difflib.SequenceMatcher.find_longest_match()

find_longest_match(alo, ahi, blo, bhi)

Find longest matching block in a[alo:ahi] and b[blo:bhi].

If isjunk was omitted or None, find_longest_match() returns (i, j, k) such that a[i:i+k] is equal to b[j:j+k], where alo <= i <= i+k <= ahi and blo <= j <= j+k <= bhi. For all (i', j', k') meeting those conditions, the additional conditions k >= k', i <= i', and if i == i', j <= j' are also met. 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 was provided, 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, this returns (alo, blo, 0).

This method returns a named tuple Match(a, b, size).