DaveM wrote:
On Sun, 27 Jul 2008 16:57:14 +0200, "Diez B. Roggisch" <[EMAIL PROTECTED]>
You'll have guessed, I'm sure, that I'm not a professional programmer. This
was the third rewrite of a program to match candidate groups to examiners on
a three day course I run, necessitated on this occasion by a change in the
structure of the course. I originally learnt python as I wrote, and some of
the early code was ugly and verbose, so once the current rewrite was working
I took the opportunity to tidy the code up and document it (yes, I know, but
as I said, I'm an amateur). The list concatenation was an itch I couldn't
scratch:
temp = []
for value in sessexam.values():
temp.extend(value)
c_exam = [name for name in set(temp)] #See what I mean about verbose?
c_exam.sort()
return c_exam
Six lines just didn't feel like it ought to be the best way to do something
so simple. I liked the attempt below better, but was foolish enough to time
it, so that was the end of that.
return sorted(list(set(reduce(lambda x, y: x+y, sessexam.values()))))
The current version (below) is a compromise, but I still feel there _ought_
to be a simple one word command to join multiple lists.
There is, as others have posted, but if you are going to dump the lists
into a set, there is no need to concatenate them together first, and it
is better not to. Just dump them directly into set.
a = list(set(itertools.chain(*sessexam.values())))
This is a pretty good way of skipping the intermediate long list. Another:
a = set()
for l in sessexam.values():
a.update(set(l))
If course, if sessexam.values() does not need to be ordered and returns
sets instead, the set call in not needed.
a.sort() #As I write I'm wondering if I really need it sorted. Hmm...
return a
If you want all in one line...
return sorted(set(itertools.chain(*sessexam.values())))
There is no virtue to calling list then .sort, since that is what sorted
basically does.
def sorted(iterable):
tem = list(iterable)
tem.sort()
return tem
tjr
tjr
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