787 lines
29 KiB
Python
Executable File
787 lines
29 KiB
Python
Executable File
#! /usr/bin/env python
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# Backported to Python 1.5.2 for the ViewCVS project by pf@artcom-gmbh.de
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# 24-Dec-2001, original version "stolen" from Python-2.1.1
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"""
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Module difflib -- helpers for computing deltas between objects.
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Function get_close_matches(word, possibilities, n=3, cutoff=0.6):
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Use SequenceMatcher to return list of the best "good enough" matches.
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word is a sequence for which close matches are desired (typically a
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string).
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possibilities is a list of sequences against which to match word
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(typically a list of strings).
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Optional arg n (default 3) is the maximum number of close matches to
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return. n must be > 0.
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Optional arg cutoff (default 0.6) is a float in [0, 1]. Possibilities
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that don't score at least that similar to word are ignored.
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The best (no more than n) matches among the possibilities are returned
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in a list, sorted by similarity score, most similar first.
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>>> get_close_matches("appel", ["ape", "apple", "peach", "puppy"])
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['apple', 'ape']
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>>> import keyword
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>>> get_close_matches("wheel", keyword.kwlist)
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['while']
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>>> get_close_matches("apple", keyword.kwlist)
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[]
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>>> get_close_matches("accept", keyword.kwlist)
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['except']
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Class SequenceMatcher
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SequenceMatcher is a flexible class for comparing pairs of sequences of any
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type, so long as the sequence elements are hashable. The basic algorithm
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predates, and is a little fancier than, an algorithm published in the late
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1980's by Ratcliff and Obershelp under the hyperbolic name "gestalt pattern
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matching". The basic idea is to find the longest contiguous matching
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subsequence that contains no "junk" elements (R-O doesn't address junk).
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The same idea is then applied recursively to the pieces of the sequences to
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the left and to the right of the matching subsequence. This does not yield
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minimal edit sequences, but does tend to yield matches that "look right"
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to people.
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Example, comparing two strings, and considering blanks to be "junk":
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>>> s = SequenceMatcher(lambda x: x == " ",
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... "private Thread currentThread;",
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... "private volatile Thread currentThread;")
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>>>
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.ratio() returns a float in [0, 1], measuring the "similarity" of the
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sequences. As a rule of thumb, a .ratio() value over 0.6 means the
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sequences are close matches:
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>>> print round(s.ratio(), 3)
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0.866
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>>>
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If you're only interested in where the sequences match,
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.get_matching_blocks() is handy:
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>>> for block in s.get_matching_blocks():
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... print "a[%d] and b[%d] match for %d elements" % block
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a[0] and b[0] match for 8 elements
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a[8] and b[17] match for 6 elements
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a[14] and b[23] match for 15 elements
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a[29] and b[38] match for 0 elements
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Note that the last tuple returned by .get_matching_blocks() is always a
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dummy, (len(a), len(b), 0), and this is the only case in which the last
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tuple element (number of elements matched) is 0.
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If you want to know how to change the first sequence into the second, use
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.get_opcodes():
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>>> for opcode in s.get_opcodes():
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... print "%6s a[%d:%d] b[%d:%d]" % opcode
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equal a[0:8] b[0:8]
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insert a[8:8] b[8:17]
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equal a[8:14] b[17:23]
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equal a[14:29] b[23:38]
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See Tools/scripts/ndiff.py for a fancy human-friendly file differencer,
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which uses SequenceMatcher both to view files as sequences of lines, and
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lines as sequences of characters.
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See also function get_close_matches() in this module, which shows how
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simple code building on SequenceMatcher can be used to do useful work.
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Timing: Basic R-O is cubic time worst case and quadratic time expected
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case. SequenceMatcher is quadratic time for the worst case and has
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expected-case behavior dependent in a complicated way on how many
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elements the sequences have in common; best case time is linear.
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SequenceMatcher methods:
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__init__(isjunk=None, a='', b='')
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Construct a SequenceMatcher.
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Optional arg isjunk is None (the default), or a one-argument function
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that takes a sequence element and returns true iff the element is junk.
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None is equivalent to passing "lambda x: 0", i.e. no elements are
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considered to be junk. For example, pass
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lambda x: x in " \\t"
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if you're comparing lines as sequences of characters, and don't want to
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synch up on blanks or hard tabs.
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Optional arg a is the first of two sequences to be compared. By
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default, an empty string. The elements of a must be hashable.
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Optional arg b is the second of two sequences to be compared. By
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default, an empty string. The elements of b must be hashable.
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set_seqs(a, b)
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Set the two sequences to be compared.
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>>> s = SequenceMatcher()
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>>> s.set_seqs("abcd", "bcde")
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>>> s.ratio()
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0.75
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set_seq1(a)
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Set the first sequence to be compared.
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The second sequence to be compared is not changed.
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>>> s = SequenceMatcher(None, "abcd", "bcde")
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>>> s.ratio()
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0.75
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>>> s.set_seq1("bcde")
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>>> s.ratio()
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1.0
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>>>
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SequenceMatcher computes and caches detailed information about the
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second sequence, so if you want to compare one sequence S against many
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sequences, use .set_seq2(S) once and call .set_seq1(x) repeatedly for
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each of the other sequences.
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See also set_seqs() and set_seq2().
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set_seq2(b)
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Set the second sequence to be compared.
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The first sequence to be compared is not changed.
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>>> s = SequenceMatcher(None, "abcd", "bcde")
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>>> s.ratio()
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0.75
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>>> s.set_seq2("abcd")
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>>> s.ratio()
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1.0
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>>>
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SequenceMatcher computes and caches detailed information about the
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second sequence, so if you want to compare one sequence S against many
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sequences, use .set_seq2(S) once and call .set_seq1(x) repeatedly for
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each of the other sequences.
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See also set_seqs() and set_seq1().
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find_longest_match(alo, ahi, blo, bhi)
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Find longest matching block in a[alo:ahi] and b[blo:bhi].
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If isjunk is not defined:
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Return (i,j,k) such that a[i:i+k] is equal to b[j:j+k], where
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alo <= i <= i+k <= ahi
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blo <= j <= j+k <= bhi
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and for all (i',j',k') meeting those conditions,
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k >= k'
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i <= i'
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and if i == i', j <= j'
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In other words, of all maximal matching blocks, return one that starts
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earliest in a, and of all those maximal matching blocks that start
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earliest in a, return the one that starts earliest in b.
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>>> s = SequenceMatcher(None, " abcd", "abcd abcd")
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>>> s.find_longest_match(0, 5, 0, 9)
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(0, 4, 5)
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If isjunk is defined, first the longest matching block is determined as
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above, but with the additional restriction that no junk element appears
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in the block. Then that block is extended as far as possible by
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matching (only) junk elements on both sides. So the resulting block
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never matches on junk except as identical junk happens to be adjacent
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to an "interesting" match.
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Here's the same example as before, but considering blanks to be junk.
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That prevents " abcd" from matching the " abcd" at the tail end of the
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second sequence directly. Instead only the "abcd" can match, and
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matches the leftmost "abcd" in the second sequence:
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>>> s = SequenceMatcher(lambda x: x==" ", " abcd", "abcd abcd")
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>>> s.find_longest_match(0, 5, 0, 9)
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(1, 0, 4)
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If no blocks match, return (alo, blo, 0).
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>>> s = SequenceMatcher(None, "ab", "c")
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>>> s.find_longest_match(0, 2, 0, 1)
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(0, 0, 0)
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get_matching_blocks()
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Return list of triples describing matching subsequences.
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Each triple is of the form (i, j, n), and means that
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a[i:i+n] == b[j:j+n]. The triples are monotonically increasing in i
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and in j.
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The last triple is a dummy, (len(a), len(b), 0), and is the only triple
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with n==0.
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>>> s = SequenceMatcher(None, "abxcd", "abcd")
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>>> s.get_matching_blocks()
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[(0, 0, 2), (3, 2, 2), (5, 4, 0)]
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get_opcodes()
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Return list of 5-tuples describing how to turn a into b.
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Each tuple is of the form (tag, i1, i2, j1, j2). The first tuple has
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i1 == j1 == 0, and remaining tuples have i1 == the i2 from the tuple
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preceding it, and likewise for j1 == the previous j2.
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The tags are strings, with these meanings:
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'replace': a[i1:i2] should be replaced by b[j1:j2]
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'delete': a[i1:i2] should be deleted.
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Note that j1==j2 in this case.
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'insert': b[j1:j2] should be inserted at a[i1:i1].
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Note that i1==i2 in this case.
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'equal': a[i1:i2] == b[j1:j2]
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>>> a = "qabxcd"
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>>> b = "abycdf"
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>>> s = SequenceMatcher(None, a, b)
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>>> for tag, i1, i2, j1, j2 in s.get_opcodes():
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... print ("%7s a[%d:%d] (%s) b[%d:%d] (%s)" %
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... (tag, i1, i2, a[i1:i2], j1, j2, b[j1:j2]))
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delete a[0:1] (q) b[0:0] ()
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equal a[1:3] (ab) b[0:2] (ab)
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replace a[3:4] (x) b[2:3] (y)
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equal a[4:6] (cd) b[3:5] (cd)
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insert a[6:6] () b[5:6] (f)
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ratio()
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Return a measure of the sequences' similarity (float in [0,1]).
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Where T is the total number of elements in both sequences, and M is the
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number of matches, this is 2,0*M / T. Note that this is 1 if the
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sequences are identical, and 0 if they have nothing in common.
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.ratio() is expensive to compute if you haven't already computed
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.get_matching_blocks() or .get_opcodes(), in which case you may want to
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try .quick_ratio() or .real_quick_ratio() first to get an upper bound.
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>>> s = SequenceMatcher(None, "abcd", "bcde")
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>>> s.ratio()
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0.75
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>>> s.quick_ratio()
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0.75
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>>> s.real_quick_ratio()
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1.0
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quick_ratio()
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Return an upper bound on .ratio() relatively quickly.
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This isn't defined beyond that it is an upper bound on .ratio(), and
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is faster to compute.
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real_quick_ratio():
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Return an upper bound on ratio() very quickly.
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This isn't defined beyond that it is an upper bound on .ratio(), and
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is faster to compute than either .ratio() or .quick_ratio().
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"""
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TRACE = 0
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class SequenceMatcher:
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def __init__(self, isjunk=None, a='', b=''):
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"""Construct a SequenceMatcher.
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Optional arg isjunk is None (the default), or a one-argument
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function that takes a sequence element and returns true iff the
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element is junk. None is equivalent to passing "lambda x: 0", i.e.
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no elements are considered to be junk. For example, pass
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lambda x: x in " \\t"
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if you're comparing lines as sequences of characters, and don't
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want to synch up on blanks or hard tabs.
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Optional arg a is the first of two sequences to be compared. By
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default, an empty string. The elements of a must be hashable. See
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also .set_seqs() and .set_seq1().
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Optional arg b is the second of two sequences to be compared. By
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default, an empty string. The elements of b must be hashable. See
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also .set_seqs() and .set_seq2().
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"""
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# Members:
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# a
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# first sequence
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# b
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# second sequence; differences are computed as "what do
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# we need to do to 'a' to change it into 'b'?"
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# b2j
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# for x in b, b2j[x] is a list of the indices (into b)
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# at which x appears; junk elements do not appear
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# b2jhas
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# b2j.has_key
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# fullbcount
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# for x in b, fullbcount[x] == the number of times x
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# appears in b; only materialized if really needed (used
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# only for computing quick_ratio())
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# matching_blocks
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# a list of (i, j, k) triples, where a[i:i+k] == b[j:j+k];
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# ascending & non-overlapping in i and in j; terminated by
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# a dummy (len(a), len(b), 0) sentinel
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# opcodes
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# a list of (tag, i1, i2, j1, j2) tuples, where tag is
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# one of
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# 'replace' a[i1:i2] should be replaced by b[j1:j2]
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# 'delete' a[i1:i2] should be deleted
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# 'insert' b[j1:j2] should be inserted
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# 'equal' a[i1:i2] == b[j1:j2]
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# isjunk
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# a user-supplied function taking a sequence element and
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# returning true iff the element is "junk" -- this has
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# subtle but helpful effects on the algorithm, which I'll
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# get around to writing up someday <0.9 wink>.
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# DON'T USE! Only __chain_b uses this. Use isbjunk.
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# isbjunk
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# for x in b, isbjunk(x) == isjunk(x) but much faster;
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# it's really the has_key method of a hidden dict.
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# DOES NOT WORK for x in a!
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self.isjunk = isjunk
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self.a = self.b = None
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self.set_seqs(a, b)
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def set_seqs(self, a, b):
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"""Set the two sequences to be compared.
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>>> s = SequenceMatcher()
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>>> s.set_seqs("abcd", "bcde")
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>>> s.ratio()
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0.75
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"""
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self.set_seq1(a)
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self.set_seq2(b)
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def set_seq1(self, a):
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"""Set the first sequence to be compared.
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The second sequence to be compared is not changed.
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>>> s = SequenceMatcher(None, "abcd", "bcde")
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>>> s.ratio()
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0.75
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>>> s.set_seq1("bcde")
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>>> s.ratio()
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1.0
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>>>
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SequenceMatcher computes and caches detailed information about the
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second sequence, so if you want to compare one sequence S against
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many sequences, use .set_seq2(S) once and call .set_seq1(x)
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repeatedly for each of the other sequences.
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See also set_seqs() and set_seq2().
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"""
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if a is self.a:
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return
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self.a = a
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self.matching_blocks = self.opcodes = None
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def set_seq2(self, b):
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"""Set the second sequence to be compared.
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The first sequence to be compared is not changed.
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>>> s = SequenceMatcher(None, "abcd", "bcde")
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>>> s.ratio()
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0.75
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>>> s.set_seq2("abcd")
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>>> s.ratio()
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1.0
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>>>
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SequenceMatcher computes and caches detailed information about the
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second sequence, so if you want to compare one sequence S against
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many sequences, use .set_seq2(S) once and call .set_seq1(x)
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repeatedly for each of the other sequences.
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See also set_seqs() and set_seq1().
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"""
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if b is self.b:
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return
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self.b = b
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self.matching_blocks = self.opcodes = None
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self.fullbcount = None
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self.__chain_b()
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# For each element x in b, set b2j[x] to a list of the indices in
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# b where x appears; the indices are in increasing order; note that
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# the number of times x appears in b is len(b2j[x]) ...
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# when self.isjunk is defined, junk elements don't show up in this
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# map at all, which stops the central find_longest_match method
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# from starting any matching block at a junk element ...
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# also creates the fast isbjunk function ...
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# note that this is only called when b changes; so for cross-product
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# kinds of matches, it's best to call set_seq2 once, then set_seq1
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# repeatedly
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def __chain_b(self):
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# Because isjunk is a user-defined (not C) function, and we test
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# for junk a LOT, it's important to minimize the number of calls.
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# Before the tricks described here, __chain_b was by far the most
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# time-consuming routine in the whole module! If anyone sees
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# Jim Roskind, thank him again for profile.py -- I never would
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# have guessed that.
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# The first trick is to build b2j ignoring the possibility
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# of junk. I.e., we don't call isjunk at all yet. Throwing
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# out the junk later is much cheaper than building b2j "right"
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# from the start.
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b = self.b
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self.b2j = b2j = {}
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self.b2jhas = b2jhas = b2j.has_key
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for i in xrange(len(b)):
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elt = b[i]
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if b2jhas(elt):
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b2j[elt].append(i)
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else:
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b2j[elt] = [i]
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# Now b2j.keys() contains elements uniquely, and especially when
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# the sequence is a string, that's usually a good deal smaller
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# than len(string). The difference is the number of isjunk calls
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# saved.
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isjunk, junkdict = self.isjunk, {}
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if isjunk:
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for elt in b2j.keys():
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if isjunk(elt):
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junkdict[elt] = 1 # value irrelevant; it's a set
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del b2j[elt]
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# Now for x in b, isjunk(x) == junkdict.has_key(x), but the
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# latter is much faster. Note too that while there may be a
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# lot of junk in the sequence, the number of *unique* junk
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# elements is probably small. So the memory burden of keeping
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# this dict alive is likely trivial compared to the size of b2j.
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self.isbjunk = junkdict.has_key
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def find_longest_match(self, alo, ahi, blo, bhi):
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"""Find longest matching block in a[alo:ahi] and b[blo:bhi].
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If isjunk is not defined:
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Return (i,j,k) such that a[i:i+k] is equal to b[j:j+k], where
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alo <= i <= i+k <= ahi
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blo <= j <= j+k <= bhi
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and for all (i',j',k') meeting those conditions,
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k >= k'
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i <= i'
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and if i == i', j <= j'
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In other words, of all maximal matching blocks, return one that
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starts earliest in a, and of all those maximal matching blocks that
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|
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)
|
|
"""
|
|
|
|
# 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.isbjunk
|
|
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 xrange(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
|
|
|
|
# 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
|
|
|
|
if TRACE:
|
|
print "get_matching_blocks", alo, ahi, blo, bhi
|
|
print " returns", besti, bestj, bestsize
|
|
return 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.
|
|
|
|
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)]
|
|
"""
|
|
|
|
if self.matching_blocks is not None:
|
|
return self.matching_blocks
|
|
self.matching_blocks = []
|
|
la, lb = len(self.a), len(self.b)
|
|
self.__helper(0, la, 0, lb, self.matching_blocks)
|
|
self.matching_blocks.append( (la, lb, 0) )
|
|
if TRACE:
|
|
print '*** matching blocks', self.matching_blocks
|
|
return self.matching_blocks
|
|
|
|
# builds list of matching blocks covering a[alo:ahi] and
|
|
# b[blo:bhi], appending them in increasing order to answer
|
|
|
|
def __helper(self, alo, ahi, blo, bhi, answer):
|
|
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 alo < i and blo < j:
|
|
self.__helper(alo, i, blo, j, answer)
|
|
answer.append(x)
|
|
if i+k < ahi and j+k < bhi:
|
|
self.__helper(i+k, ahi, j+k, bhi, answer)
|
|
|
|
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 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 = reduce(lambda sum, triple: sum + triple[-1],
|
|
self.get_matching_blocks(), 0)
|
|
return 2.0 * 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.has_key, 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 2.0 * 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 2.0 * min(la, lb) / (la + lb)
|
|
|
|
def get_close_matches(word, possibilities, n=3, cutoff=0.6):
|
|
"""Use SequenceMatcher to return list of the best "good enough" matches.
|
|
|
|
word is a sequence for which close matches are desired (typically a
|
|
string).
|
|
|
|
possibilities is a list of sequences against which to match word
|
|
(typically a list of strings).
|
|
|
|
Optional arg n (default 3) is the maximum number of close matches to
|
|
return. n must be > 0.
|
|
|
|
Optional arg cutoff (default 0.6) is a float in [0, 1]. Possibilities
|
|
that don't score at least that similar to word are ignored.
|
|
|
|
The best (no more than n) matches among the possibilities are returned
|
|
in a list, sorted by similarity score, most similar first.
|
|
|
|
>>> get_close_matches("appel", ["ape", "apple", "peach", "puppy"])
|
|
['apple', 'ape']
|
|
>>> import keyword
|
|
>>> get_close_matches("wheel", keyword.kwlist)
|
|
['while']
|
|
>>> get_close_matches("apple", keyword.kwlist)
|
|
[]
|
|
>>> get_close_matches("accept", keyword.kwlist)
|
|
['except']
|
|
"""
|
|
|
|
if not n > 0:
|
|
raise ValueError("n must be > 0: " + `n`)
|
|
if not 0.0 <= cutoff <= 1.0:
|
|
raise ValueError("cutoff must be in [0.0, 1.0]: " + `cutoff`)
|
|
result = []
|
|
s = SequenceMatcher()
|
|
s.set_seq2(word)
|
|
for x in possibilities:
|
|
s.set_seq1(x)
|
|
if s.real_quick_ratio() >= cutoff and \
|
|
s.quick_ratio() >= cutoff and \
|
|
s.ratio() >= cutoff:
|
|
result.append((s.ratio(), x))
|
|
# Sort by score.
|
|
result.sort()
|
|
# Retain only the best n.
|
|
result = result[-n:]
|
|
# Move best-scorer to head of list.
|
|
result.reverse()
|
|
# Strip scores.
|
|
# Python 2.x list comprehensions: return [x for score, x in result]
|
|
return_result = []
|
|
for score, x in result:
|
|
return_result.append(x)
|
|
return return_result
|
|
|
|
def _test():
|
|
import doctest, difflib
|
|
return doctest.testmod(difflib)
|
|
|
|
if __name__ == "__main__":
|
|
_test()
|