-
-
Notifications
You must be signed in to change notification settings - Fork 34.2k
Expand file tree
/
Copy pathdifflib.py
More file actions
2056 lines (1763 loc) · 81.4 KB
/
difflib.py
File metadata and controls
2056 lines (1763 loc) · 81.4 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
"""
Module difflib -- helpers for computing deltas between objects.
Function get_close_matches(word, possibilities, n=3, cutoff=0.6):
Use SequenceMatcher to return list of the best "good enough" matches.
Function context_diff(a, b):
For two lists of strings, return a delta in context diff format.
Function ndiff(a, b):
Return a delta: the difference between `a` and `b` (lists of strings).
Function restore(delta, which):
Return one of the two sequences that generated an ndiff delta.
Function unified_diff(a, b):
For two lists of strings, return a delta in unified diff format.
Class SequenceMatcher:
A flexible class for comparing pairs of sequences of any type.
Class Differ:
For producing human-readable deltas from sequences of lines of text.
Class HtmlDiff:
For producing HTML side by side comparison with change highlights.
"""
__all__ = ['get_close_matches', 'ndiff', 'restore', 'SequenceMatcher',
'Differ','IS_CHARACTER_JUNK', 'IS_LINE_JUNK', 'context_diff',
'unified_diff', 'diff_bytes', 'HtmlDiff', 'Match']
from heapq import nlargest as _nlargest
from collections import namedtuple as _namedtuple
from types import GenericAlias
Match = _namedtuple('Match', 'a b size')
def _calculate_ratio(matches, length):
if length:
return 2.0 * matches / length
return 1.0
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 syncing 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.
"""
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 = {}
for i, elt in enumerate(b):
indices = b2j.setdefault(elt, [])
indices.append(i)
# 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]
# 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=0, ahi=None, blo=0, bhi=None):
"""Find longest matching block in a[alo:ahi] and b[blo:bhi].
By default it will find the longest match in the entirety of a and b.
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__
if ahi is None:
ahi = len(a)
if bhi is None:
bhi = len(b)
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)
__class_getitem__ = classmethod(GenericAlias)
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 as _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: %r" % (n,))
if not 0.0 <= cutoff <= 1.0:
raise ValueError("cutoff must be in [0.0, 1.0]: %r" % (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))
# Move the best scorers to head of list
result = _nlargest(n, result)
# Strip scores for the best n matches
return [x for score, x in result]
def _keep_original_ws(s, tag_s):
"""Replace whitespace with the original whitespace characters in `s`"""
return ''.join(
c if tag_c == " " and c.isspace() else tag_c
for c, tag_c in zip(s, tag_s)
)
class Differ:
r"""
Differ is a class for comparing sequences of lines of text, and
producing human-readable differences or deltas. Differ uses
SequenceMatcher both to compare sequences of lines, and to compare
sequences of characters within similar (near-matching) lines.
Each line of a Differ delta begins with a two-letter code:
'- ' line unique to sequence 1
'+ ' line unique to sequence 2
' ' line common to both sequences
'? ' line not present in either input sequence
Lines beginning with '? ' attempt to guide the eye to intraline
differences, and were not present in either input sequence. These lines
can be confusing if the sequences contain tab characters.
Note that Differ makes no claim to produce a *minimal* diff. To the
contrary, minimal diffs are often counter-intuitive, because they synch
up anywhere possible, sometimes accidental matches 100 pages apart.
Restricting synch points to contiguous matches preserves some notion of
locality, at the occasional cost of producing a longer diff.
Example: Comparing two texts.
First we set up the texts, sequences of individual single-line strings
ending with newlines (such sequences can also be obtained from the
`readlines()` method of file-like objects):
>>> text1 = ''' 1. Beautiful is better than ugly.
... 2. Explicit is better than implicit.
... 3. Simple is better than complex.
... 4. Complex is better than complicated.
... '''.splitlines(keepends=True)
>>> len(text1)
4
>>> text1[0][-1]
'\n'
>>> text2 = ''' 1. Beautiful is better than ugly.
... 3. Simple is better than complex.
... 4. Complicated is better than complex.
... 5. Flat is better than nested.
... '''.splitlines(keepends=True)
Next we instantiate a Differ object:
>>> d = Differ()
Note that when instantiating a Differ object we may pass functions to
filter out line and character 'junk'. See Differ.__init__ for details.
Finally, we compare the two:
>>> result = list(d.compare(text1, text2))
'result' is a list of strings, so let's pretty-print it:
>>> from pprint import pprint as _pprint
>>> _pprint(result)
[' 1. Beautiful is better than ugly.\n',
'- 2. Explicit is better than implicit.\n',
'- 3. Simple is better than complex.\n',
'+ 3. Simple is better than complex.\n',
'? ++\n',
'- 4. Complex is better than complicated.\n',
'? ^ ---- ^\n',
'+ 4. Complicated is better than complex.\n',
'? ++++ ^ ^\n',
'+ 5. Flat is better than nested.\n']
As a single multi-line string it looks like this:
>>> print(''.join(result), end="")
1. Beautiful is better than ugly.
- 2. Explicit is better than implicit.
- 3. Simple is better than complex.
+ 3. Simple is better than complex.
? ++
- 4. Complex is better than complicated.
? ^ ---- ^
+ 4. Complicated is better than complex.
? ++++ ^ ^
+ 5. Flat is better than nested.
"""
def __init__(self, linejunk=None, charjunk=None):
"""
Construct a text differencer, with optional filters.
The two optional keyword parameters are for filter functions:
- `linejunk`: A function that should accept a single string argument,
and return true iff the string is junk. The module-level function
`IS_LINE_JUNK` may be used to filter out lines without visible
characters, except for at most one splat ('#'). It is recommended
to leave linejunk None; the underlying SequenceMatcher class has
an adaptive notion of "noise" lines that's better than any static
definition the author has ever been able to craft.
- `charjunk`: A function that should accept a string of length 1. The
module-level function `IS_CHARACTER_JUNK` may be used to filter out
whitespace characters (a blank or tab; **note**: bad idea to include
newline in this!). Use of IS_CHARACTER_JUNK is recommended.
"""
self.linejunk = linejunk
self.charjunk = charjunk
def compare(self, a, b):
r"""
Compare two sequences of lines; generate the resulting delta.
Each sequence must contain individual single-line strings ending with
newlines. Such sequences can be obtained from the `readlines()` method
of file-like objects. The delta generated also consists of newline-
terminated strings, ready to be printed as-is via the writelines()
method of a file-like object.
Example:
>>> print(''.join(Differ().compare('one\ntwo\nthree\n'.splitlines(True),
... 'ore\ntree\nemu\n'.splitlines(True))),
... end="")
- one
? ^
+ ore
? ^
- two
- three
? -
+ tree
+ emu
"""
cruncher = SequenceMatcher(self.linejunk, a, b)
for tag, alo, ahi, blo, bhi in cruncher.get_opcodes():
if tag == 'replace':
g = self._fancy_replace(a, alo, ahi, b, blo, bhi)
elif tag == 'delete':
g = self._dump('-', a, alo, ahi)
elif tag == 'insert':
g = self._dump('+', b, blo, bhi)
elif tag == 'equal':
g = self._dump(' ', a, alo, ahi)
else:
raise ValueError('unknown tag %r' % (tag,))
yield from g
def _dump(self, tag, x, lo, hi):
"""Generate comparison results for a same-tagged range."""
for i in range(lo, hi):
yield '%s %s' % (tag, x[i])
def _plain_replace(self, a, alo, ahi, b, blo, bhi):
assert alo < ahi and blo < bhi
# dump the shorter block first -- reduces the burden on short-term
# memory if the blocks are of very different sizes
if bhi - blo < ahi - alo:
first = self._dump('+', b, blo, bhi)
second = self._dump('-', a, alo, ahi)
else:
first = self._dump('-', a, alo, ahi)
second = self._dump('+', b, blo, bhi)
for g in first, second:
yield from g
def _fancy_replace(self, a, alo, ahi, b, blo, bhi):
r"""
When replacing one block of lines with another, search the blocks
for *similar* lines; the best-matching pair (if any) is used as a
synch point, and intraline difference marking is done on the
similar pair. Lots of work, but often worth it.
Example:
>>> d = Differ()
>>> results = d._fancy_replace(['abcDefghiJkl\n'], 0, 1,
... ['abcdefGhijkl\n'], 0, 1)
>>> print(''.join(results), end="")
- abcDefghiJkl
? ^ ^ ^
+ abcdefGhijkl
? ^ ^ ^
"""
# don't synch up unless the lines have a similarity score of at
# least cutoff; best_ratio tracks the best score seen so far
best_ratio, cutoff = 0.74, 0.75
cruncher = SequenceMatcher(self.charjunk)
eqi, eqj = None, None # 1st indices of equal lines (if any)
# search for the pair that matches best without being identical
# (identical lines must be junk lines, & we don't want to synch up
# on junk -- unless we have to)
for j in range(blo, bhi):
bj = b[j]
cruncher.set_seq2(bj)
for i in range(alo, ahi):
ai = a[i]
if ai == bj:
if eqi is None:
eqi, eqj = i, j
continue
cruncher.set_seq1(ai)
# computing similarity is expensive, so use the quick
# upper bounds first -- have seen this speed up messy
# compares by a factor of 3.
# note that ratio() is only expensive to compute the first
# time it's called on a sequence pair; the expensive part
# of the computation is cached by cruncher
if cruncher.real_quick_ratio() > best_ratio and \
cruncher.quick_ratio() > best_ratio and \
cruncher.ratio() > best_ratio:
best_ratio, best_i, best_j = cruncher.ratio(), i, j
if best_ratio < cutoff:
# no non-identical "pretty close" pair
if eqi is None:
# no identical pair either -- treat it as a straight replace
yield from self._plain_replace(a, alo, ahi, b, blo, bhi)
return
# no close pair, but an identical pair -- synch up on that
best_i, best_j, best_ratio = eqi, eqj, 1.0
else:
# there's a close pair, so forget the identical pair (if any)
eqi = None
# a[best_i] very similar to b[best_j]; eqi is None iff they're not
# identical
# pump out diffs from before the synch point
yield from self._fancy_helper(a, alo, best_i, b, blo, best_j)
# do intraline marking on the synch pair
aelt, belt = a[best_i], b[best_j]
56D3
if eqi is None:
# pump out a '-', '?', '+', '?' quad for the synched lines
atags = btags = ""
cruncher.set_seqs(aelt, belt)
for tag, ai1, ai2, bj1, bj2 in cruncher.get_opcodes():
la, lb = ai2 - ai1, bj2 - bj1
if tag == 'replace':
atags += '^' * la
btags += '^' * lb
elif tag == 'delete':
atags += '-' * la
elif tag == 'insert':
btags += '+' * lb
elif tag == 'equal':
atags += ' ' * la
btags += ' ' * lb
else:
raise ValueError('unknown tag %r' % (tag,))
yield from self._qformat(aelt, belt, atags, btags)
else:
# the synch pair is identical
yield ' ' + aelt
# pump out diffs from after the synch point
yield from self._fancy_helper(a, best_i+1, ahi, b, best_j+1, bhi)
def _fancy_helper(self, a, alo, ahi, b, blo, bhi):
g = []
if alo < ahi:
if blo < bhi:
g = self._fancy_replace(a, alo, ahi, b, blo, bhi)
else:
g = self._dump('-', a, alo, ahi)
elif blo < bhi:
g = self._dump('+', b, blo, bhi)
yield from g
def _qformat(self, aline, bline, atags, btags):
r"""