python实现毕达哥拉斯模糊零和博弈

谁借莪１个温暖的怀抱￠ 2022-10-06 10:00 183阅读 0赞

一篇一区论文《A new approach for emergency decision‐making based on zero‐sum game withPythagorean fuzzy uncertain linguistic variables》的独立重复实验。  
基本思想是通过PFULPWA算子进行矩阵集结，标准化矩阵，解模糊，之后通过线性规划求出纳什均衡解。代码如下，关键步骤有注释。

import pandas as pd#AHP方法实现
    import numpy as np
    from scipy import optimize as op
    np.set_printoptions(suppress=True)
    def getdef(n):
        df = pd.read_excel('D:\study\\test\data\\t10.xlsx',sheet_name=n)#地址中包含t会被转义，加双斜杠即可
        return df
    def getmatrix(df,n):
        df= df.loc[:, n:n + 3]
        return df.T.reset_index(drop=True).T#使用转置消除列元素的索引成一个新的矩阵，不然切片形式会对后面的引用造成很大的困扰。
    def getg(df):
        df2= (df[0]+df[1])*(df[2]*df[2]+1-df[3]*df[3])/24#计算每一行的得分函数
        return df2
    def gett(g,t):#调用得分函数计算算子中的t
        t1= 1
        t2= g[t]
        t3= g[t]*g[t]
        t4= g[t]*g[t+4]*g[t+8]
        t5= g[t]*g[t+4]*g[t+8]*g[t+12]
        tsum= t1+t2+t3+t4+t5#求出t1到t5的元素相乘，将作为分母后面使用
        expon= [t1/tsum,t2/tsum,t3/tsum,t4/tsum,t5/tsum]
        expon1= np.array(expon)#计算五个t的值，用数列表达，将与后面的五个元素对应相乘
        return expon1
    def getfirst(df,t,i,j):#df是每四列元素组成的新矩阵，g是新df计算出的得分函数，t是对于g而言的参数下标，i是对df列的选择，j是对切片后起始元素的选择。
        sub=(df[i].values.tolist())
        subnumber= np.array(sub[j::4])#第i列每隔四个构成的数列，对应于t相乘。
        st1_l=sum(subnumber*t)
        return st1_l
    def getsecond(df,t,i,j):
        sub=(df[i].values.tolist())
        subnumber= np.array(sub[j::4])
        st1_r=sum(subnumber*t)
        return st1_r
    def getthird(df,t,i,j):
        sub=(df[i].values.tolist())#
        subnumber= np.array(sub[j::4])
        miudiv= np.power(1-subnumber*subnumber,t)
        miupro= np.prod(miudiv)#隶属度的最终结果
        return 1 - miupro
    def getforth(df,t,i,j):
        sub=(df[i].values.tolist())#
        subnumber= np.array(sub[j::4])#切片第四列元素
        miudiv= np.power(subnumber,t)
        miupro= np.prod(miudiv)#隶属度的最终结果
        return miupro
    def getlinnery(df):#求属性纳什均衡的方法
        c=np.array([1,1,1,1,1,1])
        A_ub=np.array(df)
        B_ub=np.array([0,0,0,0])
        A_eq=np.array([[1,1,1,1,1,0]])
        B_eq=np.array([1])
        x1=(0,1)
        x2=(0,1)
        x3=(0,1)
        x4=(0,1)
        x5=(0,1)
        x6=(0,1)
        res=op.linprog(c,A_ub,B_ub,A_eq,B_eq,bounds=(x1,x2,x3,x4,x5,x6))
        return res
    def getlinerx(df):#求方案纳什均衡的方法，其实区别只有矩阵的转置，但是其他数据都得改变因此重新写方法
        c=np.array([1,1,1,1,1])
        A_ub=np.array(df)
        B_ub=np.array([0,0,0,0,0])
        A_eq=np.array([[1,1,1,1,0]])
        B_eq=np.array([1])
        x1=(0,1)
        x2=(0,1)
        x3=(0,1)
        x4=(0,1)
        x5=(0,1)
        res=op.linprog(-c,A_ub,B_ub,A_eq,B_eq,bounds=(x1,x2,x3,x4,x5))
        return res
    if __name__ == '__main__':
        df = getdef(0)
        print('总方案矩阵为\n',df)
        df1 =getmatrix(df, 1)
        df2 =getmatrix(df, 5)
        df3 =getmatrix(df, 9)
        df4 =getmatrix(df, 13)
        df5 =getmatrix(df, 17)
        df3[0] = df3[0].apply( lambda x:6-x)
        df3[1] = df3[1].apply( lambda x:6-x)
        df4[0] = df4[0].apply( lambda x:6-x)
        df4[1] = df4[1].apply( lambda x:6-x)
        df5[0] = df5[0].apply( lambda x:6-x)
        df5[1] = df5[1].apply( lambda x:6-x)#这里是统一成本型和收益型属性，因为工作量不大，直接使用匿名函数进行修改
        g1=[getg(df1),getg(df2),getg(df3),getg(df4),getg(df5)]#这里开始就是冗长的每每四列元素进行不同的运算，因为序号太多暂时不会套循环，就本办法分五次实现。
        t1=[gett(g1[0],0),gett(g1[0],1),gett(g1[0],2),gett(g1[0],3)]
        q11=pd.Series([getfirst(df1,t1[0],0,0),getfirst(df1,t1[1],0,1),getfirst(df1,t1[2],0,2),getfirst(df1,t1[3],0,3)],name='ul')
        q12=pd.Series([getsecond(df1,t1[0],1,0),getsecond(df1,t1[1],1,1),getsecond(df1,t1[2],1,2),getsecond(df1,t1[3],1,3)],name='ur')
        q13= pd.Series([getthird(df1,t1[0],2,0),getthird(df1,t1[1],2,1),getthird(df1,t1[2],2,2),getthird(df1,t1[3],2,3)],name='fl')
        q14=pd.Series([getforth(df1,t1[0],3,0),getforth(df1,t1[1],3,1),getforth(df1,t1[2],3,2),getforth(df1,t1[3],3,3)],name='fr')
        t2=[gett(g1[1],0),gett(g1[1],1),gett(g1[1],2),gett(g1[1],3)]
        q21=pd.Series([getfirst(df2,t2[0],0,0),getfirst(df2,t2[1],0,1),getfirst(df2,t2[2],0,2),getfirst(df2,t2[3],0,3)],name='ul')
        q22=pd.Series([getsecond(df2,t2[0],1,0),getsecond(df2,t2[1],1,1),getsecond(df2,t2[2],1,2),getsecond(df2,t2[3],1,3)],name='ur')
        q23= pd.Series([getthird(df2,t2[0],2,0),getthird(df2,t2[1],2,1),getthird(df2,t2[2],2,2),getthird(df2,t2[3],2,3)],name='fl')
        q24=pd.Series([getforth(df2,t2[0],3,0),getforth(df2,t2[1],3,1),getforth(df2,t2[2],3,2),getforth(df2,t2[3],3,3)],name='fr')
        t3=[gett(g1[2],0),gett(g1[2],1),gett(g1[2],2),gett(g1[2],3)]
        q31=pd.Series([getfirst(df3,t3[0],0,0),getfirst(df3,t3[1],0,1),getfirst(df3,t3[2],0,2),getfirst(df3,t3[3],0,3)],name='ul')
        q32=pd.Series([getsecond(df3,t3[0],1,0),getsecond(df3,t3[1],1,1),getsecond(df3,t3[2],1,2),getsecond(df3,t3[3],1,3)],name='ur')
        q33= pd.Series([getthird(df3,t3[0],2,0),getthird(df3,t3[1],2,1),getthird(df3,t3[2],2,2),getthird(df3,t3[3],2,3)],name='fl')
        q34=pd.Series([getforth(df3,t3[0],3,0),getforth(df3,t3[1],3,1),getforth(df3,t3[2],3,2),getforth(df3,t3[3],3,3)],name='fr')
        t4=[gett(g1[3],0),gett(g1[3],1),gett(g1[3],2),gett(g1[3],3)]
        q41=pd.Series([getfirst(df4,t4[0],0,0),getfirst(df4,t4[1],0,1),getfirst(df4,t4[2],0,2),getfirst(df4,t4[3],0,3)],name='ul')
        q42=pd.Series([getsecond(df4,t4[0],1,0),getsecond(df4,t4[1],1,1),getsecond(df4,t4[2],1,2),getsecond(df4,t4[3],1,3)],name='ur')
        q43= pd.Series([getthird(df4,t4[0],2,0),getthird(df4,t4[1],2,1),getthird(df4,t4[2],2,2),getthird(df4,t4[3],2,3)],name='fl')
        q44=pd.Series([getforth(df4,t4[0],3,0),getforth(df4,t4[1],3,1),getforth(df4,t4[2],3,2),getforth(df4,t4[3],3,3)],name='fr')
        t5=[gett(g1[4],0),gett(g1[4],1),gett(g1[4],2),gett(g1[4],3)]
        q51=pd.Series([getfirst(df5,t5[0],0,0),getfirst(df5,t5[1],0,1),getfirst(df5,t5[2],0,2),getfirst(df5,t5[3],0,3)],name='ul')
        q52=pd.Series([getsecond(df5,t5[0],1,0),getsecond(df5,t5[1],1,1),getsecond(df5,t5[2],1,2),getsecond(df5,t5[3],1,3)],name='ur')
        q53= pd.Series([getthird(df5,t5[0],2,0),getthird(df5,t5[1],2,1),getthird(df5,t5[2],2,2),getthird(df5,t5[3],2,3)],name='fl')
        q54=pd.Series([getforth(df5,t5[0],3,0),getforth(df5,t5[1],3,1),getforth(df5,t5[2],3,2),getforth(df5,t5[3],3,3)],name='fr')
        f=pd.concat([q11, q12, q13, q14, q21, q22, q23, q24,q31, q32, q33, q34, q41, q42, q43, q44,q51, q52, q53, q54], axis=1)
        print('算子集结矩阵为\n',f)#这里输出的就是完整的矩阵，因为数据较大显示不完整，可以考虑新建一个excel导出数据进行查看。
        # f.to_excel('D:\study\\test\data\\t11.xlsx',sheet_name="sheet2")
        w= [0.05,0.47,0.11,0.14,0.25]#这楼里是开始实现矩阵标准化，因为很简单没有单独写函数
        f['ul']=f['ul']*w
        f['ur']=f['ur']*w
        f['fl']=np.sqrt(1-np.power(1-f['fl']*f['fl'],w))
        f['fr']=np.power(f['fr'],w)
        print('标准化矩阵为\n',f)
        f=f.T.reset_index(drop=True).T#这里要去模糊化，还可以用到之前定义的函数
        f=pd.concat([getg(getmatrix(f, 0)),getg(getmatrix(f, 4)),getg(getmatrix(f, 8)),getg(getmatrix(f, 12)),getg(getmatrix(f, 16))], axis=1)
        print('去模糊化矩阵为\n',f)
        fa=f.copy(deep=True)#这里新建两个副本用来计算线性规划求解纳什均衡
        fa['5'] = -1
        print(getlinnery(fa))#实现对属性的纳什均衡
        fc=f.copy(deep=True)
        fc.loc['4']=[-1,-1,-1,-1,-1]
        print(getlinerx(fc.T.applymap(lambda x:-x)))#实现对方案的纳什均衡。
    
    实现结果
    ![在这里插入图片描述](https://img-blog.csdnimg.cn/20210612232354743.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDAxNDgzNg==,size_16,color_FFFFFF,t_70)
    ![在这里插入图片描述](https://img-blog.csdnimg.cn/20210612232406692.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDAxNDgzNg==,size_16,color_FFFFFF,t_70)
    论文结果
    ![在这里插入图片描述](https://img-blog.csdnimg.cn/20210612232510618.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDAxNDgzNg==,size_16,color_FFFFFF,t_70)
    ![在这里插入图片描述](https://img-blog.csdnimg.cn/2021061223253068.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80NDAxNDgzNg==,size_16,color_FFFFFF,t_70)
    github仓库在
    https://github.com/rivendelltom/decision-making-study