dict generator question

[Simon Mullis]

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).

Thanks in advance

SM


--
Simon Mullis

[marek.rocki@wp.pl]

Re: dict generator question

Simon Mullis napisa³(a):

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.

It seems to me that the "count" you're looking for is the number of
elements from l whose first 3 characters are the same as the v[0:3]
thing. So you may try:

>>dict_of_count s = dict((v[0:3], sum(1 for x in l if x[:3] == v[:3])) for v in l)

But this isn't particularly efficient. The 'canonical way' to
construct such histograms/frequency counts in python is probably by
using defaultdict:

>>dict_of_count s = collections.def aultdict(int)
>>for x in l:
>> dict_of_counts[x[:3]] += 1

Regards,
Marek

[pruebauno@latinmail.com]

Re: dict generator question

On Sep 18, 10:54 am, "Simon Mullis" <si...@mullis.c o.ukwrote:

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).
>
Thanks in advance
>
SM
>
--
Simon Mullis

3 lines:

from collections import defaultdict
dd=defaultdict( int)
for x in l: dd[x[0:3]]+=1

[George Sakkis]

Re: dict generator question

On Sep 18, 10:54 am, "Simon Mullis" <si...@mullis.c o.ukwrote:

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.spli t('.',2)[:2])] += 1
print major2count

HTH,
George

[pruebauno@latinmail.com]

Re: dict generator question

On Sep 18, 10:54 am, "Simon Mullis" <si...@mullis.c o.ukwrote:

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).
>
Thanks in advance
>
SM
>
--
Simon Mullis

Considering 3 identical "simultpost " solutions I'd say:
"one obvious way to do it" FTW :-)

[Simon Mullis]

Re: dict generator question

Haha!

Thanks for all of the suggestions... (I love this list!)

SM

2008/9/18 <pruebauno@lati nmail.com>:

On Sep 18, 10:54 am, "Simon Mullis" <si...@mullis.c o.ukwrote:

>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_coun ts = { "1.1" : "1",
> "1.2" : "2",
> "1.3" : "2" }
>>
>Something like:
>>
>dict_of_coun ts = 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).
>>
>Thanks in advance
>>
>SM
>>
>--
>Simon Mullis

>
Considering 3 identical "simultpost " solutions I'd say:
"one obvious way to do it" FTW :-)
--

http://mail.python.org/mailman/listinfo/python-list

>

--
Simon Mullis
_______________ __
simon@mullis.co .uk

[Paul Rubin]

Re: dict generator question

"Simon Mullis" <simon@mullis.c o.ukwrites:

l = [ "1.1.1.1", "1.2.2.2", "1.2.2.3", "1.3.1.2", "1.3.4.5"]
...
dict_of_counts = dict([(v[0:3], "count") for v in l])

Untested:

def major_version(v ersion_string):
"convert '1.2.3.2' to '1.2'"
return '.'.join(versio n_string.split( '.')[:2])

dict_of_counts = defaultdict(int )

for x in l:
dict_of_counts[major_version(l )] += 1