值
list_min=np.array(list_min)
max_list = [] #存放第i个评价对象与最大值的距离
min_list = [] #存放第i个评价对象与最小值的距离
answer_list=[] #存放评价对象的未归一化得分
for k in range(0,np.size(answer,axis = 1)): #遍历每一列数据
max_sum = 0
min_sum = 0
for q in range(0,9): #有四个指标
max_sum +=w[q]*np.power(answer[q,k]-list_max[q],2) #按每一列计算Di+
min_sum +=w[q]*np.power(answer[q,k]-list_min[q],2) #按每一列计算Di-
max_list.append(pow(max_sum,0.5))
min_list.append(pow(min_sum,0.5))
answer_list.append(min_list[k]/ (min_list[k] + max_list[k])) #套用计算得分的公式 Si = (Di-) / ((Di+) +(Di-))
max_sum = 0
min_sum = 0
answer = np.array(answer_list) #得分归一化
return (answer / np.sum(answer))
def main():
file = 'D:\\kong\\数学建模\\数据汇总.xlsx'
answer1 = read(file) # 读取文件
answer2 = answer1.copy()
for i in range(0,9): # 按照不同的列，根据不同的指标转换为极大型指标，因为只有四列
if i == 3: # 中间型指标
answer1[i] = dataDirection_2(answer1[i],0)
elif i == 1 or i == 2 or i == 8: # 极小型指标
answer1[i] = dataDirection_1(answer1[i])
else:# 本来就是极大型指标，不用转换
answer1[i] = answer1[i]
answer3 = temp2(answer1) # 正向数组标准化
w=entropy(answer2)#计算权重
answer4 = temp3(answer3,w) #topsis
data = pd.DataFrame(answer4) # 计算得分
print(data)
#将得分输出到excel表格中
writer = pd.ExcelWriter('D:\\kong\\数学建模\\结果.xlsx') # 写入Excel文件
data.to_excel(writer, '得分', float_format='%.5f') # ‘page_1’是写入excel的sheet名
writer.save()
writer.close()
if __name__=='__main__':
main()
注意：我用的是pycharm,直接导入下载xlrd会出现pycharm no module named xlrd的问题
解决方法：
https://www.jianshu.com/p/cd2aea13b6bc
|