Minimum Edit Distance in Python

Minimum Edit Distance algorithm

Minimum Edit Distance algorithm

This is my quick-and-dirty implementation of the Minimum Edit Distance algorithm in Python.

class EditDistance(object):

    def __init__(self):
        pass

    def costInsert(self, s):
        return 1

    def costDelete(self, s):
        return 1

    def costSubstitution(self, target, source):
        return (0 if target == source else 2)

    def minEditDistance(self, target, source):
        n = len(target)
        m = len(source)
        assert n > 0 and m > 0

        distance = np.zeros((n+1, m+1))
        backPt = np.zeros((n+1, m+1), dtype=int)
        opCount = np.zeros((n+1, m+1), dtype=int)
        backPt[0, 1:] = 3
        backPt[1:, 0] = 1
        opCount[0, 1:] = xrange(1, m+1)
        opCount[1:, 0] = xrange(1, n+1)

        for i in xrange(1, n+1):
            distance[i, 0] = distance[i-1, 0] + self.costInsert(target[i-1])
        for j in xrange(1, m+1):
            distance[0, j] = distance[0, j-1] + self.costDelete(source[j-1])

        for i in xrange(1, n+1):
            for j in xrange(1, m+1):
                d = [distance[i-1, j] + self.costInsert(target[i-1]), \
                    distance[i-1, j-1] + self.costSubstitution(source[j-1], target[i-1]), \
                    distance[i, j-1] + self.costDelete(source[j-1])]
                distance[i, j] = min(d)
                op = [opCount[i-1, j] + 1, \
                    opCount[i-1, j-1] + (0 if source[j-1] == target[i-1] else 1), \
                    opCount[i, j-1] + 1]
                backPt[i, j] = 1 + np.argmin(op)
                opCount[i, j] = min(op)

        # backtrace
        counts = [0, 0, 0]
        alignedStrings = [[], []]
        i = n
        j = m

        while i != 0 or j != 0:
            pt = backPt[i, j]
            assert pt in [1, 2, 3]
            if pt == 1:
                counts[pt-1] += 1
                alignedStrings[0].append(target[i-1])
                alignedStrings[1].append('*')
                i -= 1
            elif pt == 2:
                counts[pt-1] += (0 if target[i-1] == source[j-1] else 1)
                alignedStrings[0].append(target[i-1])
                alignedStrings[1].append(source[j-1])
                i -= 1
                j -= 1
            else:
                counts[pt-1] += 1
                alignedStrings[0].append('*')
                alignedStrings[1].append(source[j-1])
                j -= 1
        alignedStrings = [s[::-1] for s in alignedStrings]
        return (counts, distance[n, m], alignedStrings)

Note that I am using weights of 1, 1 and 2 for the cost of insertion, deletion and substitution. Standard packages might use other values (like 7, 7, 10 in HTK), but it is easy to change.

The algorithm can be found in the image. In the Python implementation, I already modified it to compute the back-traced minimum distance alignment. Since there might be several alignments that have the same edit distance, I select the one that minimize the sum of deletion, insertion and substitution operations. This is eventually corresponding to the alignment that minimizes Word Error Rate. The “best” alignment is also returned, so this implementation can be a good demonstration for studying this algorithm.
As always, the full source code can be found on github.

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