optimization question

[Terry Reedy]

"Andrew Koenig" <ark at research.att.com> wrote in message
news:yu99d6sp2gel.fsf at europa.research.att.com...
> >> def eqsub(s, i, j, t):
> >> return (len(t) == j-i) and s[i:j] == t
> >>
> >> which avoids building the substrings unless necessary.
> >>
>
> Bernard> Wouldn't the following avoid it altogether?
>
> Bernard>     return (len(t) == j-i) and (s.find(t,i) != -1)
>
> Yes, it looks like it would.  On the other hand, I might want
> to generalize the technique to sequences other than strings...

def subseqmatch(seq, i, j, sub):
  if len(sub) != j-i: return False
  for item in sub:
    if item != seq[i]: return False
    i += 1
  return True

print subseqmatch('0123456789', 3, 6, '345'),\
      subseqmatch([0,1,2,3,4,5,6], 3, 6, [3,4,5]),\
      subseqmatch((0,1,2,3,4,5,6), 3, 6, [3,4,6]) # miss

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Terry J. Reedy