Re: dict generator question

[Gerard flanagan]

Simon Mullis 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" }

[...]
data = [ "1.1.1.1", "1.2.2.2", "1.2.2.3", "1.3.1.2", "1.3.4.5"]

from itertools import groupby

datadict = \
dict((k, len(list(g))) for k,g in groupby(data, lambda s: s[:3]))
print datadict

[bearophileHUGS]

Steven D'Aprano:

>I'm sorry, I don't recognise leniter(). Did I miss something?<

I have removed the docstring/doctests:

def leniter(iterato r):
if hasattr(iterato r, "__len__"):
return len(iterator)
nelements = 0
for _ in iterator:
nelements += 1
return nelements

>it doesn't work for arbitrary iterables, only sequences (lazy or otherwise)<

I don't understand well.

>Since you're generating the entire length anyway, len(list(iterab le)) is more readable and almost as efficient for most practical cases.<

I don't agree, len(list()) creates an actual list, with lot of GC
activity.

>But the expected semantics of __len__ is that it is expected to return an int, and do it quickly with minimal effort. Methods that do something else are an abuse of __len__ and should be treated as a bug.<

I see. In the past I have read similar positions in discussions
regarding API of data structures in D, so this may be right, and this
fault may be enough to kill my proposal. But I'll keep using
leniter().

Bye,
bearophile