I checked ‘spam’ vs. ham (2)
‘hasddddddddddddddddam’ vs ‘ham’ (18)
Here is the full updated code:
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def dameraulevenshtein(seq1, seq2):
“”"Calculate the Damerau-Levenshtein distance between sequences.

This distance is the number of additions, deletions, substitutions,
and transpositions needed to transform the first sequence into the
second. Although generally used with strings, any sequences of
comparable objects will work.

Transpositions are exchanges of *consecutive* characters; all other
operations are self-explanatory.

This implementation is O(N*M) time and O(M) space, for N and M the
lengths of the two sequences.

>>> dameraulevenshtein(‘ba’, ‘abc’)
2
>>> dameraulevenshtein(‘fee’, ‘deed’)
2

It works with arbitrary sequences too:
>>> dameraulevenshtein(‘abcd’, ['b', 'a', 'c', 'd', 'e'])
2
“”"
# codesnippet:D0DE4716-B6E6-4161-9219-2903BF8F547F
# Conceptually, this is based on a len(seq1) + 1 * len(seq2) + 1 matrix.
# However, only the current and two previous rows are needed at once,
# so we only store those.
oneago = None
thisrow = list(range(1, len(seq2) + 1)) + [0]
for x in range(len(seq1)):
# Python lists wrap around for negative indices, so put the
# leftmost column at the *end* of the list. This matches with
# the zero-indexed strings and saves extra calculation.
twoago, oneago, thisrow = oneago, thisrow, [0] * len(seq2) + [x + 1]
for y in range(len(seq2)):
delcost = oneago[y] + 1
addcost = thisrow[y - 1] + 1
subcost = oneago[y - 1] + (seq1[x] != seq2[y])
thisrow[y] = min(delcost, addcost, subcost)
# This block deals with transpositions
if (x > 0 and y > 0 and seq1[x] == seq2[y - 1]
and seq1[x-1] == seq2[y] and seq1[x] != seq2[y]):
thisrow[y] = min(thisrow[y], twoago[y - 2] + 1)
return thisrow[len(seq2) - 1]

Both algorithms give the correct answer for that pair: delete H, U, R, replace B with K, transpose O and H, replace P with Z. Six steps.

but it’s not as nice as the one in difflib.