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& @ C implementation of PatienceSequenceMatcher Isolate change clusters by eliminating ranges with no changes.
Return a list of groups with upto n lines of context.
Each group is in the same format as returned by get_opcodes().
>>> from pprint import pprint
>>> a = 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(PatienceSequenceMatcher(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)]]
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 = PatienceSequenceMatcher(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)
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 = PatienceSequenceMatcher(None, "abxcd", "abcd")
>>> s.get_matching_blocks()
[(0, 0, 2), (3, 2, 2), (5, 4, 0)]
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