计算Dynamic Time Warping遇ValueError：输入向量需为1-D的问题求助

问题原因 #

scipy提供的euclidean距离函数要求输入必须是1维数组，但fastdtw在计算过程中，会将序列中的每个单独元素作为参数传入距离函数。当你的序列是由标量组成的列表（比如x = [7,1,2,5,9]）时，每个元素是单个数值而非1维数组，就会触发这个错误。

两种解决方案 #

方案一：自定义支持标量的欧氏距离函数

自己实现一个简单的欧氏距离函数，既兼容标量输入，也支持数组输入：

def custom_euclidean(a, b):
    return np.abs(a - b)

调用fastdtw时替换原有的euclidean函数即可：

dtw_distance, warp_path = fastdtw(x, y, dist=custom_euclidean)

方案二：将输入序列转换为二维数组

把每个标量元素包装成单元素的二维数组，让scipy的euclidean函数可以正确识别：

x = np.array([7, 1, 2, 5, 9]).reshape(-1, 1)
y = np.array([1, 8, 0, 4, 4, 2, 0]).reshape(-1, 1)

之后直接使用原有的euclidean函数调用fastdtw即可。

修改后的完整代码示例（方案一） #

import pandas as pd
import numpy as np

# Plotting Packages
import matplotlib.pyplot as plt
import seaborn as sbn

import matplotlib as mpl
mpl.rcParams['figure.dpi'] = 150
savefig_options = dict(format="png", dpi=150, bbox_inches="tight")

# Computation packages
from scipy.spatial.distance import euclidean
from fastdtw import fastdtw

def compute_euclidean_distance_matrix(x, y) -> np.array:
    """Calculate distance matrix
    This method calcualtes the pairwise Euclidean distance between two sequences.
    The sequences can have different lengths.
    """
    dist = np.zeros((len(y), len(x)))
    for i in range(len(y)):
        for j in range(len(x)):
            dist[i,j] = (x[j]-y[i])**2
    return dist


def compute_accumulated_cost_matrix(x, y) -> np.array:
    """Compute accumulated cost matrix for warp path using Euclidean distance
    """
    distances = compute_euclidean_distance_matrix(x, y)

    # Initialization
    cost = np.zeros((len(y), len(x)))
    cost[0,0] = distances[0,0]
    
    for i in range(1, len(y)):
        cost[i, 0] = distances[i, 0] + cost[i-1, 0]  
        
    for j in range(1, len(x)):
        cost[0, j] = distances[0, j] + cost[0, j-1]  

    # Accumulated warp path cost
    for i in range(1, len(y)):
        for j in range(1, len(x)):
            cost[i, j] = min(
                cost[i-1, j],    # insertion
                cost[i, j-1],    # deletion
                cost[i-1, j-1]   # match
            ) + distances[i, j] 
            
    return cost

# 自定义兼容标量的欧氏距离函数
def custom_euclidean(a, b):
    return np.abs(a - b)

# Create two sequences
x = [7, 1, 2, 5, 9]
y = [1, 8, 0, 4, 4, 2, 0]
# 使用自定义距离函数调用fastdtw
dtw_distance, warp_path = fastdtw(x, y, dist=custom_euclidean)
cost_matrix = compute_accumulated_cost_matrix(x, y)

# 输出结果验证
print(f"DTW距离: {dtw_distance}")
print(f"Warp路径: {warp_path}")

内容的提问来源于stack exchange，提问作者Jab