在 knn 回归模型中使用 levenshtein 距离

Question

我有兴趣使用 KNN 回归模型预测我的蛋白质的稳定性，但是我想使用作为我的蛋白质嵌入计算的 levenshtein 距离而不是序列本身，作为我的模型的输入变量。我希望根据每个序列的 levenshtein 距离，使用 k 个最近邻的训练误差来衡量每个实例的不一致性。

如果可能的话，我不希望对它们进行热编码。

我的输入表如下所示：

new_host  sequence    expression
FALSE     AQVPYGVS-   0.039267878
FALSE     ASVPYGVSI   0.039267878
FALSE     STNL-GSGR   0.261456561
FALSE     NLYGSGL-R   0.265188519
FALSE     SLGP-NLYG   0.419680588
FALSE     A-SLGTTNG   0.145710993

对于我的输出，我不确定我是否真的可以得到每个特定的距离，以及这是否会用于我的 KNN 回归模型的输入。

我计算 levenshtein 距离的函数：

def levenshtein(seq1, seq2):
    size_x = len(seq1) + 1
    size_y = len(seq2) + 1
    matrix = np.zeros ((size_x, size_y))
    for x in range(size_x):
        matrix [x, 0] = x
    for y in range(size_y):
        matrix [0, y] = y

    for x in range(1, size_x):
        for y in range(1, size_y):
            if seq1[x-1] == seq2[y-1]:
                matrix [x,y] = min(
                    matrix[x-1, y] + 1,
                    matrix[x-1, y-1],
                    matrix[x, y-1] + 1
                )
            else:
                matrix [x,y] = min(
                    matrix[x-1,y] + 1,
                    matrix[x-1,y-1] + 1,
                    matrix[x,y-1] + 1
                )
    print (matrix)
    return (matrix[size_x - 1, size_y - 1])

我的函数来计算每个实例的不合格：

  def nonconformity(self, z, Z):
        """Return k-Nearest Neighbours (kNN) nonconformity measure.
        
        Parameters
        ----------
        z : array-like, shape (n_features,)
            Test vector, where n_features is the number of features.
        Z : array-like, shape (n_samples, n_features)
            Training vectors, where n_samples is the number of samples,
            n_features is the number of features.
        Returns
        -------
        r : float
            kNN nonconformity measure on z with respect to Z.
        """
        # Take the k smallest distances between z rows and zn and sum them.
        dist = cdist(Z, [z])[:,0]
        r = np.sort(dist)[:self.k].sum()
    
        return r

期望的输出（不确定这是否可能）

new_host  sequence    expression   Levenshtein Distance
FALSE     AQVPYGVS-   0.039267878    0
FALSE     ASVPYGVSI   0.039267878    0.1
FALSE     STNL-GSGR   0.261456561    0.2 
FALSE     NLYGSGL-R   0.265188519    0.2 
FALSE     SLGP-NLYG   0.419680588    0.4
FALSE     A-SLGTTNG   0.145710993    0.5

Answer 1

据我了解，您需要找到每个字符串与所有字符串的距离。但是这样做会失败，就好像序列的长度是 n 和序列长度是 m 然后 O(n*n*m*m) 复杂性一样。您可以尝试使用虚拟字符串（如 origin:"---------"）来计算所有序列的 Levenshtein 距离。

import pandas as pd
import numpy as np

sequence = ["AQVPYGVS-", "ASVPYGVSI", "STNL-GSGR", "NLYGSGL-R", "SLGP-NLYG", "A-SLGTTNG"]   
expression = [0.039267878, 0.039267878, 0.145710993, 0.419680588, 0.265188519, 0.261456561]

def levenshtein(seq1, seq2):
    size_x = len(seq1) + 1
    size_y = len(seq2) + 1
    matrix = np.zeros ((size_x, size_y))
    for x in range(size_x):
        matrix [x, 0] = x
    for y in range(size_y):
        matrix [0, y] = y

    for x in range(1, size_x):
        for y in range(1, size_y):
            if seq1[x-1] == seq2[y-1]:
                matrix [x,y] = min(
                    matrix[x-1, y] + 1,
                    matrix[x-1, y-1],
                    matrix[x, y-1] + 1
                )
            else:
                matrix [x,y] = min(
                    matrix[x-1,y] + 1,
                    matrix[x-1,y-1] + 1,
                    matrix[x,y-1] + 1
                )
    # print (matrix)
    return (matrix[size_x - 1, size_y - 1])


seq_distances = np.empty((0, len(sequence)), int)
for seq1 in range(0, len(sequence)):
    seq1_dist = []
    for seq2 in range(0, len(sequence)):
        seq1_dist.append(levenshtein(sequence[seq1], sequence[seq2]))    
    seq_distances = np.vstack((seq_distances, np.array(seq1_dist)))

print(seq_distances)

# Something of this sort can be used.
dist = distance.cdist(seq_distances, [seq_distances[0]])
r = np.sort(dist, axis=0)[:k].sum()

输出：

[[0. 2. 8. 8. 8. 8.]
 [2. 0. 8. 8. 8. 8.]
 [8. 8. 0. 5. 7. 8.]
 [8. 8. 5. 0. 7. 9.]
 [8. 8. 7. 7. 0. 6.]
 [8. 8. 8. 9. 6. 0.]]