DTW (Dynamic Time Warping) 算法基于动态规划的思想，可以衡量两个长度不一致时间序列的相似度，由日本学者Itakura提出。
DTW
案例：
假设我们有三个时间序列，分别是
ts_a = [1,5,8,10,56,21,32,8]
ts_b = [1,5,8,10,23,56,21,32,8]
ts_c = [1,3,6,9,16,29,31,32,33]
ts_a与ts_b和ts_c的长度不一样，现在需要知道ts_a与ts_b和ts_c哪个更相似，通过观察，我们可以清楚的看出ts_a与ts_b的相似度更高。使用DTW相似度解决该问题的代码如下：
import sys
import numpy as np
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
"""
d=lambda x, y: abs(x - y)
max_warping_window = 10000
# 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 = sys.maxsize * np.ones((M, N))
# Initialize the first row and column
cost[0, 0] = d(ts_a[0], ts_b[0])
for i in range(1, M):
cost[i, 0] = cost[i - 1, 0] + d(ts_a[i], ts_b[0])
for j in range(1, N):
cost[0, j] = cost[0, j - 1] + d(ts_a[0], ts_b[j])
# Populate rest of cost matrix within window
for i in range(1, M):
for j in range(max(1, i - max_warping_window),
min(N, i + max_warping_window)):
choices = cost[i - 1, j - 1], cost[i, j - 1], cost[i - 1, j]
cost[i, j] = min(choices) + d(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
ts_a = [1,5,8,10,56,21,32,8]
ts_b = [1,5,8,10,23,56,21,32,8]
ts_c = [1,3,6,9,16,29,31,32,33]
# 调用cal_dtw_distance计算dtw相似度
dtw_ab = cal_dtw_distance(ts_a, ts_b)
dtw_ac = cal_dtw_distance(ts_a, ts_c)
print("ts_a与ts_b的dtw相似度为 %2.f，\nts_a与ts_c的dtw相似度为 %2.f。" % (dtw_ab, dtw_ac))
if dtw_ab < dtw_ac:
print("ts_a与ts_b 更相似！")
else:
print("ts_a与ts_c 更相似！")
ts_a与ts_b的dtw相似度为 13，
ts_a与ts_c的dtw相似度为 71。
ts_a与ts_b 更相似！
从执行结果可以看出，使用DTW相似度计算的结果与我们观察到的结果一致。