DTW算法采用欧氏距离得到相似度

import numpy as np

def do(ts_a, ts_b):

    ts_c = ts_a - ts_b
    ts_c = ts_c[np.newaxis,:]
    osd = np.linalg.norm(ts_c, ord=None, axis=1, keepdims=True)
    return osd

def cal_dtw_distance(ts_a, ts_b):
    """Returns the DTW similarity distance between two 2-D
    timeseries numpy arrays.

    Arguments
    ---------
    ts_a, ts_b : array of shape [n_samples, n_timepoints]
        Two arrays containing n_samples of timeseries data
        whose DTW distance between each sample of A and B
        will be compared

    d : DistanceMetric object (default = abs(x-y))
        the distance measure used for A_i - B_j in the
        DTW dynamic programming function

    Returns
    -------
    DTW distance between A and B
    """
    # Create cost matrix via broadcasting with large int
    ts_a, ts_b = np.array(ts_a), np.array(ts_b)
    M, N = len(ts_a), len(ts_b)
    cost = np.ones((M, N))

    # Initialize the first row and column
    cost[0, 0] = do(ts_a[0], ts_b[0])
    for i in range(1, M):
        cost[i, 0] = cost[i - 1, 0] + do(ts_a[i], ts_b[0])

    for j in range(1, N):
        cost[0, j] = cost[0, j - 1] + do(ts_a[0], ts_b[j])

    # Populate rest of cost matrix within window
    for i in range(1, M):
        for j in range(1,N):
            choices = cost[i - 1, j - 1], cost[i, j - 1], cost[i - 1, j]
            cost[i, j] = min(choices) + do(ts_a[i], ts_b[j])

    # Return DTW distance given window
    return cost[-1, -1]



if __name__ == "__main__":
    # 案例:判断ts_a与ts_b和ts_c哪个更相似
    path1 = './m1_a2_list.npy'
    data1 = np.load(path1)
    ts_a1 = data1[0]
    ts_b1 = data1[1]
    print(ts_a1)

    path2 = './m2_a2_list.npy'
    data2 = np.load(path2)
    ts_a2 = data2[0]

    # 调用cal_dtw_distance计算dtw相似度
    dtw_ab = cal_dtw_distance(ts_a1, ts_a2)

    print('ts_a与ts_b的dtw相似度为 : ',dtw_ab)

 

你可能感兴趣的:(DTW算法采用欧氏距离得到相似度)