Gerard flanagan wrote:
George Sakkis wrote:
..
Note that this works correctly only if the versions are already sorted
by major version.
Yes, I should have mentioned it. Here's a fuller example below. There's
maybe better ways of sorting version numbers, but this is what I do.
Indeed, your sort takes George's objection too litterally, what's needed for a
correct endresult is only that major versions be grouped together, and this is
most simply obtained by sorting the input data in (default) string order, is it
not ?
data = [ "1.2.2.2", "1.2.2.3", "1.3.1.2", "1.1.1.1", "1.3.14.5",
"1.3.21.6" ]
from itertools import groupby
import re
RXBUILDSORT = re.compile(r'\d+|[a-zA-Z]')
def versionsort(s):
key = []
for part in RXBUILDSORT.findall(s.lower()):
try:
key.append(int(part))
except ValueError:
key.append(ord(part))
return tuple(key)
data.sort(key=versionsort)
print data
datadict = \
dict((k, len(list(g))) for k,g in groupby(data, lambda s: s[:3]))
print datadict
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