计算不相等字符串python的最小编辑距离

Question

我正在尝试以2的替换成本实现最小编辑距离。以下是我到目前为止的代码。它对于长度相等的字符串效果很好，但是对于不相等的字符串会产生错误。请纠正我哪里错了

def med(source, target):
#     if len(x) > len(y):
#         print("insode if")
#         source, target = y, x
print(len(source), len(target))
cost = [[0 for inner in range(len(source)+1)] for outer in 
range(len(target)+1)]

global backtrace 
backtrace = [[0 for inner in range(len(source)+1)] for outer in 
range(len(target)+1)]
global SUB
global INS
global DEL

for i in range(0,len(target)+1):
    cost[i][0] = i

for j in range(0,len(source)+1):
    cost[0][j] = j

for i in range(1,len(target)+1):
    for j in range(1,len(source)+1):
        if source[i-1]==target[j-1]:
            cost[i][j] = cost[i-1][j-1] 
        else:
            deletion = cost[i-1][j]+1
            insertion = cost[i][j-1]+1
            substitution = cost[i-1][j-1]+2
            cost[i][j] = min(insertion,deletion,substitution)

            if cost[i][j] == substitution:
                backtrace[i][j] = SUB
            elif cost[i][j] == insertion:
                backtrace[i][j] = INS
            else:
                backtrace[i][j] = DEL


return cost[i][j]

med("levenshtein","levels")

我得到的错误是：

---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
<ipython-input-26-86bf20ea27c7> in <module>()
 49     return cost[i][j]
 50 
---> 51 med("levenshtein","levels")

<ipython-input-26-86bf20ea27c7> in med(source, target)
 31     for i in range(1,len(target)+1):
 32         for j in range(1,len(source)+1):
---> 33             if source[i-1]==target[j-1]:
 34                 cost[i][j] = cost[i-1][j-1]
 35             else:

IndexError: string index out of range

Answer 1

对于不同长度的字符串，cost和backtrace的索引不匹配。

可以通过在每一步仅更新一个numpy m * n arr来实现具有2个替换成本的最小编辑距离。

根据算法， 下面的代码将完成这项工作。

def minimumEditDistance(first, second): 

    #Creating numpy ndarray( initialized with 0 of dimension of size of both strings

    matrix = np.zeros((len(first)+1,len(second)+1), dtype=np.int)


    # Cross relation loop through each character of each string with each other and
    # fill the respective index of matrxi (row,column)

    for i in range(len(first)+1): 
        for j in range(len(second)+1): 

            #First doing the boundary value analysis, if first or second string is empty so directly adding insertion cost
            if i == 0:  
                matrix[i][j] = j  
            #Second case
            elif j == 0: 
                matrix[i][j] = i
            else: 
                matrix[i][j] = min(matrix[i][j-1] + 1,  
                                   matrix[i-1][j] + 1,        
                                   matrix[i-1][j-1] + 2 if first[i-1] != second[j-1] else matrix[i-1][j-1] + 0)     
                                   # Adjusted the cost accordinly, insertion = 1, deletion=1 and substitution=2
    return matrix[len(first)][len(second)]  # Returning the final

输出

>>>print(minimumEditDistance('levenshtein','levels'))
7
>>>print(minimumEditDistance('levenshtein','levenshtien'))
0