基于python的数学建模---多模糊评价

权重 ak的确定——频数统计法

选取正整数p的方法

画箱形图取1/4与3/4的距离（IQR） ceil（）取整

代码：

import numpy as np
def frequency(matrix,p):
    '''
    频数统计法确定权重
    :param matrix: 因素矩阵
    :param p: 分组数
    :return: 权重向量
    '''
    A = np.zeros((matrix.shape[0]))
    for i in range(0, matrix.shape[0]):
        ## 根据频率确定频数区间列表
        row = list(matrix[i, :])
        maximum = max(row)
        minimum = min(row)
        gap = (maximum - minimum) / p
        row.sort()
        group = []
        item = minimum
        while(item < maximum):
            group.append([item, item + gap])
            item = item + gap
        print(group)
        # 初始化一个数据字典，便于记录频数
        dataDict = {}
        for k in range(0, len(group)):
            dataDict[str(k)] = 0
        # 判断本行的每个元素在哪个区间内，并记录频数
        for j in range(0, matrix.shape[1]):
            for k in range(0, len(group)):
             if(matrix[k, j] >= group[k][0]):
                 dataDict[str(k)] = dataDict[str(k)] + 1
             break
        print(dataDict)
        # 取出最大频数对应的key，并以此为索引求组中值
        index = int(max(dataDict,key=dataDict.get))
        mid = (group[index][0] + group[index][1]) / 2
        print(mid)
        A[i] = mid
    A = A / sum(A[:]) # 归一化
    return A

权重 ak的确定——模糊层次分析法

代码：

import numpy as np
 
 
def AHP(matrix):
    if isConsist(matrix):
        lam, x = np.linalg.eig(matrix)
        return x[0] / sum(x[0][:])
    else:
        print("一致性检验未通过")
        return None
 
 
def isConsist(matrix):
    '''
    :param matrix: 成对比较矩阵
    :return:    通过一致性检验则返回true，否则返回false
    '''
    n = np.shape(matrix)[0]
    a, b = np.linalg.eig(matrix)
    maxlam = a[0].real
    CI = (maxlam - n) / (n - 1)
    RI = [0, 0, 0.58, 0.9, 1.12, 1.24, 1.32, 1.41, 1.45]
    CR = CI / RI[n - 1]
    if CR < 0.1:
        return True, CI, RI[n - 1]
    else:
        return False, None, None

import numpy as np


def appraise(criterionMatrix, targetMatrixs, relationMatrixs):
    '''
    :param criterionMatrix: 准则层权重矩阵
    :param targetMatrix:    指标层权重矩阵列表
    :param relationMatrixs: 关系矩阵列表
    :return:
    '''
    R = np.zeros((criterionMatrix.shape[1], relationMatrixs[0].shape[1]))
    for index in range(0, len(targetMatrixs)):
        row = mul_mymin_operator(targetMatrixs[index], relationMatrixs[index])
        R[index] = row
    B = mul_mymin_operator(criterionMatrix, R)
    return B / sum(B[:])


def mul_mymin_operator(A, R):
    B = np.zeros(1, R.shape[1])
    for column in range(1, R.shape[1]):
        list = []
        for row in range(1, R.shape[0]):
            list = list.append(A[row] * R[row, column])
        B[0, column] = mymin(list)
    return B


def mymin(list):
    global temp
    for index in range(1, len(list)):
        if index == 1:
            temp = min(1, list[0] + list[1])
        else:
            temp = min(1, temp + list[index])
    return temp