difflib.py

"""
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.

Function context_diff(a, b):
    For two lists of strings, return a delta in context diff format.

Function ndiff(a, b):
    Return a delta: the difference between `a` and `b` (lists of strings).

Function restore(delta, which):
    Return one of the two sequences that generated an ndiff delta.

Function unified_diff(a, b):
    For two lists of strings, return a delta in unified diff format.

Class SequenceMatcher:
    A flexible class for comparing pairs of sequences of any type.

Class Differ:
    For producing human-readable deltas from sequences of lines of text.

Class HtmlDiff:
    For producing HTML side by side comparison with change highlights.
"""

__all__ = ['get_close_matches', 'ndiff', 'restore', 'SequenceMatcher',
           'Differ','IS_CHARACTER_JUNK', 'IS_LINE_JUNK', 'context_diff',
           'unified_diff', 'diff_bytes', 'HtmlDiff', 'Match']

from heapq import nlargest as _nlargest
from collections import namedtuple as _namedtuple
from types import GenericAlias

Match = _namedtuple('Match', 'a b size')

def _calculate_ratio(matches, length):
    if length:
        return 2.0 * matches / length
    return 1.0

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.

    SequenceMatcher tries to compute a "human-friendly diff" between two
    sequences.  Unlike e.g. UNIX(tm) diff, the fundamental notion is the
    longest *contiguous* & junk-free matching subsequence.  That's what
    catches peoples' eyes.  The Windows(tm) windiff has another interesting
    notion, pairing up elements that appear uniquely in each sequence.
    That, and the method here, appear to yield more intuitive difference
    reports than does diff.  This method appears to be the least vulnerable
    to syncing up on blocks of "junk lines", though (like blank lines in
    ordinary text files, or maybe "<P>" lines in HTML files).  That may be
    because this is the only method of the 3 that has a *concept* of
    "junk" <wink>.

    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(), 2))
    0.87
    >>>

    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 21 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[8:29] b[17:38]

    See the Differ class for a fancy human-friendly file differencer, which
    uses SequenceMatcher both to compare sequences of lines, and to compare
    sequences of characters within similar (near-matching) lines.

    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 for the worst case and has
    expected-case behavior dependent in a complicated way on how many
    elements the sequences have in common; best case time is linear.
    """

    def __init__(self, isjunk=None, a='', b='', autojunk=True):
        """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 example, 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 b must be hashable. See
        also .set_seqs() and .set_seq2().

        Optional arg autojunk should be set to False to disable the
        "automatic junk heuristic" that treats popular elements as junk
        (see module documentation for more information).
        """

        # 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 and popular elements do not appear
        # 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 "in self.bjunk".
        # bjunk
        #      the items in b for which isjunk is True.
        # bpopular
        #      nonjunk items in b treated as junk by the heuristic (if used).

        self.isjunk = isjunk
        self.a = self.b = None
        self.autojunk = autojunk
        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 ...
    # b2j also does not contain entries for "popular" elements, meaning
    # elements that account for more than 1 + 1% of the total elements, and
    # when the sequence is reasonably large (>= 200 elements); this can
    # be viewed as an adaptive notion of semi-junk, and yields an enormous
    # speedup when, e.g., comparing program files with hundreds of
    # instances of "return NULL;" ...
    # 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 = {}

        for i, elt in enumerate(b):
            indices = b2j.setdefault(elt, [])
            indices.append(i)

        # Purge junk elements
        self.bjunk = junk = set()
        isjunk = self.isjunk
        if isjunk:
            for elt in b2j.keys():
                if isjunk(elt):
                    junk.add(elt)
            for elt in junk: # separate loop avoids separate list of keys
                del b2j[elt]

        # Purge popular elements that are not junk
        self.bpopular = popular = set()
        n = len(b)
        if self.autojunk and n >= 200:
            ntest = n // 100 + 1
            for elt, idxs in b2j.items():
                if len(idxs) > ntest:
                    popular.add(elt)
            for elt in popular: # ditto; as fast for 1% deletion
                del b2j[elt]

    def find_longest_match(self, alo=0, ahi=None, blo=0, bhi=None):
        """Find longest matching block in a[alo:ahi] and b[blo:bhi].

        By default it will find the longest match in the entirety of a and b.

        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)
        Match(a=0, b=4, size=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)
        Match(a=1, b=0, size=4)

        If no blocks match, return (alo, blo, 0).

        >>> s = SequenceMatcher(None, "ab", "c")
        >>> s.find_longest_match(0, 2, 0, 1)
        Match(a=0, b=0, size=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.bjunk.__contains__
        if ahi is None:
            ahi = len(a)
        if bhi is None:
            bhi = len(b)
        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 range(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

        # Extend the best by non-junk elements on each end.  In particular,
        # "popular" non-junk elements aren't in b2j, which greatly speeds
        # the inner loop above, but also means "the best" match so far
        # doesn't contain any junk *or* popular non-junk elements.
        while besti > alo and bestj > blo and \
              not 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 \
              not isbjunk(b[bestj+bestsize]) and \
              a[besti+bestsize] == b[bestj+bestsize]:
            bestsize += 1

        # 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

        return Match(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.  New in Python 2.5, it's also guaranteed that if
        (i, j, n) and (i', j', n') are adjacent triples in the list, and
        the second is not the last triple in the list, then i+n != i' or
        j+n != j'.  IOW, adjacent triples never describe adjacent equal
        blocks.

        The last triple is a dummy, (len(a), len(b), 0), and is the only
        triple with n==0.

        >>> s = SequenceMatcher(None, "abxcd", "abcd")
        >>> list(s.get_matching_blocks())
        [Match(a=0, b=0, size=2), Match(a=3, b=2, size=2), Match(a=5, b=4, size=0)]
        """

        if self.matching_blocks is not None:
            return self.matching_blocks
        la, lb = len(self.a), len(self.b)

        # This is most naturally expressed as a recursive algorithm, but
        # at least one user bumped into extreme use cases that exceeded
        # the recursion limit on their box.  So, now we maintain a list
        # ('queue`) of blocks we still need to look at, and append partial
        # results to `matching_blocks` in a loop; the matches are sorted
        # at the end.
        queue = [(0, la, 0, lb)]
        matching_blocks = []
        while queue:
            alo, ahi, blo, bhi = queue.pop()
            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 k is 0, there was no matching block
                matching_blocks.append(x)
                if alo < i and blo < j:
                    queue.append((alo, i, blo, j))
                if i+k < ahi and j+k < bhi:
                    queue.append((i+k, ahi, j+k, bhi))
        matching_blocks.sort()

        # It's possible that we have adjacent equal blocks in the
        # matching_blocks list now.  Starting with 2.5, this code was added
        # to collapse them.
        i1 = j1 = k1 = 0
        non_adjacent = []
        for i2, j2, k2 in matching_blocks:
            # Is this block adjacent to i1, j1, k1?
            if i1 + k1 == i2 and j1 + k1 == j2:
                # Yes, so collapse them -- this just increases the length of
                # the first block by the length of the second, and the first
                # block so lengthened remains the block to compare against.
                k1 += k2
            else:
                # Not adjacent.  Remember the first block (k1==0 means it's
                # the dummy we started with), and make the second block the
                # new block to compare against.
                if k1:
                    non_adjacent.append((i1, j1, k1))
                i1, j1, k1 = i2, j2, k2
        if k1:
            non_adjacent.append((i1, j1, k1))

        non_adjacent.append( (la, lb, 0) )
        self.matching_blocks = list(map(Match._make, non_adjacent))
        return self.matching_blocks

    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 get_grouped_opcodes(self, n=3):
        """ Isolate change clusters by eliminating ranges with no changes.

        Return a generator of groups with up to n lines of context.
        Each group is in the same format as returned by get_opcodes().

        >>> from pprint import pprint
        >>> a = list(map(str, range(1,40)))
        >>> b = a[:]
        >>> b[8:8] = ['i']     # Make an insertion
        >>> b[20] += 'x'       # Make a replacement
        >>> b[23:28] = []      # Make a deletion
        >>> b[30] += 'y'       # Make another replacement
        >>> pprint(list(SequenceMatcher(None,a,b).get_grouped_opcodes()))
        [[('equal', 5, 8, 5, 8), ('insert', 8, 8, 8, 9), ('equal', 8, 11, 9, 12)],
         [('equal', 16, 19, 17, 20),
          ('replace', 19, 20, 20, 21),
          ('equal', 20, 22, 21, 23),
          ('delete', 22, 27, 23, 23),
          ('equal', 27, 30, 23, 26)],
         [('equal', 31, 34, 27, 30),
          ('replace', 34, 35, 30, 31),
          ('equal', 35, 38, 31, 34)]]
        """

        codes = self.get_opcodes()
        if not codes:
            codes = [("equal", 0, 1, 0, 1)]
        # Fixup leading and trailing groups if they show no changes.
        if codes[0][0] == 'equal':
            tag, i1, i2, j1, j2 = codes[0]
            codes[0] = tag, max(i1, i2-n), i2, max(j1, j2-n), j2
        if codes[-1][0] == 'equal':
            tag, i1, i2, j1, j2 = codes[-1]
            codes[-1] = tag, i1, min(i2, i1+n), j1, min(j2, j1+n)

        nn = n + n
        group = []
        for tag, i1, i2, j1, j2 in codes:
            # End the current group and start a new one whenever
            # there is a large range with no changes.
            if tag == 'equal' and i2-i1 > nn:
                group.append((tag, i1, min(i2, i1+n), j1, min(j2, j1+n)))
                yield group
                group = []
                i1, j1 = max(i1, i2-n), max(j1, j2-n)
            group.append((tag, i1, i2, j1 ,j2))
        if group and not (len(group)==1 and group[0][0] == 'equal'):
            yield group

    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 = sum(triple[-1] for triple in self.get_matching_blocks())
        return _calculate_ratio(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.__contains__, 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 _calculate_ratio(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 _calculate_ratio(min(la, lb), la + lb)

    __class_getitem__ = classmethod(GenericAlias)


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 as _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: %r" % (n,))
    if not 0.0 <= cutoff <= 1.0:
        raise ValueError("cutoff must be in [0.0, 1.0]: %r" % (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))

    # Move the best scorers to head of list
    result = _nlargest(n, result)
    # Strip scores for the best n matches
    return [x for score, x in result]


def _keep_original_ws(s, tag_s):
    """Replace whitespace with the original whitespace characters in `s`"""
    return ''.join(
        c if tag_c == " " and c.isspace() else tag_c
        for c, tag_c in zip(s, tag_s)
    )



class Differ:
    r"""
    Differ is a class for comparing sequences of lines of text, and
    producing human-readable differences or deltas.  Differ uses
    SequenceMatcher both to compare sequences of lines, and to compare
    sequences of characters within similar (near-matching) lines.

    Each line of a Differ delta begins with a two-letter code:

        '- '    line unique to sequence 1
        '+ '    line unique to sequence 2
        '  '    line common to both sequences
        '? '    line not present in either input sequence

    Lines beginning with '? ' attempt to guide the eye to intraline
    differences, and were not present in either input sequence.  These lines
    can be confusing if the sequences contain tab characters.

    Note that Differ makes no claim to produce a *minimal* diff.  To the
    contrary, minimal diffs are often counter-intuitive, because they synch
    up anywhere possible, sometimes accidental matches 100 pages apart.
    Restricting synch points to contiguous matches preserves some notion of
    locality, at the occasional cost of producing a longer diff.

    Example: Comparing two texts.

    First we set up the texts, sequences of individual single-line strings
    ending with newlines (such sequences can also be obtained from the
    `readlines()` method of file-like objects):

    >>> text1 = '''  1. Beautiful is better than ugly.
    ...   2. Explicit is better than implicit.
    ...   3. Simple is better than complex.
    ...   4. Complex is better than complicated.
    ... '''.splitlines(keepends=True)
    >>> len(text1)
    4
    >>> text1[0][-1]
    '\n'
    >>> text2 = '''  1. Beautiful is better than ugly.
    ...   3.   Simple is better than complex.
    ...   4. Complicated is better than complex.
    ...   5. Flat is better than nested.
    ... '''.splitlines(keepends=True)

    Next we instantiate a Differ object:

    >>> d = Differ()

    Note that when instantiating a Differ object we may pass functions to
    filter out line and character 'junk'.  See Differ.__init__ for details.

    Finally, we compare the two:

    >>> result = list(d.compare(text1, text2))

    'result' is a list of strings, so let's pretty-print it:

    >>> from pprint import pprint as _pprint
    >>> _pprint(result)
    ['    1. Beautiful is better than ugly.\n',
     '-   2. Explicit is better than implicit.\n',
     '-   3. Simple is better than complex.\n',
     '+   3.   Simple is better than complex.\n',
     '?     ++\n',
     '-   4. Complex is better than complicated.\n',
     '?            ^                     ---- ^\n',
     '+   4. Complicated is better than complex.\n',
     '?           ++++ ^                      ^\n',
     '+   5. Flat is better than nested.\n']

    As a single multi-line string it looks like this:

    >>> print(''.join(result), end="")
        1. Beautiful is better than ugly.
    -   2. Explicit is better than implicit.
    -   3. Simple is better than complex.
    +   3.   Simple is better than complex.
    ?     ++
    -   4. Complex is better than complicated.
    ?            ^                     ---- ^
    +   4. Complicated is better than complex.
    ?           ++++ ^                      ^
    +   5. Flat is better than nested.
    """

    def __init__(self, linejunk=None, charjunk=None):
        """
        Construct a text differencer, with optional filters.

        The two optional keyword parameters are for filter functions:

        - `linejunk`: A function that should accept a single string argument,
          and return true iff the string is junk. The module-level function
          `IS_LINE_JUNK` may be used to filter out lines without visible
          characters, except for at most one splat ('#').  It is recommended
          to leave linejunk None; the underlying SequenceMatcher class has
          an adaptive notion of "noise" lines that's better than any static
          definition the author has ever been able to craft.

        - `charjunk`: A function that should accept a string of length 1. The
          module-level function `IS_CHARACTER_JUNK` may be used to filter out
          whitespace characters (a blank or tab; **note**: bad idea to include
          newline in this!).  Use of IS_CHARACTER_JUNK is recommended.
        """

        self.linejunk = linejunk
        self.charjunk = charjunk

    def compare(self, a, b):
        r"""
        Compare two sequences of lines; generate the resulting delta.

        Each sequence must contain individual single-line strings ending with
        newlines. Such sequences can be obtained from the `readlines()` method
        of file-like objects.  The delta generated also consists of newline-
        terminated strings, ready to be printed as-is via the writelines()
        method of a file-like object.

        Example:

        >>> print(''.join(Differ().compare('one\ntwo\nthree\n'.splitlines(True),
        ...                                'ore\ntree\nemu\n'.splitlines(True))),
        ...       end="")
        - one
        ?  ^
        + ore
        ?  ^
        - two
        - three
        ?  -
        + tree
        + emu
        """

        cruncher = SequenceMatcher(self.linejunk, a, b)
        for tag, alo, ahi, blo, bhi in cruncher.get_opcodes():
            if tag == 'replace':
                g = self._fancy_replace(a, alo, ahi, b, blo, bhi)
            elif tag == 'delete':
                g = self._dump('-', a, alo, ahi)
            elif tag == 'insert':
                g = self._dump('+', b, blo, bhi)
            elif tag == 'equal':
                g = self._dump(' ', a, alo, ahi)
            else:
                raise ValueError('unknown tag %r' % (tag,))

            yield from g

    def _dump(self, tag, x, lo, hi):
        """Generate comparison results for a same-tagged range."""
        for i in range(lo, hi):
            yield '%s %s' % (tag, x[i])

    def _plain_replace(self, a, alo, ahi, b, blo, bhi):
        assert alo < ahi and blo < bhi
        # dump the shorter block first -- reduces the burden on short-term
        # memory if the blocks are of very different sizes
        if bhi - blo < ahi - alo:
            first  = self._dump('+', b, blo, bhi)
            second = self._dump('-', a, alo, ahi)
        else:
            first  = self._dump('-', a, alo, ahi)
            second = self._dump('+', b, blo, bhi)

        for g in first, second:
            yield from g

    def _fancy_replace(self, a, alo, ahi, b, blo, bhi):
        r"""
        When replacing one block of lines with another, search the blocks
        for *similar* lines; the best-matching pair (if any) is used as a
        synch point, and intraline difference marking is done on the
        similar pair. Lots of work, but often worth it.

        Example:

        >>> d = Differ()
        >>> results = d._fancy_replace(['abcDefghiJkl\n'], 0, 1,
        ...                            ['abcdefGhijkl\n'], 0, 1)
        >>> print(''.join(results), end="")
        - abcDefghiJkl
        ?    ^  ^  ^
        + abcdefGhijkl
        ?    ^  ^  ^
        """

        # don't synch up unless the lines have a similarity score of at
        # least cutoff; best_ratio tracks the best score seen so far
        best_ratio, cutoff = 0.74, 0.75
        cruncher = SequenceMatcher(self.charjunk)
        eqi, eqj = None, None   # 1st indices of equal lines (if any)

        # search for the pair that matches best without being identical
        # (identical lines must be junk lines, & we don't want to synch up
        # on junk -- unless we have to)
        for j in range(blo, bhi):
            bj = b[j]
            cruncher.set_seq2(bj)
            for i in range(alo, ahi):
                ai = a[i]
                if ai == bj:
                    if eqi is None:
                        eqi, eqj = i, j
                    continue
                cruncher.set_seq1(ai)
                # computing similarity is expensive, so use the quick
                # upper bounds first -- have seen this speed up messy
                # compares by a factor of 3.
                # note that ratio() is only expensive to compute the first
                # time it's called on a sequence pair; the expensive part
                # of the computation is cached by cruncher
                if cruncher.real_quick_ratio() > best_ratio and \
                      cruncher.quick_ratio() > best_ratio and \
                      cruncher.ratio() > best_ratio:
                    best_ratio, best_i, best_j = cruncher.ratio(), i, j
        if best_ratio < cutoff:
            # no non-identical "pretty close" pair
            if eqi is None:
                # no identical pair either -- treat it as a straight replace
                yield from self._plain_replace(a, alo, ahi, b, blo, bhi)
                return
            # no close pair, but an identical pair -- synch up on that
            best_i, best_j, best_ratio = eqi, eqj, 1.0
        else:
            # there's a close pair, so forget the identical pair (if any)
            eqi = None

        # a[best_i] very similar to b[best_j]; eqi is None iff they're not
        # identical

        # pump out diffs from before the synch point
        yield from self._fancy_helper(a, alo, best_i, b, blo, best_j)

        # do intraline marking on the synch pair
        aelt, belt = a[best_i], b[best_j]
        if eqi is None:
            # pump out a '-', '?', '+', '?' quad for the synched lines
            atags = btags = ""
            cruncher.set_seqs(aelt, belt)
            for tag, ai1, ai2, bj1, bj2 in cruncher.get_opcodes():
                la, lb = ai2 - ai1, bj2 - bj1
                if tag == 'replace':
                    atags += '^' * la
                    btags += '^' * lb
                elif tag == 'delete':
                    atags += '-' * la
                elif tag == 'insert':
                    btags += '+' * lb
                elif tag == 'equal':
                    atags += ' ' * la
                    btags += ' ' * lb
                else:
                    raise ValueError('unknown tag %r' % (tag,))
            yield from self._qformat(aelt, belt, atags, btags)
        else:
            # the synch pair is identical
            yield '  ' + aelt

        # pump out diffs from after the synch point
        yield from self._fancy_helper(a, best_i+1, ahi, b, best_j+1, bhi)

    def _fancy_helper(self, a, alo, ahi, b, blo, bhi):
        g = []
        if alo < ahi:
            if blo < bhi:
                g = self._fancy_replace(a, alo, ahi, b, blo, bhi)
            else:
                g = self._dump('-', a, alo, ahi)
        elif blo < bhi:
            g = self._dump('+', b, blo, bhi)

        yield from g

    def _qformat(self, aline, bline, atags, btags):
        r"""