python官方库difflib的类SequenceMatcher
功能：比较文本的距离
去掉junk字符
__chain_b()
首先创建字典self.b2j，以字符为key，出现的位置放在一个list中当作元素，然后去掉key为junk的元素ratio():
计算所有匹配片段的长度之和T，然后2*T/(len(a)+len(b))
结果在[0,1]，相同的时候返回1，没有相同片段返回0get_matching_blocks():
依次获取最长的相同段，然后分裂下去，最后返回所有相同的段。find_longest_match():
上面方法中用来查找最长相同段的子方法。
如果有多段最长的段长度一样，返回比较靠前的。
>>> s = SequenceMatcher(None, " abcd", "abcd abcd")
>>> s.find_longest_match(0, 5, 0, 9)
Match(a=0, b=4, size=5)
如果将空格当作junk
>>> s = SequenceMatcher(lambda x: x==" ", " abcd", "abcd abcd")
>>> s.find_longest_match(0, 5, 0, 9)
Match(a=1, b=0, size=4)
看代码不是很懂？？
class SequenceMatcher:
"""
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 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 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.
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 5-tuples 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.
"""
def __init__(self, isjunk=None, a='', b='', autojunk=True):
"""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).
"""
# Members:
# a
# first sequence
# b
# second sequence; differences are computed as "what do
# we need to do to 'a' to change it into 'b'?"
# b2j
# for x in b, b2j[x] is a list of the indices (into b)
# at which x appears; junk and popular elements do not appear
# fullbcount
# for x in b, fullbcount[x] == the number of times x
# appears in b; only materialized if really needed (used
# only for computing quick_ratio())
# matching_blocks
# a list of (i, j, k) triples, where a[i:i+k] == b[j:j+k];
# ascending & non-overlapping in i and in j; terminated by
# a dummy (len(a), len(b), 0) sentinel
# opcodes
# a list of (tag, i1, i2, j1, j2) tuples, where tag is
# one of
# 'replace' a[i1:i2] should be replaced by b[j1:j2]
# 'delete' a[i1:i2] should be deleted
# 'insert' b[j1:j2] should be inserted
# 'equal' a[i1:i2] == b[j1:j2]
# isjunk
# a user-supplied function taking a sequence element and
# returning true iff the element is "junk" -- this has
# subtle but helpful effects on the algorithm, which I'll
# get around to writing up someday <0.9 wink>.
# DON'T USE! Only __chain_b uses this. Use "in self.bjunk".
# bjunk
# the items in b for which isjunk is True.
# bpopular
# nonjunk items in b treated as junk by the heuristic (if used).
self.isjunk = isjunk
self.a = self.b = None
self.autojunk = autojunk
self.set_seqs(a, b)
def set_seqs(self, a, b):
"""Set the two sequences to be compared.
>>> s = SequenceMatcher()
>>> s.set_seqs("abcd", "bcde")
>>> s.ratio()
0.75
"""
self.set_seq1(a)
self.set_seq2(b)
def 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().
"""
if a is self.a:
return
self.a = a
self.matching_blocks = self.opcodes = None
def 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().
"""
if b is self.b:
return
self.b = b
self.matching_blocks = self.opcodes = None
self.fullbcount = None
self.__chain_b()
# For each element x in b, set b2j[x] to a list of the indices in
# b where x appears; the indices are in increasing order; note that
# the number of times x appears in b is len(b2j[x]) ...
# when self.isjunk is defined, junk elements don't show up in this
# map at all, which stops the central find_longest_match method
# from starting any matching block at a junk element ...
# b2j also does not contain entries for "popular" elements, meaning
# elements that account for more than 1 + 1% of the total elements, and
# when the sequence is reasonably large (>= 200 elements); this can
# be viewed as an adaptive notion of semi-junk, and yields an enormous
# speedup when, e.g., comparing program files with hundreds of
# instances of "return NULL;" ...
# note that this is only called when b changes; so for cross-product
# kinds of matches, it's best to call set_seq2 once, then set_seq1
# repeatedly
def __chain_b(self):
# Because isjunk is a user-defined (not C) function, and we test
# for junk a LOT, it's important to minimize the number of calls.
# Before the tricks described here, __chain_b was by far the most
# time-consuming routine in the whole module! If anyone sees
# Jim Roskind, thank him again for profile.py -- I never would
# have guessed that.
# The first trick is to build b2j ignoring the possibility
# of junk. I.e., we don't call isjunk at all yet. Throwing
# out the junk later is much cheaper than building b2j "right"
# from the start.
b = self.b
self.b2j = b2j = {}
# 经过这个循环之后，b2j是一个将b分解为 字符：出现位置的列表
# 比如 b = 'aab' --> b2j = {'a':[0,1],'b':[2]}
for i, elt in enumerate(b):
# 取出字典对应内容，然后增加内容
indices = b2j.setdefault(elt, [])
indices.append(i)
# 找到所有的junk，然后将他们放在junk，最后将b2j这个字典中的junk字符都删除掉
# Purge junk elements
self.bjunk = junk = set()
isjunk = self.isjunk
if isjunk:
for elt in b2j.keys():
if isjunk(elt):
junk.add(elt)
for elt in junk: # separate loop avoids separate list of keys
del b2j[elt]
# 若b个数大于200个，那么找出超过1%的元素，然后从b2j中删掉
# Purge popular elements that are not junk
self.bpopular = popular = set()
n = len(b)
if self.autojunk and n >= 200:
ntest = n // 100 + 1
for elt, idxs in b2j.items():
if len(idxs) > ntest:
popular.add(elt)
for elt in popular: # ditto; as fast for 1% deletion
del b2j[elt]
def 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)
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)
"""
# CAUTION: stripping common prefix or suffix would be incorrect.
# E.g.,
# ab
# acab
# Longest matching block is "ab", but if common prefix is
# stripped, it's "a" (tied with "b"). UNIX(tm) diff does so
# strip, so ends up claiming that ab is changed to acab by
# inserting "ca" in the middle. That's minimal but unintuitive:
# "it's obvious" that someone inserted "ac" at the front.
# Windiff ends up at the same place as diff, but by pairing up
# the unique 'b's and then matching the first two 'a's.
a, b, b2j, isbjunk = self.a, self.b, self.b2j, self.bjunk.__contains__
besti, bestj, bestsize = alo, blo, 0
# find longest junk-free match
# during an iteration of the loop, j2len[j] = length of longest
# junk-free match ending with a[i-1] and b[j]
j2len = {}
nothing = []
for i in range(alo, ahi):
# look at all instances of a[i] in b; note that because
# b2j has no junk keys, the loop is skipped if a[i] is junk
j2lenget = j2len.get
newj2len = {}
for j in b2j.get(a[i], nothing):
# a[i] matches b[j]
if j < blo:
continue
if j >= bhi:
break
k = newj2len[j] = j2lenget(j-1, 0) + 1
if k > bestsize:
besti, bestj, bestsize = i-k+1, j-k+1, k
j2len = newj2len
# Extend the best by non-junk elements on each end. In particular,
# "popular" non-junk elements aren't in b2j, which greatly speeds
# the inner loop above, but also means "the best" match so far
# doesn't contain any junk *or* popular non-junk elements.
while besti > alo and bestj > blo and \
not isbjunk(b[bestj-1]) and \
a[besti-1] == b[bestj-1]:
besti, bestj, bestsize = besti-1, bestj-1, bestsize+1
while besti+bestsize < ahi and bestj+bestsize < bhi and \
not isbjunk(b[bestj+bestsize]) and \
a[besti+bestsize] == b[bestj+bestsize]:
bestsize += 1
# Now that we have a wholly interesting match (albeit possibly
# empty!), we may as well suck up the matching junk on each
# side of it too. Can't think of a good reason not to, and it
# saves post-processing the (possibly considerable) expense of
# figuring out what to do with it. In the case of an empty
# interesting match, this is clearly the right thing to do,
# because no other kind of match is possible in the regions.
while besti > alo and bestj > blo and \
isbjunk(b[bestj-1]) and \
a[besti-1] == b[bestj-1]:
besti, bestj, bestsize = besti-1, bestj-1, bestsize+1
while besti+bestsize < ahi and bestj+bestsize < bhi and \
isbjunk(b[bestj+bestsize]) and \
a[besti+bestsize] == b[bestj+bestsize]:
bestsize = bestsize + 1
return Match(besti, bestj, bestsize)
def 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")
>>> 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)]
"""
if self.matching_blocks is not None:
return self.matching_blocks
la, lb = len(self.a), len(self.b)
# This is most naturally expressed as a recursive algorithm, but
# at least one user bumped into extreme use cases that exceeded
# the recursion limit on their box. So, now we maintain a list
# ('queue`) of blocks we still need to look at, and append partial
# results to `matching_blocks` in a loop; the matches are sorted
# at the end.
queue = [(0, la, 0, lb)]
matching_blocks = []
while queue:
alo, ahi, blo, bhi = queue.pop()
i, j, k = x = self.find_longest_match(alo, ahi, blo, bhi)
# a[alo:i] vs b[blo:j] unknown
# a[i:i+k] same as b[j:j+k]
# a[i+k:ahi] vs b[j+k:bhi] unknown
if k: # if k is 0, there was no matching block
matching_blocks.append(x)
if alo < i and blo < j:
queue.append((alo, i, blo, j))
if i+k < ahi and j+k < bhi:
queue.append((i+k, ahi, j+k, bhi))
matching_blocks.sort()
# It's possible that we have adjacent equal blocks in the
# matching_blocks list now. Starting with 2.5, this code was added
# to collapse them.
i1 = j1 = k1 = 0
non_adjacent = []
for i2, j2, k2 in matching_blocks:
# Is this block adjacent to i1, j1, k1?
if i1 + k1 == i2 and j1 + k1 == j2:
# Yes, so collapse them -- this just increases the length of
# the first block by the length of the second, and the first
# block so lengthened remains the block to compare against.
k1 += k2
else:
# Not adjacent. Remember the first block (k1==0 means it's
# the dummy we started with), and make the second block the
# new block to compare against.
if k1:
non_adjacent.append((i1, j1, k1))
i1, j1, k1 = i2, j2, k2
if k1:
non_adjacent.append((i1, j1, k1))
non_adjacent.append( (la, lb, 0) )
self.matching_blocks = list(map(Match._make, non_adjacent))
return self.matching_blocks
def get_opcodes(self):
"""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)
insert a[6:6] () b[5:6] (f)
"""
if self.opcodes is not None:
return self.opcodes
i = j = 0
self.opcodes = answer = []
for ai, bj, size in self.get_matching_blocks():
# invariant: we've pumped out correct diffs to change
# a[:i] into b[:j], and the next matching block is
# a[ai:ai+size] == b[bj:bj+size]. So we need to pump
# out a diff to change a[i:ai] into b[j:bj], pump out
# the matching block, and move (i,j) beyond the match
tag = ''
if i < ai and j < bj:
tag = 'replace'
elif i < ai:
tag = 'delete'
elif j < bj:
tag = 'insert'
if tag:
answer.append( (tag, i, ai, j, bj) )
i, j = ai+size, bj+size
# the list of matching blocks is terminated by a
# sentinel with size 0
if size:
answer.append( ('equal', ai, i, bj, j) )
return answer
def get_grouped_opcodes(self, n=3):
""" 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)]]
"""
codes = self.get_opcodes()
if not codes:
codes = [("equal", 0, 1, 0, 1)]
# Fixup leading and trailing groups if they show no changes.
if codes[0][0] == 'equal':
tag, i1, i2, j1, j2 = codes[0]
codes[0] = tag, max(i1, i2-n), i2, max(j1, j2-n), j2
if codes[-1][0] == 'equal':
tag, i1, i2, j1, j2 = codes[-1]
codes[-1] = tag, i1, min(i2, i1+n), j1, min(j2, j1+n)
nn = n + n
group = []
for tag, i1, i2, j1, j2 in codes:
# End the current group and start a new one whenever
# there is a large range with no changes.
if tag == 'equal' and i2-i1 > nn:
group.append((tag, i1, min(i2, i1+n), j1, min(j2, j1+n)))
yield group
group = []
i1, j1 = max(i1, i2-n), max(j1, j2-n)
group.append((tag, i1, i2, j1 ,j2))
if group and not (len(group)==1 and group[0][0] == 'equal'):
yield group
def 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
"""
matches = sum(triple[-1] for triple in self.get_matching_blocks())
return _calculate_ratio(matches, len(self.a) + len(self.b))
def 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.
"""
# viewing a and b as multisets, set matches to the cardinality
# of their intersection; this counts the number of matches
# without regard to order, so is clearly an upper bound
if self.fullbcount is None:
self.fullbcount = fullbcount = {}
for elt in self.b:
fullbcount[elt] = fullbcount.get(elt, 0) + 1
fullbcount = self.fullbcount
# avail[x] is the number of times x appears in 'b' less the
# number of times we've seen it in 'a' so far ... kinda
avail = {}
availhas, matches = avail.__contains__, 0
for elt in self.a:
if availhas(elt):
numb = avail[elt]
else:
numb = fullbcount.get(elt, 0)
avail[elt] = numb - 1
if numb > 0:
matches = matches + 1
return _calculate_ratio(matches, len(self.a) + len(self.b))
def 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().
"""
la, lb = len(self.a), len(self.b)
# can't have more matches than the number of elements in the
# shorter sequence
return _calculate_ratio(min(la, lb), la + lb)
