Python中的字符串距离矩阵

Question

如何在 Python 中计算字符串的 Levenshtein 距离矩阵

              str1    str2    str3    str4    ...     strn
      str1    0.8     0.4     0.6     0.1     ...     0.2
      str2    0.4     0.7     0.5     0.1     ...     0.1
      str3    0.6     0.5     0.6     0.1     ...     0.1
      str4    0.1     0.1     0.1     0.5     ...     0.6
      .       .       .       .       .       ...     .
      .       .       .       .       .       ...     .
      .       .       .       .       .       ...     .
      strn    0.2     0.1     0.1     0.6     ...     0.7

使用距离函数，我们可以计算两个单词之间的距离。但在这里我有 1 个包含 n 个字符串的列表。我想计算距离矩阵，然后我想对单词进行聚类。

Answer 1

只需使用接受自定义指标的pdist 版本。

Y = pdist(X, levensthein)

对于levensthein，您可以按照 Tanu 的建议使用 Rosettacode 的实现

如果您想要一个完整的平方矩阵，只需在结果上使用squareform：

Y = scipy.spatial.distance.squareform(Y)

Answer 2

你可以这样做

from Levenshtein import distance
import numpy as np
from time import time

def get_distance_matrix(str_list):
    """ Construct a levenshtein distance matrix for a list of strings"""
    dist_matrix = np.zeros(shape=(len(str_list), len(str_list)))
    t0 = time()
    print "Starting to build distance matrix. This will iterate from 0 till ", len(str_list) 
    for i in range(0, len(str_list)):
        print i
        for j in range(i+1, len(str_list)):
                dist_matrix[i][j] = distance(str_list[i], str_list[j]) 
    for i in range(0, len(str_list)):
        for j in range(0, len(str_list)):
            if i == j:
                dist_matrix[i][j] = 0 
            elif i > j:
                dist_matrix[i][j] = dist_matrix[j][i]
    t1 = time()
    print "took", (t1-t0), "seconds"
    return dist_matrix

str_list = ["analyze", "analyse", "analysis", "analyst"]
get_distance_matrix(str_list)

Starting to build distance matrix. This will iterate from 0 till  4
0
1
2
3
took 0.000197887420654 seconds
>>> array([[ 0.,  1.,  3.,  2.],
   [ 1.,  0.,  2.,  1.],
   [ 3.,  2.,  0.,  2.],
   [ 2.,  1.,  2.,  0.]])

Answer 3

这是我的代码

import pandas as pd
from Levenshtein import distance
import numpy as np

Target = ['Tree','Trip','Treasure','Nothingtodo']

List1 = Target
List2 = Target

Matrix = np.zeros((len(List1),len(List2)),dtype=np.int)

for i in range(0,len(List1)):
  for j in range(0,len(List2)):
      Matrix[i,j] = distance(List1[i],List2[j])

print Matrix

[[ 0  2  4 11]
 [ 2  0  6 10]
 [ 4  6  0 11]
 [11 10 11  0]]