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Module difflib ::
Class SequenceMatcher


InsensitiveSequenceMatcher
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 (RO 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 "humanfriendly diff" between two sequences. Unlike e.g. UNIX(tm) diff, the fundamental notion is the longest *contiguous* & junkfree 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 synching 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 humanfriendly file differencer, which uses SequenceMatcher both to compare sequences of lines, and to compare sequences of characters within similar (nearmatching) 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 RO is cubic time worst case and quadratic time expected case. SequenceMatcher is quadratic time for the worst case and has expectedcase behavior dependent in a complicated way on how many elements the sequences have in common; best case time is linear. Methods: __init__(isjunk=None, a='', b='') Construct a SequenceMatcher. set_seqs(a, b) Set the two sequences to be compared. set_seq1(a) Set the first sequence to be compared. set_seq2(b) Set the second sequence to be compared. find_longest_match(alo, ahi, blo, bhi) Find longest matching block in a[alo:ahi] and b[blo:bhi]. get_matching_blocks() Return list of triples describing matching subsequences. get_opcodes() Return list of 5tuples describing how to turn a into b. ratio() Return a measure of the sequences' similarity (float in [0,1]). quick_ratio() Return an upper bound on .ratio() relatively quickly. real_quick_ratio() Return an upper bound on ratio() very quickly.
Method Summary  

Construct a SequenceMatcher.  
Find longest matching block in a[alo:ahi] and b[blo:bhi].  
Isolate change clusters by eliminating ranges with no changes.  
Return list of triples describing matching subsequences.  
Return list of 5tuples describing how to turn a into b.  
Return an upper bound on ratio() relatively quickly.  
Return a measure of the sequences' similarity (float in [0,1]).  
Return an upper bound on ratio() very quickly.  
Set the first sequence to be compared.  
Set the second sequence to be compared.  
Set the two sequences to be compared.  
__chain_b(self)

Method Details 

__init__(self,
isjunk=None,
a='',
b='')

find_longest_match(self, alo, ahi, blo, bhi)Find longest matching block in a[alo:ahi] and b[blo:bhi]. 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) (0, 4, 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) (1, 0, 4) If no blocks match, return (alo, blo, 0). >>> s = SequenceMatcher(None, "ab", "c") >>> s.find_longest_match(0, 2, 0, 1) (0, 0, 0) 
get_grouped_opcodes(self, n=3)Isolate change clusters by eliminating ranges with no changes. Return a generator 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(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)]] 
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") >>> s.get_matching_blocks() [(0, 0, 2), (3, 2, 2), (5, 4, 0)] 
get_opcodes(self)Return list of 5tuples 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) 
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. 
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 
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(). 
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(). 
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(). 
set_seqs(self, a, b)Set the two sequences to be compared.>>> s = SequenceMatcher() >>> s.set_seqs("abcd", "bcde") >>> s.ratio() 0.75 
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