对于标准化和有限差分法的具体要求，简单便于操作就可以

首先，对于标准化，我们可以采用Z-score标准化方法，即将每个元素减去该行或该列的均值，再除以该行或该列的标准差。这样，每个元素的值都将落在以0为中心的正负值间。为了保证标准化后的元素值都不小于1，我们可以对每个元素加上1。最后，为了转化为方阵，我们可以添加对角线元素为1的单位矩阵，然后将其与原矩阵相加。

其次，对于有限差分法，我们可以采用一阶有限差分法，即将每个元素的增长率设为该元素与其右侧元素之差除以时间间隔。为了避免分母为0，我们可以将分母设为一个极小值，比如0.0001。

下面是代码实现，您可以根据自己的数据进行调整。假设矩阵A和B均为m行n列的矩阵，时间序列为行向量，指标为列向量。

import numpy as np

# 标准化
def normalize(matrix):
    row_mean = np.mean(matrix, axis=1, keepdims=True)
    row_std = np.std(matrix, axis=1, keepdims=True)
    col_mean = np.mean(matrix, axis=0, keepdims=True)
    col_std = np.std(matrix, axis=0, keepdims=True)
    matrix = (matrix - row_mean) / row_std
    matrix = (matrix - col_mean) / col_std
    matrix += 1
    return matrix

# 转化为方阵
def to_square(matrix):
    m, n = matrix.shape
    if m == n:
        return matrix
    elif m > n:
        diff = m - n
        return np.hstack((matrix, np.zeros((m, diff)))) + np.eye(m)
    else:
        diff = n - m
        return np.vstack((matrix, np.zeros((diff, n)))).T + np.eye(n)

# 有限差分法
def finite_difference(matrix, time_series):
    diff = np.diff(matrix, axis=1) / np.diff(time_series)
    diff = np.hstack((diff, np.zeros((matrix.shape[0], 1))))
    diff = np.nan_to_num(diff, nan=0, posinf=0, neginf=0) + 0.0001 # 避免分母为0
    return diff

# 示例数据
A = np.random.rand(5, 3) * 100
B = np.random.rand(5, 4) * 100
time_series = np.array([2010, 2011, 2012, 2013, 2014])

# 标准化
A_norm = normalize(A)
B_norm = normalize(B)

# 转化为方阵
n = max(A_norm.shape[1], B_norm.shape[1])
A_square = to_square(A_norm)
B_square = to_square(B_norm)

# 有限差分法
A_diff = finite_difference(A_square, time_series)
B_diff = finite_difference(B_square, time_series)

# 计算相互作用的增长率
growth_rate = np.dot(A_diff, B_diff.T)