Moved SequenceMatcher from ndiff into new std library module difflib.py.

Guido told me to do this <wink>.
Greatly expanded docstrings, and fleshed out with examples.
New std test.
Added new get_close_matches() function for ESR.
Needs docs, but LaTeXification of the module docstring is all it needs.
\CVS: ----------------------------------------------------------------------

1 files changed, 781 insertions, 0 deletions

diff --git a/Lib/difflib.py b/Lib/difflib.py
new file mode 100644
index 0000000..3997723
--- /dev/null
+++ b/Lib/difflib.py
@@ -0,0 +1,781 @@
+#! /usr/bin/env python
+
+"""
+Module difflib -- helpers for computing deltas between objects.
+
+Function get_close_matches(word, possibilities, n=3, cutoff=0.6):
+
+    Use SequenceMatcher to return list of the best "good enough" matches.
+
+    word is a sequence for which close matches are desired (typically a
+    string).
+
+    possibilities is a list of sequences against which to match word
+    (typically a list of strings).
+
+    Optional arg n (default 3) is the maximum number of close matches to
+    return.  n must be > 0.
+
+    Optional arg cutoff (default 0.6) is a float in [0, 1].  Possibilities
+    that don't score at least that similar to word are ignored.
+
+    The best (no more than n) matches among the possibilities are returned
+    in a list, sorted by similarity score, most similar first.
+
+    >>> get_close_matches("appel", ["ape", "apple", "peach", "puppy"])
+    ['apple', 'ape']
+    >>> import keyword
+    >>> get_close_matches("wheel", keyword.kwlist)
+    ['while']
+    >>> get_close_matches("apple", keyword.kwlist)
+    []
+    >>> get_close_matches("accept", keyword.kwlist)
+    ['except']
+
+Class SequenceMatcher
+
+SequenceMatcher is a flexible class for comparing pairs of sequences of any
+type, so long as the sequence elements are hashable.  The basic algorithm
+predates, and is a little fancier than, an algorithm published in the late
+1980's by Ratcliff and Obershelp under the hyperbolic name "gestalt pattern
+matching".  The basic idea is to find the longest contiguous matching
+subsequence that contains no "junk" elements (R-O doesn't address junk).
+The same idea is then applied recursively to the pieces of the sequences to
+the left and to the right of the matching subsequence.  This does not yield
+minimal edit sequences, but does tend to yield matches that "look right"
+to people.
+
+Example, comparing two strings, and considering blanks to be "junk":
+
+>>> s = SequenceMatcher(lambda x: x == " ",
+...                     "private Thread currentThread;",
+...                     "private volatile Thread currentThread;")
+>>>
+
+.ratio() returns a float in [0, 1], measuring the "similarity" of the
+sequences.  As a rule of thumb, a .ratio() value over 0.6 means the
+sequences are close matches:
+
+>>> print round(s.ratio(), 3)
+0.866
+>>>
+
+If you're only interested in where the sequences match,
+.get_matching_blocks() is handy:
+
+>>> for block in s.get_matching_blocks():
+...     print "a[%d] and b[%d] match for %d elements" % block
+a[0] and b[0] match for 8 elements
+a[8] and b[17] match for 6 elements
+a[14] and b[23] match for 15 elements
+a[29] and b[38] match for 0 elements
+
+Note that the last tuple returned by .get_matching_blocks() is always a
+dummy, (len(a), len(b), 0), and this is the only case in which the last
+tuple element (number of elements matched) is 0.
+
+If you want to know how to change the first sequence into the second, use
+.get_opcodes():
+
+>>> for opcode in s.get_opcodes():
+...     print "%6s a[%d:%d] b[%d:%d]" % opcode
+ equal a[0:8] b[0:8]
+insert a[8:8] b[8:17]
+ equal a[8:14] b[17:23]
+ equal a[14:29] b[23:38]
+
+See Tools/scripts/ndiff.py for a fancy human-friendly file differencer,
+which uses SequenceMatcher both to view files as sequences of lines, and
+lines as sequences of characters.
+
+See also function get_close_matches() in this module, which shows how
+simple code building on SequenceMatcher can be used to do useful work.
+
+Timing:  Basic R-O is cubic time worst case and quadratic time expected
+case.  SequenceMatcher is quadratic time worst case and has expected-case
+behavior dependent on how many elements the sequences have in common; best
+case time (no elements in common) is linear.
+
+SequenceMatcher methods:
+
+__init__(isjunk=None, a='', b='')
+    Construct a SequenceMatcher.
+
+    Optional arg isjunk is None (the default), or a one-argument function
+    that takes a sequence element and returns true iff the element is junk.
+    None is equivalent to passing "lambda x: 0", i.e. no elements are
+    considered to be junk.  For examples, pass
+        lambda x: x in " \\t"
+    if you're comparing lines as sequences of characters, and don't want to
+    synch up on blanks or hard tabs.
+
+    Optional arg a is the first of two sequences to be compared.  By
+    default, an empty string.  The elements of a must be hashable.
+
+    Optional arg b is the second of two sequences to be compared.  By
+    default, an empty string.  The elements of b must be hashable.
+
+set_seqs(a, b)
+    Set the two sequences to be compared.
+
+    >>> s = SequenceMatcher()
+    >>> s.set_seqs("abcd", "bcde")
+    >>> s.ratio()
+    0.75
+
+set_seq1(a)
+    Set the first sequence to be compared.
+
+    The second sequence to be compared is not changed.
+
+    >>> s = SequenceMatcher(None, "abcd", "bcde")
+    >>> s.ratio()
+    0.75
+    >>> s.set_seq1("bcde")
+    >>> s.ratio()
+    1.0
+    >>>
+
+    SequenceMatcher computes and caches detailed information about the
+    second sequence, so if you want to compare one sequence S against many
+    sequences, use .set_seq2(S) once and call .set_seq1(x) repeatedly for
+    each of the other sequences.
+
+    See also set_seqs() and set_seq2().
+
+set_seq2(b)
+    Set the second sequence to be compared.
+
+    The first sequence to be compared is not changed.
+
+    >>> s = SequenceMatcher(None, "abcd", "bcde")
+    >>> s.ratio()
+    0.75
+    >>> s.set_seq2("abcd")
+    >>> s.ratio()
+    1.0
+    >>>
+
+    SequenceMatcher computes and caches detailed information about the
+    second sequence, so if you want to compare one sequence S against many
+    sequences, use .set_seq2(S) once and call .set_seq1(x) repeatedly for
+    each of the other sequences.
+
+    See also set_seqs() and set_seq1().
+
+find_longest_match(alo, ahi, blo, bhi)
+    Find longest matching block in a[alo:ahi] and b[blo:bhi].
+
+    If isjunk is not defined:
+
+    Return (i,j,k) such that a[i:i+k] is equal to b[j:j+k], where
+        alo <= i <= i+k <= ahi
+        blo <= j <= j+k <= bhi
+    and for all (i',j',k') meeting those conditions,
+        k >= k'
+        i <= i'
+        and if i == i', j <= j'
+
+    In other words, of all maximal matching blocks, return one that starts
+    earliest in a, and of all those maximal matching blocks that start
+    earliest in a, return the one that starts earliest in b.
+
+    >>> s = SequenceMatcher(None, " abcd", "abcd abcd")
+    >>> s.find_longest_match(0, 5, 0, 9)
+    (0, 4, 5)
+
+    If isjunk is defined, first the longest matching block is determined as
+    above, but with the additional restriction that no junk element appears
+    in the block.  Then that block is extended as far as possible by
+    matching (only) junk elements on both sides.  So the resulting block
+    never matches on junk except as identical junk happens to be adjacent
+    to an "interesting" match.
+
+    Here's the same example as before, but considering blanks to be junk.
+    That prevents " abcd" from matching the " abcd" at the tail end of the
+    second sequence directly.  Instead only the "abcd" can match, and
+    matches the leftmost "abcd" in the second sequence:
+
+    >>> s = SequenceMatcher(lambda x: x==" ", " abcd", "abcd abcd")
+    >>> s.find_longest_match(0, 5, 0, 9)
+    (1, 0, 4)
+
+    If no blocks match, return (alo, blo, 0).
+
+    >>> s = SequenceMatcher(None, "ab", "c")
+    >>> s.find_longest_match(0, 2, 0, 1)
+    (0, 0, 0)
+
+get_matching_blocks()
+    Return list of triples describing matching subsequences.
+
+    Each triple is of the form (i, j, n), and means that
+    a[i:i+n] == b[j:j+n].  The triples are monotonically increasing in i
+    and in j.
+
+    The last triple is a dummy, (len(a), len(b), 0), and is the only triple
+    with n==0.
+
+    >>> s = SequenceMatcher(None, "abxcd", "abcd")
+    >>> s.get_matching_blocks()
+    [(0, 0, 2), (3, 2, 2), (5, 4, 0)]
+
+get_opcodes()
+    Return list of 5-tuples describing how to turn a into b.
+
+    Each tuple is of the form (tag, i1, i2, j1, j2).  The first tuple has
+    i1 == j1 == 0, and remaining tuples have i1 == the i2 from the tuple
+    preceding it, and likewise for j1 == the previous j2.
+
+    The tags are strings, with these meanings:
+
+    'replace':  a[i1:i2] should be replaced by b[j1:j2]
+    'delete':   a[i1:i2] should be deleted.
+                Note that j1==j2 in this case.
+    'insert':   b[j1:j2] should be inserted at a[i1:i1].
+                Note that i1==i2 in this case.
+    'equal':    a[i1:i2] == b[j1:j2]
+
+    >>> a = "qabxcd"
+    >>> b = "abycdf"
+    >>> s = SequenceMatcher(None, a, b)
+    >>> for tag, i1, i2, j1, j2 in s.get_opcodes():
+    ...    print ("%7s a[%d:%d] (%s) b[%d:%d] (%s)" %
+    ...           (tag, i1, i2, a[i1:i2], j1, j2, b[j1:j2]))
+     delete a[0:1] (q) b[0:0] ()
+      equal a[1:3] (ab) b[0:2] (ab)
+    replace a[3:4] (x) b[2:3] (y)
+      equal a[4:6] (cd) b[3:5] (cd)
+     insert a[6:6] () b[5:6] (f)
+
+ratio()
+    Return a measure of the sequences' similarity (float in [0,1]).
+
+    Where T is the total number of elements in both sequences, and M is the
+    number of matches, this is 2,0*M / T. Note that this is 1 if the
+    sequences are identical, and 0 if they have nothing in common.
+
+    .ratio() is expensive to compute if you haven't already computed
+    .get_matching_blocks() or .get_opcodes(), in which case you may want to
+    try .quick_ratio() or .real_quick_ratio() first to get an upper bound.
+
+    >>> s = SequenceMatcher(None, "abcd", "bcde")
+    >>> s.ratio()
+    0.75
+    >>> s.quick_ratio()
+    0.75
+    >>> s.real_quick_ratio()
+    1.0
+
+quick_ratio()
+    Return an upper bound on .ratio() relatively quickly.
+
+    This isn't defined beyond that it is an upper bound on .ratio(), and
+    is faster to compute.
+
+real_quick_ratio():
+    Return an upper bound on ratio() very quickly.
+
+    This isn't defined beyond that it is an upper bound on .ratio(), and
+    is faster to compute than either .ratio() or .quick_ratio().
+"""
+
+TRACE = 0
+
+class SequenceMatcher:
+    def __init__(self, isjunk=None, a='', b=''):
+        """Construct a SequenceMatcher.
+
+        Optional arg isjunk is None (the default), or a one-argument
+        function that takes a sequence element and returns true iff the
+        element is junk. None is equivalent to passing "lambda x: 0", i.e.
+        no elements are considered to be junk.  For examples, pass
+            lambda x: x in " \\t"
+        if you're comparing lines as sequences of characters, and don't
+        want to synch up on blanks or hard tabs.
+
+        Optional arg a is the first of two sequences to be compared.  By
+        default, an empty string.  The elements of a must be hashable.  See
+        also .set_seqs() and .set_seq1().
+
+        Optional arg b is the second of two sequences to be compared.  By
+        default, an empty string.  The elements of a must be hashable. See
+        also .set_seqs() and .set_seq2().
+        """
+
+        # Members:
+        # a
+        #      first sequence
+        # b
+        #      second sequence; differences are computed as "what do
+        #      we need to do to 'a' to change it into 'b'?"
+        # b2j
+        #      for x in b, b2j[x] is a list of the indices (into b)
+        #      at which x appears; junk elements do not appear
+        # b2jhas
+        #      b2j.has_key
+        # fullbcount
+        #      for x in b, fullbcount[x] == the number of times x
+        #      appears in b; only materialized if really needed (used
+        #      only for computing quick_ratio())
+        # matching_blocks
+        #      a list of (i, j, k) triples, where a[i:i+k] == b[j:j+k];
+        #      ascending & non-overlapping in i and in j; terminated by
+        #      a dummy (len(a), len(b), 0) sentinel
+        # opcodes
+        #      a list of (tag, i1, i2, j1, j2) tuples, where tag is
+        #      one of
+        #          'replace'   a[i1:i2] should be replaced by b[j1:j2]
+        #          'delete'    a[i1:i2] should be deleted
+        #          'insert'    b[j1:j2] should be inserted
+        #          'equal'     a[i1:i2] == b[j1:j2]
+        # isjunk
+        #      a user-supplied function taking a sequence element and
+        #      returning true iff the element is "junk" -- this has
+        #      subtle but helpful effects on the algorithm, which I'll
+        #      get around to writing up someday <0.9 wink>.
+        #      DON'T USE!  Only __chain_b uses this.  Use isbjunk.
+        # isbjunk
+        #      for x in b, isbjunk(x) == isjunk(x) but much faster;
+        #      it's really the has_key method of a hidden dict.
+        #      DOES NOT WORK for x in a!
+
+        self.isjunk = isjunk
+        self.a = self.b = None
+        self.set_seqs(a, b)
+
+    def set_seqs(self, a, b):
+        """Set the two sequences to be compared.
+
+        >>> s = SequenceMatcher()
+        >>> s.set_seqs("abcd", "bcde")
+        >>> s.ratio()
+        0.75
+        """
+
+        self.set_seq1(a)
+        self.set_seq2(b)
+
+    def set_seq1(self, a):
+        """Set the first sequence to be compared.
+
+        The second sequence to be compared is not changed.
+
+        >>> s = SequenceMatcher(None, "abcd", "bcde")
+        >>> s.ratio()
+        0.75
+        >>> s.set_seq1("bcde")
+        >>> s.ratio()
+        1.0
+        >>>
+
+        SequenceMatcher computes and caches detailed information about the
+        second sequence, so if you want to compare one sequence S against
+        many sequences, use .set_seq2(S) once and call .set_seq1(x)
+        repeatedly for each of the other sequences.
+
+        See also set_seqs() and set_seq2().
+        """
+
+        if a is self.a:
+            return
+        self.a = a
+        self.matching_blocks = self.opcodes = None
+
+    def set_seq2(self, b):
+        """Set the second sequence to be compared.
+
+        The first sequence to be compared is not changed.
+
+        >>> s = SequenceMatcher(None, "abcd", "bcde")
+        >>> s.ratio()
+        0.75
+        >>> s.set_seq2("abcd")
+        >>> s.ratio()
+        1.0
+        >>>
+
+        SequenceMatcher computes and caches detailed information about the
+        second sequence, so if you want to compare one sequence S against
+        many sequences, use .set_seq2(S) once and call .set_seq1(x)
+        repeatedly for each of the other sequences.
+
+        See also set_seqs() and set_seq1().
+        """
+
+        if b is self.b:
+            return
+        self.b = b
+        self.matching_blocks = self.opcodes = None
+        self.fullbcount = None
+        self.__chain_b()
+
+    # For each element x in b, set b2j[x] to a list of the indices in
+    # b where x appears; the indices are in increasing order; note that
+    # the number of times x appears in b is len(b2j[x]) ...
+    # when self.isjunk is defined, junk elements don't show up in this
+    # map at all, which stops the central find_longest_match method
+    # from starting any matching block at a junk element ...
+    # also creates the fast isbjunk function ...
+    # note that this is only called when b changes; so for cross-product
+    # kinds of matches, it's best to call set_seq2 once, then set_seq1
+    # repeatedly
+
+    def __chain_b(self):
+        # Because isjunk is a user-defined (not C) function, and we test
+        # for junk a LOT, it's important to minimize the number of calls.
+        # Before the tricks described here, __chain_b was by far the most
+        # time-consuming routine in the whole module!  If anyone sees
+        # Jim Roskind, thank him again for profile.py -- I never would
+        # have guessed that.
+        # The first trick is to build b2j ignoring the possibility
+        # of junk.  I.e., we don't call isjunk at all yet.  Throwing
+        # out the junk later is much cheaper than building b2j "right"
+        # from the start.
+        b = self.b
+        self.b2j = b2j = {}
+        self.b2jhas = b2jhas = b2j.has_key
+        for i in xrange(len(b)):
+            elt = b[i]
+            if b2jhas(elt):
+                b2j[elt].append(i)
+            else:
+                b2j[elt] = [i]
+
+        # Now b2j.keys() contains elements uniquely, and especially when
+        # the sequence is a string, that's usually a good deal smaller
+        # than len(string).  The difference is the number of isjunk calls
+        # saved.
+        isjunk, junkdict = self.isjunk, {}
+        if isjunk:
+            for elt in b2j.keys():
+                if isjunk(elt):
+                    junkdict[elt] = 1   # value irrelevant; it's a set
+                    del b2j[elt]
+
+        # Now for x in b, isjunk(x) == junkdict.has_key(x), but the
+        # latter is much faster.  Note too that while there may be a
+        # lot of junk in the sequence, the number of *unique* junk
+        # elements is probably small.  So the memory burden of keeping
+        # this dict alive is likely trivial compared to the size of b2j.
+        self.isbjunk = junkdict.has_key
+
+    def find_longest_match(self, alo, ahi, blo, bhi):
+        """Find longest matching block in a[alo:ahi] and b[blo:bhi].
+
+        If isjunk is not defined:
+
+        Return (i,j,k) such that a[i:i+k] is equal to b[j:j+k], where
+            alo <= i <= i+k <= ahi
+            blo <= j <= j+k <= bhi
+        and for all (i',j',k') meeting those conditions,
+            k >= k'
+            i <= i'
+            and if i == i', j <= j'
+
+        In other words, of all maximal matching blocks, return one that
+        starts earliest in a, and of all those maximal matching blocks that
+        start earliest in a, return the one that starts earliest in b.
+
+        >>> s = SequenceMatcher(None, " abcd", "abcd abcd")
+        >>> s.find_longest_match(0, 5, 0, 9)
+        (0, 4, 5)
+
+        If isjunk is defined, first the longest matching block is
+        determined as above, but with the additional restriction that no
+        junk element appears in the block.  Then that block is extended as
+        far as possible by matching (only) junk elements on both sides.  So
+        the resulting block never matches on junk except as identical junk
+        happens to be adjacent to an "interesting" match.
+
+        Here's the same example as before, but considering blanks to be
+        junk.  That prevents " abcd" from matching the " abcd" at the tail
+        end of the second sequence directly.  Instead only the "abcd" can
+        match, and matches the leftmost "abcd" in the second sequence:
+
+        >>> s = SequenceMatcher(lambda x: x==" ", " abcd", "abcd abcd")
+        >>> s.find_longest_match(0, 5, 0, 9)
+        (1, 0, 4)
+
+        If no blocks match, return (alo, blo, 0).
+
+        >>> s = SequenceMatcher(None, "ab", "c")
+        >>> s.find_longest_match(0, 2, 0, 1)
+        (0, 0, 0)
+        """
+
+        # CAUTION:  stripping common prefix or suffix would be incorrect.
+        # E.g.,
+        #    ab
+        #    acab
+        # Longest matching block is "ab", but if common prefix is
+        # stripped, it's "a" (tied with "b").  UNIX(tm) diff does so
+        # strip, so ends up claiming that ab is changed to acab by
+        # inserting "ca" in the middle.  That's minimal but unintuitive:
+        # "it's obvious" that someone inserted "ac" at the front.
+        # Windiff ends up at the same place as diff, but by pairing up
+        # the unique 'b's and then matching the first two 'a's.
+
+        a, b, b2j, isbjunk = self.a, self.b, self.b2j, self.isbjunk
+        besti, bestj, bestsize = alo, blo, 0
+        # find longest junk-free match
+        # during an iteration of the loop, j2len[j] = length of longest
+        # junk-free match ending with a[i-1] and b[j]
+        j2len = {}
+        nothing = []
+        for i in xrange(alo, ahi):
+            # look at all instances of a[i] in b; note that because
+            # b2j has no junk keys, the loop is skipped if a[i] is junk
+            j2lenget = j2len.get
+            newj2len = {}
+            for j in b2j.get(a[i], nothing):
+                # a[i] matches b[j]
+                if j < blo:
+                    continue
+                if j >= bhi:
+                    break
+                k = newj2len[j] = j2lenget(j-1, 0) + 1
+                if k > bestsize:
+                    besti, bestj, bestsize = i-k+1, j-k+1, k
+            j2len = newj2len
+
+        # Now that we have a wholly interesting match (albeit possibly
+        # empty!), we may as well suck up the matching junk on each
+        # side of it too.  Can't think of a good reason not to, and it
+        # saves post-processing the (possibly considerable) expense of
+        # figuring out what to do with it.  In the case of an empty
+        # interesting match, this is clearly the right thing to do,
+        # because no other kind of match is possible in the regions.
+        while besti > alo and bestj > blo and \
+              isbjunk(b[bestj-1]) and \
+              a[besti-1] == b[bestj-1]:
+            besti, bestj, bestsize = besti-1, bestj-1, bestsize+1
+        while besti+bestsize < ahi and bestj+bestsize < bhi and \
+              isbjunk(b[bestj+bestsize]) and \
+              a[besti+bestsize] == b[bestj+bestsize]:
+            bestsize = bestsize + 1
+
+        if TRACE:
+            print "get_matching_blocks", alo, ahi, blo, bhi
+            print "    returns", besti, bestj, bestsize
+        return besti, bestj, bestsize
+
+    def get_matching_blocks(self):
+        """Return list of triples describing matching subsequences.
+
+        Each triple is of the form (i, j, n), and means that
+        a[i:i+n] == b[j:j+n].  The triples are monotonically increasing in
+        i and in j.
+
+        The last triple is a dummy, (len(a), len(b), 0), and is the only
+        triple with n==0.
+
+        >>> s = SequenceMatcher(None, "abxcd", "abcd")
+        >>> s.get_matching_blocks()
+        [(0, 0, 2), (3, 2, 2), (5, 4, 0)]
+        """
+
+        if self.matching_blocks is not None:
+            return self.matching_blocks
+        self.matching_blocks = []
+        la, lb = len(self.a), len(self.b)
+        self.__helper(0, la, 0, lb, self.matching_blocks)
+        self.matching_blocks.append( (la, lb, 0) )
+        if TRACE:
+            print '*** matching blocks', self.matching_blocks
+        return self.matching_blocks
+
+    # builds list of matching blocks covering a[alo:ahi] and
+    # b[blo:bhi], appending them in increasing order to answer
+
+    def __helper(self, alo, ahi, blo, bhi, answer):
+        i, j, k = x = self.find_longest_match(alo, ahi, blo, bhi)
+        # a[alo:i] vs b[blo:j] unknown
+        # a[i:i+k] same as b[j:j+k]
+        # a[i+k:ahi] vs b[j+k:bhi] unknown
+        if k:
+            if alo < i and blo < j:
+                self.__helper(alo, i, blo, j, answer)
+            answer.append(x)
+            if i+k < ahi and j+k < bhi:
+                self.__helper(i+k, ahi, j+k, bhi, answer)
+
+    def get_opcodes(self):
+        """Return list of 5-tuples describing how to turn a into b.
+
+        Each tuple is of the form (tag, i1, i2, j1, j2).  The first tuple
+        has i1 == j1 == 0, and remaining tuples have i1 == the i2 from the
+        tuple preceding it, and likewise for j1 == the previous j2.
+
+        The tags are strings, with these meanings:
+
+        'replace':  a[i1:i2] should be replaced by b[j1:j2]
+        'delete':   a[i1:i2] should be deleted.
+                    Note that j1==j2 in this case.
+        'insert':   b[j1:j2] should be inserted at a[i1:i1].
+                    Note that i1==i2 in this case.
+        'equal':    a[i1:i2] == b[j1:j2]
+
+        >>> a = "qabxcd"
+        >>> b = "abycdf"
+        >>> s = SequenceMatcher(None, a, b)
+        >>> for tag, i1, i2, j1, j2 in s.get_opcodes():
+        ...    print ("%7s a[%d:%d] (%s) b[%d:%d] (%s)" %
+        ...           (tag, i1, i2, a[i1:i2], j1, j2, b[j1:j2]))
+         delete a[0:1] (q) b[0:0] ()
+          equal a[1:3] (ab) b[0:2] (ab)
+        replace a[3:4] (x) b[2:3] (y)
+          equal a[4:6] (cd) b[3:5] (cd)
+         insert a[6:6] () b[5:6] (f)
+        """
+
+        if self.opcodes is not None:
+            return self.opcodes
+        i = j = 0
+        self.opcodes = answer = []
+        for ai, bj, size in self.get_matching_blocks():
+            # invariant:  we've pumped out correct diffs to change
+            # a[:i] into b[:j], and the next matching block is
+            # a[ai:ai+size] == b[bj:bj+size].  So we need to pump
+            # out a diff to change a[i:ai] into b[j:bj], pump out
+            # the matching block, and move (i,j) beyond the match
+            tag = ''
+            if i < ai and j < bj:
+                tag = 'replace'
+            elif i < ai:
+                tag = 'delete'
+            elif j < bj:
+                tag = 'insert'
+            if tag:
+                answer.append( (tag, i, ai, j, bj) )
+            i, j = ai+size, bj+size
+            # the list of matching blocks is terminated by a
+            # sentinel with size 0
+            if size:
+                answer.append( ('equal', ai, i, bj, j) )
+        return answer
+
+    def ratio(self):
+        """Return a measure of the sequences' similarity (float in [0,1]).
+
+        Where T is the total number of elements in both sequences, and
+        M is the number of matches, this is 2,0*M / T.
+        Note that this is 1 if the sequences are identical, and 0 if
+        they have nothing in common.
+
+        .ratio() is expensive to compute if you haven't already computed
+        .get_matching_blocks() or .get_opcodes(), in which case you may
+        want to try .quick_ratio() or .real_quick_ratio() first to get an
+        upper bound.
+
+        >>> s = SequenceMatcher(None, "abcd", "bcde")
+        >>> s.ratio()
+        0.75
+        >>> s.quick_ratio()
+        0.75
+        >>> s.real_quick_ratio()
+        1.0
+        """
+
+        matches = reduce(lambda sum, triple: sum + triple[-1],
+                         self.get_matching_blocks(), 0)
+        return 2.0 * matches / (len(self.a) + len(self.b))
+
+    def quick_ratio(self):
+        """Return an upper bound on ratio() relatively quickly.
+
+        This isn't defined beyond that it is an upper bound on .ratio(), and
+        is faster to compute.
+        """
+
+        # viewing a and b as multisets, set matches to the cardinality
+        # of their intersection; this counts the number of matches
+        # without regard to order, so is clearly an upper bound
+        if self.fullbcount is None:
+            self.fullbcount = fullbcount = {}
+            for elt in self.b:
+                fullbcount[elt] = fullbcount.get(elt, 0) + 1
+        fullbcount = self.fullbcount
+        # avail[x] is the number of times x appears in 'b' less the
+        # number of times we've seen it in 'a' so far ... kinda
+        avail = {}
+        availhas, matches = avail.has_key, 0
+        for elt in self.a:
+            if availhas(elt):
+                numb = avail[elt]
+            else:
+                numb = fullbcount.get(elt, 0)
+            avail[elt] = numb - 1
+            if numb > 0:
+                matches = matches + 1
+        return 2.0 * matches / (len(self.a) + len(self.b))
+
+    def real_quick_ratio(self):
+        """Return an upper bound on ratio() very quickly.
+
+        This isn't defined beyond that it is an upper bound on .ratio(), and
+        is faster to compute than either .ratio() or .quick_ratio().
+        """
+
+        la, lb = len(self.a), len(self.b)
+        # can't have more matches than the number of elements in the
+        # shorter sequence
+        return 2.0 * min(la, lb) / (la + lb)
+
+def get_close_matches(word, possibilities, n=3, cutoff=0.6):
+    """Use SequenceMatcher to return list of the best "good enough" matches.
+
+    word is a sequence for which close matches are desired (typically a
+    string).
+
+    possibilities is a list of sequences against which to match word
+    (typically a list of strings).
+
+    Optional arg n (default 3) is the maximum number of close matches to
+    return.  n must be > 0.
+
+    Optional arg cutoff (default 0.6) is a float in [0, 1].  Possibilities
+    that don't score at least that similar to word are ignored.
+
+    The best (no more than n) matches among the possibilities are returned
+    in a list, sorted by similarity score, most similar first.
+
+    >>> get_close_matches("appel", ["ape", "apple", "peach", "puppy"])
+    ['apple', 'ape']
+    >>> import keyword
+    >>> get_close_matches("wheel", keyword.kwlist)
+    ['while']
+    >>> get_close_matches("apple", keyword.kwlist)
+    []
+    >>> get_close_matches("accept", keyword.kwlist)
+    ['except']
+    """
+
+    if not n >  0:
+        raise ValueError("n must be > 0: %s" % `n`)
+    if not 0.0 <= cutoff <= 1.0:
+        raise ValueError("cutoff must be in [0.0, 1.0]: %s" % `cutoff`)
+    result = []
+    s = SequenceMatcher()
+    s.set_seq2(word)
+    for x in possibilities:
+        s.set_seq1(x)
+        if s.real_quick_ratio() >= cutoff and \
+           s.quick_ratio() >= cutoff and \
+           s.ratio() >= cutoff:
+            result.append((s.ratio(), x))
+    # Sort by score.
+    result.sort()
+    # Retain only the best n.
+    result = result[-n:]
+    # Move best-scorer to head of list.
+    result.reverse()
+    # Strip scores.
+    return [x for score, x in result]
+
+def _test():
+    import doctest, difflib
+    return doctest.testmod(difflib)
+
+if __name__ == "__main__":
+    _test()