3

我有一个程序可以告诉我两个字符串之间的距离，它工作得很好。

例如

word 1 = hello
word 2 = hi

从一个到另一个的成本为 5（用e替换i是 2，并且有 3 个插入）。

基本上，插入成本为 1，删除成本为 1，替换成本为 2。单词也可以在字符串中打乱以降低成本。

我需要一种方法来记住在什么时候发生了什么操作，以便我可以显示对齐方式。

例如

wax
S M S(substitute move substitute, cost of 4)
and

有什么想法或提示吗？

import sys
from sys import stdout


def  minEditDist(target, source):

    # Length of the target strings set to variables
    n = len(target)
    m = len(source)

    distance = [[0 for i in range(m+1)] for j in range(n+1)]

    for i in range(1,n+1):
        distance[i][0] = distance[i-1][0] + insertCost(target[i-1])

    for j in range(1,m+1):
        distance[0][j] = distance[0][j-1] + deleteCost(source[j-1])


    for i in range(1,n+1):
        for j in range(1,m+1):
           distance[i][j] = min(distance[i-1][j]+1,
                                distance[i][j-1]+1,
                                distance[i-1][j-1]+subCost(source[j-1],target[i-1]))

    # Return the minimum distance using all the table cells
    return distance[i][j]

def subCost(x,y):
    if x == y:
        return 0
    else:
        return 2

def insertCost(x):
    return 1

def deleteCost(x):
    return 1

# User inputs the strings for comparison
# Commented out here because cloud9 won't take input like this
# word1 = raw_input("Enter A Word: ")
# word2 = raw_input("Enter The Second Word: ")
word1 = "wax"
word2 = "and"
word1x = word1
word2x = word2
# Reassign variables to words with stripped right side whitespace
word1x = word1x.strip()
word2x = word2x.strip()

if(len(word1) > len(word2)):
    range_num = len(word1)
else:
    range_num = len(word2)

# Display the minimum distance between the two specified strings
print "The minimum edit distance between S1 and S2 is: ", minEditDist(word1x,word2x), "!"
print (word1x)
print (word2x)
4

2 回答 2

1

你可以从这样的事情开始。

我已经为“S”添加了正确的数据。

path = []

def  minEditDist(target, source):

    # Length of the target strings set to variables
    n = len(target)
    m = len(source)

    distance = [[0 for i in range(m+1)] for j in range(n+1)]

    for i in range(1,n+1):
        distance[i][0] = distance[i-1][0] + insertCost(target[i-1])

    for j in range(1,m+1):
        distance[0][j] = distance[0][j-1] + deleteCost(source[j-1])


    for i in range(1,n+1):
        for j in range(1,m+1):
           sc = subCost(source[j-1],target[i-1])
           distance[i][j] = min(distance[i-1][j]+1,
                                distance[i][j-1]+1,
                                distance[i-1][j-1]+sc)
           if distance[i-1][j]+1 > distance[i-1][j-1]+sc and distance[i][j-1]+1 > distance[i-1][j-1]+sc:
               path.append("S");

    print path

    # Return the minimum distance using all the table cells
    return distance[i][j]

def subCost(x,y):
    if x == y:
        return 0
    else:
        return 2

def insertCost(x):
    path.append("I")
    return 1

def deleteCost(x):
    path.append("D")
    return 1
于 2013-01-09T08:58:12.200 回答
1

您正在计算Levenshtein 距离（或者更好的是加权 Levenshtein 距离，因为您的操作成本不同：I/ D=> 1，M=> 2）。

要获得操作的顺序，一种常见的方法是进行某种回溯。

考虑以下方法backtrace*：

...
# Return the minimum distance using all the table cells
def backtrace(i, j):
    if i>0 and j>0 and distance[i-1][j-1] + 2 == distance[i][j]:
        return backtrace(i-1, j-1) + "S"
    if i>0 and j>0 and distance[i-1][j-1] == distance[i][j]:
        return backtrace(i-1, j-1) + "M"
    if i>0 and distance[i-1][j] + 1 == distance[i][j]:
        return backtrace(i-1, j) + "D"
    if j>0 and distance[i][j-1] + 1 == distance[i][j]:
        return backtrace(i, j-1) + "I"
    return ""

return distance[i][j], backtrace(i, j)

我将它作为嵌套方法添加到您的方法中，因此我不必将您的距离矩阵distance作为参数传递给它。

现在你的脚本输出

S1 和 S2 之间的最小编辑距离为：(4, 'SMS') !

另请注意，如果您想在 python 中使用 Levenshtein 距离，谷歌代码上有一个名为 pylevenshtein的快速实现

* 可能不是 100% 准确 :-)

于 2013-01-09T09:54:22.243 回答