Levenshtein-Damerau, found originally in my sublime text 2 gist account (giststhebearbear) This is a memory/time optimized solution of the Levenshtein-Damerou distance algorithm. There are some non idiomatic things in the code, but it is pure standard python and can be used freely by anyone.

gistfile1.py

def leven(s1, s2, maxDistance):
    #  get smallest string so our rows are minimized
    s1, s2 = (s1, s2) if len(s1) <= len(s2) else (s2, s1)
    #  set lengths
    l1, l2 = len(s1), len(s2)
 
    #  We are simulatng an NM matrix where n is the longer string
    #  and m is the shorter string. By doing this we can minimize
    #  memory usage to O(M).
    #  Since we are simulating the matrix we only maintain two rows
    #  at a time the current row and the previous rows.
    #  A move from the current cell looking at the cell before it indicates
    #  consideration of an insert operation.
    #  A move from the current cell looking at the cell above it indicates
    #  consideration of a deletion
    #  Both operations are cost 1
    #  A move from the current cell to the cell up and to the left indicates
    #  an edit operation of 0 cost for a matching character and a 1 cost for
    #  a non matching characters
    #  no row has been previously computed yet, set empty row
    #  Since this is also a Damerou-Levenshtein calculation transposition
    #  costs will be taken into account. These look back 2 characters to
    #  determine optimal cost based on a possible transposition
    #  example: aei -> aie with levensthein has a cost of 2
    #  match a, change e->i change i->e => aie
    #  Damarau-Levenshtein has a cost of 1
    #  match a, transpose ei to ie => aie
    transpositionRow = None
    prevRow = None
 
    #  build first leven matrix row
    #  The first row represents transformation from an empty string
    #  to the shorter string making it static [0-n]
    #  since this row is static we can set it as
    #  curRow and start computation at the second row or index 1
    curRow = range(0, l1 + 1)]
 
    # use second length to loop through all the rows being built
    # we start at row one
    for rowNum in range(1, l2 + 1):
        #  set transposition, previous, and current
        #  because the rowNum always increments by one
        #  we can use rowNum to set the value representing
        #  the first column which is indicitive of transforming TO
        #  the empty string from our longer string
        #  transposition row maintains an extra row so that it is possible
        #  for us to apply Damarou's formula
        transpositionRow, prevRow, curRow = prevRow, curRow, [rowNum] + [0] * l1
 
        #  consider if we have passed the max distance if all paths through
        #  the transposition row are larger than the max we can stop calculating
        #  distance and return the last element in that row and return the max
        if transpositionRow:
            if not any(cellValue < maxDistance for cellValue in transpositionRow):
                return maxDistance
 
        for colNum in range(1, l1 + 1):
            insertionCost = curRow[colNum - 1] + 1
            deletionCost = prevRow[colNum] + 1
            changeCost = prevRow[colNum - 1] + (0 if s1[colNum - 1] == s2[rowNum - 1] else 1)
            #  set the cell value - min distance to reach this
            #  position
            curRow[colNum] = min(insertionCost, deletionCost, changeCost)
 
            #  test for a possible transposition optimization
            #  check to see if we have at least 2 characters
            if 1 < rowNum <= colNum:
                #  test for possible transposition
                if s1[colNum - 1] == s2[colNum - 2] and s2[colNum - 1] == s1[colNum - 2]:
                    curRow[colNum] = min(curRow[colNum], transpositionRow[colNum - 2] + 1)
 
    #  the last cell of the matrix is ALWAYS the shortest distance between the two strings
    return curRow[-1]

Levenshtein-Damerau

def leven(s1, s2, maxDistance):
    #  get smallest string so our rows are minimized
    s1, s2 = (s1, s2) if len(s1) <= len(s2) else (s2, s1)
    #  set lengths
    l1, l2 = len(s1), len(s2)
 
    #  We are simulatng an NM matrix where n is the longer string
    #  and m is the shorter string. By doing this we can minimize
    #  memory usage to O(M).
    #  Since we are simulating the matrix we only maintain two rows
    #  at a time the current row and the previous rows.
    #  A move from the current cell looking at the cell before it indicates
    #  consideration of an insert operation.
    #  A move from the current cell looking at the cell above it indicates
    #  consideration of a deletion
    #  Both operations are cost 1
    #  A move from the current cell to the cell up and to the left indicates
    #  an edit operation of 0 cost for a matching character and a 1 cost for
    #  a non matching characters
    #  no row has been previously computed yet, set empty row
    #  Since this is also a Damerou-Levenshtein calculation transposition
    #  costs will be taken into account. These look back 2 characters to
    #  determine optimal cost based on a possible transposition
    #  example: aei -> aie with levensthein has a cost of 2
    #  match a, change e->i change i->e => aie
    #  Damarau-Levenshtein has a cost of 1
    #  match a, transpose ei to ie => aie
    transpositionRow = None
    prevRow = None
 
    #  build first leven matrix row
    #  The first row represents transformation from an empty string
    #  to the shorter string making it static [0-n]
    #  since this row is static we can set it as
    #  curRow and start computation at the second row or index 1
    curRow = range(0, l1 + 1)]
 
    # use second length to loop through all the rows being built
    # we start at row one
    for rowNum in range(1, l2 + 1):
        #  set transposition, previous, and current
        #  because the rowNum always increments by one
        #  we can use rowNum to set the value representing
        #  the first column which is indicitive of transforming TO
        #  the empty string from our longer string
        #  transposition row maintains an extra row so that it is possible
        #  for us to apply Damarou's formula
        transpositionRow, prevRow, curRow = prevRow, curRow, [rowNum] + [0] * l1
 
        #  consider if we have passed the max distance if all paths through
        #  the transposition row are larger than the max we can stop calculating
        #  distance and return the last element in that row and return the max
        if transpositionRow:
            if not any(cellValue < maxDistance for cellValue in transpositionRow):
                return maxDistance
 
        for colNum in range(1, l1 + 1):
            insertionCost = curRow[colNum - 1] + 1
            deletionCost = prevRow[colNum] + 1
            changeCost = prevRow[colNum - 1] + (0 if s1[colNum - 1] == s2[rowNum - 1] else 1)
            #  set the cell value - min distance to reach this
            #  position
            curRow[colNum] = min(insertionCost, deletionCost, changeCost)
 
            #  test for a possible transposition optimization
            #  check to see if we have at least 2 characters
            if 1 < rowNum <= colNum:
                #  test for possible transposition
                if s1[colNum - 1] == s2[colNum - 2] and s2[colNum - 1] == s1[colNum - 2]:
                    curRow[colNum] = min(curRow[colNum], transpositionRow[colNum - 2] + 1)
 
    #  the last cell of the matrix is ALWAYS the shortest distance between the two strings
    return curRow[-1]