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

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