In article <[EMAIL PROTECTED]>,
<[EMAIL PROTECTED]> wrote:
>hello,
>
>i'm looking for a way to have a list of number grouped by consecutive
>interval, after a search, for example :
>
>[3, 6, 7, 8, 12, 13, 15]
>
>=>
>
>[[3, 4], [6,9], [12, 14], [15, 16]]
>
>(6, not following 3, so 3 => [3:4] ; 7, 8 following 6 so 6, 7, 8 =>
>[6:9], and so on)
>
>i was able to to it without generators/yield but i think it could be
>better with them, may be do you an idea?
There are probably better ways, but this works
>
>best regards,
>
>J.
class IterInterval(object):
"""Create an iterator which, given a list of integers,
for each run of consecutive integers i...j, yields a two
element list [i, j + 1]
Singleton lists [i] return [i, i + 1]
Empty lists return None
"""
def __init__(self, seq):
self.seq = seq
self.firstval = None
self.index = 0
def __iter__(self):
# firstval = the start of a run of consecutive integers
# lastval = the last value found in the run
# nextval = the most recent value taken from the input list
if not self.firstval:
# set up the first iteration
if self.index >= len(self.seq):
# empty list, return
raise StopIteration
self.firstval = lastval = int(self.seq[self.index])
self.index += 1
while True:
if self.index >= len(self.seq):
# list exhausted, output the last value read
yield [self.firstval, lastval + 1]
raise StopIteration
nextval = int(self.seq[self.index])
self.index += 1
if nextval == lastval + 1:
lastval = nextval
continue
else:
# end of run - output the run, reset for next call
yield [self.firstval, lastval + 1]
self.firstval = lastval = nextval
continue
if __name__ == '__main__':
for l in [[3, 6, 7, 8, 12, 13, 15], [2], []]:
print l, "=>", [lst for lst in IterInterval(l)]
/usr/home/jes% python interval.py
[3, 6, 7, 8, 12, 13, 15] => [[3, 4], [6, 9], [12, 14], [15, 16]]
[3] => [[3, 4]]
[] => []
--
Jim Segrave ([EMAIL PROTECTED])
--
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