[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?
>
> best regards,
>
a list comprehension/itertools version (this won't work with an empty
list):
from itertools import groupby
a = [3, 6, 7, 8, 12, 13, 15]
result = [[3, 4], [6,9], [12, 14], [15, 16]]
b = [ list(g)[0] for k,g in groupby(range(a[0],a[-1]+2), lambda x:
x in a)]
c = [ b[i:i+2] for i in range(0,len(a),2) ]
assert c == result
Gerard
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