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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.
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    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(), 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 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.
    Nc�`�|
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)a!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).
        N)�isjunk�
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        >>> s = SequenceMatcher()
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        N)�set_seq1�set_seq2)r!rrs   rr zSequenceMatcher.set_seqs�s��	
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        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().
        N)r�matching_blocks�opcodes)r!rs  rr$zSequenceMatcher.set_seq1�s(��*
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)aMSet 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().
        N)rr'r(�
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|�S)aAFind 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)
        rr/)	rrr0r5�__contains__r9�range�getr
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|_|jS)aReturn 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)]
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�a|S)a[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)
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)a� 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)]]
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        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
        c
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        This isn't defined beyond that it is an upper bound on .ratio(), and
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        '- '    line unique to sequence 1
        '+ '    line unique to sequence 2
        '  '    line common to both sequences
        '? '    line not present in either input sequence

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    `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)

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    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:

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        1. Beautiful is better than ugly.
    -   2. Explicit is better than implicit.
    -   3. Simple is better than complex.
    +   3.   Simple is better than complex.
    ?     ++
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    Example:

    >>> for line in unified_diff('one two three four'.split(),
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    ...             '2005-01-26 23:30:50', '2010-04-02 10:20:52',
    ...             lineterm=''):
    ...     print(line)                 # doctest: +NORMALIZE_WHITESPACE
    --- Original        2005-01-26 23:30:50
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    ...       end="")
    *** Original
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