dict generator question

[George Sakkis]

On Sep 18, 10:54 am, "Simon Mullis" <si... at mullis.co.uk> wrote:
> Hi,
>
> Let's say I have an arbitrary list of minor software versions of an
> imaginary software product:
>
> l = [ "1.1.1.1", "1.2.2.2", "1.2.2.3", "1.3.1.2", "1.3.4.5"]
>
> I'd like to create a dict with major_version : count.
>
> (So, in this case:
>
> dict_of_counts = { "1.1" : "1",
>                    "1.2" : "2",
>                    "1.3" : "2" }
>
> Something like:
>
> dict_of_counts = dict([(v[0:3], "count") for v in l])
>
> I can't seem to figure out how to get "count", as I cannot do x += 1
> or x++ as x may or may not yet exist, and I haven't found a way to
> create default values.
>
> I'm most probably not thinking pythonically enough... (I know I could
> do this pretty easily with a couple more lines, but I'd like to
> understand if there's a way to use a dict generator for this).

Not everything has to be a one-liner; also v[0:3] is wrong if any sub-
version is greater than 9. Here's a standard idiom (in 2.5+ at least):

from collection import defaultdict
versions = [ "1.1.1.1", "1.2.2.2", "1.2.2.3", "1.3.1.2", "1.3.4.5"]

major2count = defaultdict(int)
for v in versions:
    major2count['.'.join(v.split('.',2)[:2])] += 1
print major2count

HTH,
George