On Jun 2, 10:19 am, Steve Howell <[EMAIL PROTECTED]> wrote:
> George Sakkis produced the following cookbook recipe,
> which addresses a common problem that comes up on this
> mailing list:
ISTM, this is a common mailing list problem because it is fun
to solve, not because people actually need it on a day-to-day basis.
In that spirit, it would be fun to compare several different
approaches to the same problem using re.finditer, itertools.groupby,
or the tokenize module. To get the ball rolling, here is one variant:
from itertools import groupby
def blocks(s, start, end):
def classify(c, ingroup=[0], delim={start:2, end:3}):
result = delim.get(c, ingroup[0])
ingroup[0] = result in (1, 2)
return result
return [tuple(g) for k, g in groupby(s, classify) if k == 1]
print blocks('the <quick> brown <fox> jumped', start='<', end='>')
One observation is that groupby() is an enormously flexible tool.
Given a well crafted key= function, it makes short work of almost
any data partitioning problem.
Raymond
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