计算 Pandas 数据帧中的动态时间扭曲距离 [英] Calculating Dynamic Time Warping Distance in a Pandas Data Frame

问题描述

我想计算数据帧中的动态时间扭曲 (DTW) 距离.结果必须是一个新的数据帧(距离矩阵)，其中包括每行之间的成对 dtw 距离.

对于欧几里得距离，我使用以下代码:

from scipy.spatial.distance import pdist, squareformeuclidean_dist = squareform(pdist(sample_dataframe,'euclidean'))

我需要一个类似的 DTW 代码.

提前致谢.

解决方案

有多种方法可以做到这一点.我将在下面留下两个选项.

如果你想知道欧几里得距离和 DTW 之间的区别，

I want to calculate Dynamic Time Warping (DTW) distances in a dataframe. The result must be a new dataframe (a distance matrix) which includes the pairwise dtw distances among each row.

For Euclidean Distance I use the following code:

from scipy.spatial.distance import pdist, squareform
euclidean_dist = squareform(pdist(sample_dataframe,'euclidean'))

I need a similar code for DTW.

Thanks in advance.

解决方案

There are various ways one might do that. I'll leave two options bellow.

In case one wants to know the difference between the euclidean distance and DTW, this is a good resource.

---

Option 1

Using fastdtw.

Install it with

pip install fastdtw

Then use it as following

import numpy as np from scipy.spatial.distance import euclidean

from fastdtw import fastdtw

x = np.array([[1,1], [2,2], [3,3], [4,4], [5,5]])
y = np.array([[2,2],
[3,3], [4,4]])
distance, path = fastdtw(x, y, dist=euclidean)
print(distance)

---

Option 2 (Source)

def dtw(s, t):
    n, m = len(s), len(t)
    dtw_matrix = np.zeros((n+1, m+1))
    for i in range(n+1):
        for j in range(m+1):
            dtw_matrix[i, j] = np.inf
    dtw_matrix[0, 0] = 0
    
    for i in range(1, n+1):
        for j in range(1, m+1):
            cost = abs(s[i-1] - t[j-1])
            # take last min from a square box
            last_min = np.min([dtw_matrix[i-1, j], dtw_matrix[i, j-1], dtw_matrix[i-1, j-1]])
            dtw_matrix[i, j] = cost + last_min
    return dtw_matrix 

It works like the following

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