Python - - - numpy的矩阵计算

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import numpy as np

def numpyWork1(): # 矩阵的求和运算

    one = np.array([[1, 0, 1, 3, 4, 1],
                    [2, 1, 4, 2, 3, 0],
                    [3, 5, 4, 1, 3, 2],
                    [2, 6, 3, 1, 3, 8],
                    [9, 1, 2, 1, 5, 0],
                    ])

    two = np.sum(one, axis= 0)  # 每一列求和，得到的新矩阵
    three = np.sum(one, axis= 1) # 每一行求和，得到的新矩阵

    print('two=', two)
    print('three=', three)

    return

def numpyWork2(): # 两个向量相乘

    one = np.array([3, 5, 8, 10, 9, 2, 4, 5, 1])
    two = np.array([2, 3, 4, 2, 6, 7, 9, 10, 2])

    three = np.multiply(one, two)
    print('three=', three)

    return

def numpyWork3(): # 矩阵的欧氏距离


    one = np.array([[1, 2, 3, 4, 5, 6, 7, 8, 9], [1, 2, 3, 4, 3, 9, 7, 8, 9], [1, 2, 6, 4, 8, 6, 7, 8, 9]])
    two = np.array([[2, 3, 4, 5, 6, 7, 8, 9, 10], [2, 3, 4, 8, 6, 7, 12, 9, 10], [2, 3, 4, 9, 2, 7, 8, 9, 10]])

    dist1 = np.linalg.norm(one - two) # 矩阵的欧氏距离
    dist2 = np.square(one - two)

    print('one=', one)
    print('two=', two)

    print('dist1=', dist1)  # 此处为一个数值
    print('dist2=', dist2)

    return



def EuclideanDistances(A, B): # 两个矩阵求欧氏距离 - - - 偏慢（网上提供的方法）

    BT = B.transpose()
    vecProd = A * BT
    SqA =  A.getA()**2
    sumSqA = np.matrix(np.sum(SqA, axis=1))
    sumSqAEx = np.tile(sumSqA.transpose(), (1, vecProd.shape[1]))
    SqB = B.getA()**2
    sumSqB = np.sum(SqB, axis=1)
    sumSqBEx = np.tile(sumSqB, (vecProd.shape[0], 1))
    SqED = sumSqBEx + sumSqAEx - 2*vecProd
    ED = (SqED.getA())**0.5

    return np.matrix(ED)

def EuclideanDistancesBetter(): # 矩阵与向量的欧氏距离计算速度很快，根据数学公式计算的，算起来较快，推荐使用

    one = np.array([[1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
                    [2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
                    [15, 16, 17, 18, 19, 20, 21, 22, 23, 24],
                    [1, 2, 1, 2, 1, 2, 1, 2, 1, 2],
                    ])

    two = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])

    result_ed1 = np.linalg.multi_dot([one, two])  # 两个矩阵向量的欧氏距离
    result_ed2 = 1 / (1 + result_ed1)

    print('result_ed1= ', result_ed1)  # result_ed1=  [ 385  440 1155   85]
    print('result_ed2= ', result_ed2)  # result_ed2=  [0.00259067 0.00226757 0.00086505 0.01162791]

    return


#numpyWork1(): # 矩阵的求和运算
#numpyWork2(): # 两个向量相乘
#numpyWork3() # 矩阵的欧氏距离
#EuclideanDistances(A, B): # 两个矩阵求欧氏距离 - - - 偏慢
#EuclideanDistancesBetter() # 矩阵与向量的欧氏距离计算速度很快（推荐使用）