Optional argument isjunk must be None (the default) or
a one-argument function that takes a sequence element and returns
true if and only if the element is ``junk'' and should be ignored.
Passing None for b is equivalent to passing
lambda x: 0; in other words, no elements are ignored. For
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.
The optional arguments a and b are sequences to be
compared; both default to empty strings. The elements of both
sequences must be hashable.
SequenceMatcher objects have the following methods:
SequenceMatcher computes and caches detailed information about
the second sequence, so if you want to compare one sequence against
many sequences, use set_seq2() to set the commonly used
sequence once and call set_seq1() repeatedly, once for each
of the other sequences.
Find longest matching block in a[alo:ahi]
If isjunk was omitted or None,
get_longest_match() returns (i, j,
k) such that a[i:i+k] is equal
to b[j:j+k], where
alo <= i <= i+k <= ahi and
blo <= j <= j+k <= bhi.
For all (i', j', k') meeting those
conditions, the additional conditions
k >= k',
i <= i',
and if i == i', j <= j'
are also met.
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
If isjunk was provided, 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:
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 equal to the
i2 from the preceeding tuple, and, likewise, j1 equal to
the previous j2.
The tag values are strings, with these meanings:
a[i1:i2] should be
replaced by b[j1:j2].
a[i1:i2] should be
deleted. Note that j1 == j2 in
b[j1:j2] should be
inserted at a[i1:i1].
Note that i1 == i2 in this
b[j1:j2] (the sub-sequences are
>>> 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)
Return a measure of the sequences' similarity as a float in the
range [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
1.0 if the sequences are identical, and 0.0 if they
have nothing in common.
This is expensive to compute if get_matching_blocks() or
get_opcodes() hasn't already been called, in which case you
may want to try quick_ratio() or
real_quick_ratio() first to get an upper bound.
This isn't defined beyond that it is an upper bound on
ratio(), and is faster to compute than either
ratio() or quick_ratio().
The three methods that return the ratio of matching to total characters
can give different results due to differing levels of approximation,
although quick_ratio() and real_quick_ratio() are always
at least as large as ratio():