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.
Noneis equivalent to passing
lambda x: 0, i.e. no elements are ignored. 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.
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.
If isjunk was omitted or
k) such that
a[i:i+k] is equal
alo <= i <= i+k <= ahi and
blo <= j <= j+k <= bhi.
(i', j', k') meeting those
conditions, the additional conditions
k >= k',
i <= i',
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
>>> s = SequenceMatcher(None, " abcd", "abcd abcd") >>> s.find_longest_match(0, 5, 0, 9) (0, 4, 5)
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.
' abcd' from matching the
' abcd' at the
tail end of the second sequence directly. Instead only the
'abcd' can match, and matches the leftmost
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, this returns
(alo, blo, 0).
(i, j, n), and means that
a[i:i+n] == b[j:j+n]. The triples are monotonically increasing in i and j.
The last triple is a dummy, and has the value
len(b), 0). It 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)]
(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 = "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)
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
if the sequences are identical, and
0. if they have nothing in
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.
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():
>>> s = SequenceMatcher(None, "abcd", "bcde") >>> s.ratio() 0.75 >>> s.quick_ratio() 0.75 >>> s.real_quick_ratio() 1.0
See About this document... for information on suggesting changes.