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672 lines
26 KiB
Python
Executable file
672 lines
26 KiB
Python
Executable file
#! /usr/bin/env python
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# Released to the public domain $JustDate: 3/16/98 $,
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# by Tim Peters (email tim_one@email.msn.com).
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# ndiff file1 file2 -- a human-friendly file differencer.
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# $Revision$
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# SequenceMatcher tries to compute a "human-friendly diff" between
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# two sequences (chiefly picturing a file as a sequence of lines,
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# and a line as a sequence of characters, here). Unlike UNIX(tm) diff,
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# e.g., the fundamental notion is the longest *contiguous* & junk-free
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# matching subsequence. That's what catches peoples' eyes. The
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# Windows(tm) windiff has another interesting notion, pairing up elements
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# that appear uniquely in each sequence. That, and the method here,
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# appear to yield more intuitive difference reports than does diff. This
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# method appears to be the least vulnerable to synching up on blocks
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# of "junk lines", though (like blank lines in ordinary text files,
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# or maybe "<P>" lines in HTML files). That may be because this is
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# the only method of the 3 that has a *concept* of "junk" <wink>.
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#
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# Note that ndiff makes no claim to produce a *minimal* diff. To the
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# contrary, minimal diffs are often counter-intuitive, because they
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# synch up anywhere possible, sometimes accidental matches 100 pages
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# apart. Restricting synch points to contiguous matches preserves some
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# notion of locality, at the occasional cost of producing a longer diff.
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#
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# With respect to junk, an earlier verion of ndiff simply refused to
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# *start* a match with a junk element. The result was cases like this:
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# before: private Thread currentThread;
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# after: private volatile Thread currentThread;
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# If you consider whitespace to be junk, the longest continguous match
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# not starting with junk is "e Thread currentThread". So ndiff reported
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# that "e volatil" was inserted between the 't' and the 'e' in "private".
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# While an accurate view, to people that's absurd. The current version
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# looks for matching blocks that are entirely junk-free, then extends the
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# longest one of those as far as possible but only with matching junk.
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# So now "currentThread" is matched, then extended to suck up the
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# preceding blank; then "private" is matched, and extended to suck up the
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# following blank; then "Thread" is matched; and finally ndiff reports
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# that "volatile " was inserted before "Thread". The only quibble
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# remaining is that perhaps it was really the case that " volative"
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# was inserted after "private". I can live with that <wink>.
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#
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# NOTE on the output: From an ndiff report,
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# 1) The first file can be recovered by retaining only lines that begin
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# with " " or "- ", and deleting those 2-character prefixes.
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# 2) The second file can be recovered similarly, but by retaining only
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# " " and "+ " lines.
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# 3) Lines beginning with "? " attempt to guide the eye to intraline
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# differences, and were not present in either input file.
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#
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# COROLLARY:
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# On Unix, the second file can be recovered by piping the output through
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# sed -n '/^[+ ] /s/^..//p'
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# Modifications to recover the first file are left as an exercise for
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# the reader.
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#
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# NOTE on junk: the module-level names
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# IS_LINE_JUNK
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# IS_CHARACTER_JUNK
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# can be set to any functions you like. The first one should accept
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# a single string argument, and return true iff the string is junk.
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# The default is whether the regexp r"\s*#?\s*$" matches (i.e., a
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# line without visible characters, except for at most one splat).
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# The second should accept a string of length 1 etc. The default is
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# whether the character is a blank or tab (note: bad idea to include
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# newline in this!).
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#
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# After setting those, you can call fcompare(f1name, f2name) with the
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# names of the files you want to compare. The difference report
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# is sent to stdout. Or you can call main(), which expects to find
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# (exactly) the two file names in sys.argv.
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import string
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TRACE = 0
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# define what "junk" means
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import re
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def IS_LINE_JUNK(line, pat=re.compile(r"\s*#?\s*$").match):
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return pat(line) is not None
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def IS_CHARACTER_JUNK(ch, ws=" \t"):
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return ch in ws
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del re
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class SequenceMatcher:
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def __init__(self, isjunk=None, a='', b=''):
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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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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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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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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(0, 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, returns one
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that starts earliest in a, and of all those maximal matching
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blocks that start earliest in a, returns the one that starts
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earliest in b.
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If isjunk is defined, first the longest matching block is
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determined as above, but with the additional restriction that
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no junk element appears in the block. Then that block is
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extended as far as possible by matching (only) junk elements on
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both sides. So the resulting block never matches on junk except
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as identical junk happens to be adjacent to an "interesting"
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match.
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If no blocks match, returns (alo, blo, 0).
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"""
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# CAUTION: stripping common prefix or suffix would be incorrect.
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# E.g.,
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# ab
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# acab
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# Longest matching block is "ab", but if common prefix is
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# stripped, it's "a" (tied with "b"). UNIX(tm) diff does so
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# strip, so ends up claiming that ab is changed to acab by
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# inserting "ca" in the middle. That's minimal but unintuitive:
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# "it's obvious" that someone inserted "ac" at the front.
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# Windiff ends up at the same place as diff, but by pairing up
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# the unique 'b's and then matching the first two 'a's.
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# find longest junk-free match
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a, b, b2j, isbjunk = self.a, self.b, self.b2j, self.isbjunk
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besti, bestj, bestsize = alo, blo, 0
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for i in xrange(alo, ahi):
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# check for longest match starting at a[i]
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if i + bestsize >= ahi:
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# we're too far right to get a new best
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break
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# look at all instances of a[i] in b; note that because
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# b2j has no junk keys, the loop is skipped if a[i] is junk
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for j in b2j.get(a[i], []):
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# a[i] matches b[j]
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if j < blo:
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continue
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if j + bestsize >= bhi:
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# we're too far right to get a new best, here or
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# anywhere to the right
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break
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if a[i + bestsize] != b[j + bestsize]:
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# can't be longer match; this test is not necessary
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# for correctness, but is a huge win for efficiency
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continue
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# set k to length of match
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k = 1 # a[i] == b[j] already known
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while i + k < ahi and j + k < bhi and \
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a[i+k] == b[j+k] and not isbjunk(b[j+k]):
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k = k + 1
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if k > bestsize:
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besti, bestj, bestsize = i, j, k
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if i + bestsize >= ahi:
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# only time in my life I really wanted a
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# labelled break <wink> -- we're done with
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# both loops now
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break
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# Now that we have a wholly interesting match (albeit possibly
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# empty!), we may as well suck up the matching junk on each
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# side of it too. Can't think of a good reason not to, and it
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# saves post-processing the (possibly considerable) expense of
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# figuring out what to do with it. In the case of an empty
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# interesting match, this is clearly the right thing to do,
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# because no other kind of match is possible in the regions.
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while besti > alo and bestj > blo and \
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isbjunk(b[bestj-1]) and \
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a[besti-1] == b[bestj-1]:
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besti, bestj, bestsize = besti-1, bestj-1, bestsize+1
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while besti+bestsize < ahi and bestj+bestsize < bhi and \
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isbjunk(b[bestj+bestsize]) and \
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a[besti+bestsize] == b[bestj+bestsize]:
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bestsize = bestsize + 1
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if TRACE:
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print "get_matching_blocks", alo, ahi, blo, bhi
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print " returns", besti, bestj, bestsize
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return besti, bestj, bestsize
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# A different implementation, using a binary doubling technique that
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# does far fewer element compares (trades 'em for integer compares),
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# and has n*lg n worst-case behavior. Alas, the code is much harder
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# to follow (the details are tricky!), and in most cases I've seen,
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# it takes at least 50% longer than the "clever dumb" method above;
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# probably due to creating layers of small dicts.
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# NOTE: this no longer matches the version above wrt junk; remains
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# too unpromising to update it; someday, though ...
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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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#
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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, returns one
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# that starts earliest in a, and of all those maximal matching
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# blocks that start earliest in a, returns the one that starts
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# earliest in b.
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#
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# If no blocks match, returns (alo, blo, 0).
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# """
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#
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# a, b2j = self.a, self.b2j
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# # alljs[size][i] is a set of all j's s.t. a[i:i+len] matches
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# # b[j:j+len]
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# alljs = {}
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# alljs[1] = js = {}
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# ahits = {}
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# for i in xrange(alo, ahi):
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# elt = a[i]
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# if ahits.has_key(elt):
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# js[i] = ahits[elt]
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# continue
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# if b2j.has_key(elt):
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# in_range = {}
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# for j in b2j[elt]:
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# if j >= blo:
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# if j >= bhi:
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# break
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# in_range[j] = 1
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# if in_range:
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# ahits[elt] = js[i] = in_range
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# del ahits
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# size = 1
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# while js:
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# oldsize = size
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# size = size + size
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# oldjs = js
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# alljs[size] = js = {}
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# for i in oldjs.keys():
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# # i has matches of size oldsize
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# if not oldjs.has_key(i + oldsize):
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# # can't double it
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# continue
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# second_js = oldjs[i + oldsize]
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# answer = {}
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# for j in oldjs[i].keys():
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# if second_js.has_key(j + oldsize):
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# answer[j] = 1
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# if answer:
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# js[i] = answer
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# del alljs[size]
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# size = size >> 1 # max power of 2 with a match
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# if not size:
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# return alo, blo, 0
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# besti, bestj, bestsize = alo, blo, 0
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# fatis = alljs[size].keys()
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# fatis.sort()
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# for i in fatis:
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# # figure out longest match starting at a[i]
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# totalsize = halfsize = size
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# # i has matches of len totalsize at the indices in js
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# js = alljs[size][i].keys()
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# while halfsize > 1:
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# halfsize = halfsize >> 1
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# # is there a match of len halfsize starting at
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# # i + totalsize?
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# newjs = []
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# if alljs[halfsize].has_key(i + totalsize):
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# second_js = alljs[halfsize][i + totalsize]
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# for j in js:
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# if second_js.has_key(j + totalsize):
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# newjs.append(j)
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# if newjs:
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# totalsize = totalsize + halfsize
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# js = newjs
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# if totalsize > bestsize:
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# besti, bestj, bestsize = i, min(js), totalsize
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# return besti, bestj, bestsize
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def get_matching_blocks(self):
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if self.matching_blocks is not None:
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return self.matching_blocks
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self.matching_blocks = []
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la, lb = len(self.a), len(self.b)
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self.__helper(0, la, 0, lb, self.matching_blocks)
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self.matching_blocks.append( (la, lb, 0) )
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if TRACE:
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print '*** matching blocks', self.matching_blocks
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return self.matching_blocks
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# builds list of matching blocks covering a[alo:ahi] and
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# b[blo:bhi], appending them in increasing order to answer
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def __helper(self, alo, ahi, blo, bhi, answer):
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i, j, k = x = self.find_longest_match(alo, ahi, blo, bhi)
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# a[alo:i] vs b[blo:j] unknown
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# a[i:i+k] same as b[j:j+k]
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# a[i+k:ahi] vs b[j+k:bhi] unknown
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if k:
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if alo < i and blo < j:
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self.__helper(alo, i, blo, j, answer)
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answer.append( x )
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if i+k < ahi and j+k < bhi:
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self.__helper(i+k, ahi, j+k, bhi, answer)
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def ratio(self):
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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
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M is the number of matches, this is 2*M / T.
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Note that this is 1 if the sequences are identical, and 0 if
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they have nothing in common.
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"""
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matches = reduce(lambda sum, triple: sum + triple[-1],
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self.get_matching_blocks(), 0)
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return 2.0 * matches / (len(self.a) + len(self.b))
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def quick_ratio(self):
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"""Return an upper bound on ratio() relatively quickly."""
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# viewing a and b as multisets, set matches to the cardinality
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# of their intersection; this counts the number of matches
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# without regard to order, so is clearly an upper bound
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if self.fullbcount is None:
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self.fullbcount = fullbcount = {}
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for elt in self.b:
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fullbcount[elt] = fullbcount.get(elt, 0) + 1
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fullbcount = self.fullbcount
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|
# 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"""
|
|
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_opcodes(self):
|
|
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
|
|
|
|
# meant for dumping lines
|
|
def dump(tag, x, lo, hi):
|
|
for i in xrange(lo, hi):
|
|
print tag, x[i],
|
|
|
|
# figure out which mark to stick under characters in lines that
|
|
# have changed (blank = same, - = deleted, + = inserted, ^ = replaced)
|
|
_combine = { ' ': ' ',
|
|
'. ': '-',
|
|
' .': '+',
|
|
'..': '^' }
|
|
|
|
def plain_replace(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:
|
|
dump('+', b, blo, bhi)
|
|
dump('-', a, alo, ahi)
|
|
else:
|
|
dump('-', a, alo, ahi)
|
|
dump('+', b, blo, bhi)
|
|
|
|
# When replacing one block of lines with another, this guy searches
|
|
# 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.
|
|
|
|
def fancy_replace(a, alo, ahi, b, blo, bhi):
|
|
if TRACE:
|
|
print '*** fancy_replace', alo, ahi, blo, bhi
|
|
dump('>', a, alo, ahi)
|
|
dump('<', b, blo, bhi)
|
|
|
|
# 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(IS_CHARACTER_JUNK)
|
|
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 xrange(blo, bhi):
|
|
bj = b[j]
|
|
cruncher.set_seq2(bj)
|
|
for i in xrange(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
|
|
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
|
|
if TRACE:
|
|
print '*** best_ratio', best_ratio, best_i, best_j
|
|
dump('>', a, best_i, best_i+1)
|
|
dump('<', b, best_j, best_j+1)
|
|
|
|
# pump out diffs from before the synch point
|
|
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]
|
|
if eqi is None:
|
|
# pump out a '-', '+', '?' triple 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 = atags + '.' * la
|
|
btags = btags + '.' * lb
|
|
elif tag == 'delete':
|
|
atags = atags + '.' * la
|
|
elif tag == 'insert':
|
|
btags = btags + '.' * lb
|
|
elif tag == 'equal':
|
|
atags = atags + ' ' * la
|
|
btags = btags + ' ' * lb
|
|
else:
|
|
raise ValueError, 'unknown tag ' + `tag`
|
|
la, lb = len(atags), len(btags)
|
|
if la < lb:
|
|
atags = atags + ' ' * (lb - la)
|
|
elif lb < la:
|
|
btags = btags + ' ' * (la - lb)
|
|
combined = map(lambda x,y: _combine[x+y], atags, btags)
|
|
print '-', aelt, '+', belt, '?', \
|
|
string.rstrip(string.join(combined, ''))
|
|
else:
|
|
# the synch pair is identical
|
|
print ' ', aelt,
|
|
|
|
# pump out diffs from after the synch point
|
|
fancy_helper(a, best_i+1, ahi, b, best_j+1, bhi)
|
|
|
|
def fancy_helper(a, alo, ahi, b, blo, bhi):
|
|
if alo < ahi:
|
|
if blo < bhi:
|
|
fancy_replace(a, alo, ahi, b, blo, bhi)
|
|
else:
|
|
dump('-', a, alo, ahi)
|
|
elif blo < bhi:
|
|
dump('+', b, blo, bhi)
|
|
|
|
# open a file & return the file object; gripe and return 0 if it
|
|
# couldn't be opened
|
|
def fopen(fname):
|
|
try:
|
|
return open(fname, 'r')
|
|
except IOError, detail:
|
|
print "couldn't open " + fname + ": " + `detail`
|
|
return 0
|
|
|
|
# open two files & spray the diff to stdout; return false iff a problem
|
|
def fcompare(f1name, f2name):
|
|
f1 = fopen(f1name)
|
|
f2 = fopen(f2name)
|
|
if not f1 or not f2:
|
|
return 0
|
|
|
|
a = f1.readlines(); f1.close()
|
|
b = f2.readlines(); f2.close()
|
|
|
|
cruncher = SequenceMatcher(IS_LINE_JUNK, a, b)
|
|
for tag, alo, ahi, blo, bhi in cruncher.get_opcodes():
|
|
if tag == 'replace':
|
|
fancy_replace(a, alo, ahi, b, blo, bhi)
|
|
elif tag == 'delete':
|
|
dump('-', a, alo, ahi)
|
|
elif tag == 'insert':
|
|
dump('+', b, blo, bhi)
|
|
elif tag == 'equal':
|
|
dump(' ', a, alo, ahi)
|
|
else:
|
|
raise ValueError, 'unknown tag ' + `tag`
|
|
|
|
return 1
|
|
|
|
# get file names from argv & compare; return false iff a problem
|
|
def main():
|
|
from sys import argv
|
|
if len(argv) != 3:
|
|
print 'need 2 args'
|
|
return 0
|
|
[f1name, f2name] = argv[1:3]
|
|
print '-:', f1name
|
|
print '+:', f2name
|
|
return fcompare(f1name, f2name)
|
|
|
|
if __name__ == '__main__':
|
|
if 1:
|
|
main()
|
|
else:
|
|
import profile, pstats
|
|
statf = "ndiff.pro"
|
|
profile.run("main()", statf)
|
|
stats = pstats.Stats(statf)
|
|
stats.strip_dirs().sort_stats('time').print_stats()
|
|
|