> I've stumble to find a solution to get a list from a set
> <code>
> >>> aa= ['a','b','c','f']
Creates a new list object. Binds the name ‘aa’ to that object.
> >>> aa > ['a', 'b', 'c', 'f']
Evaluates the object referenced by the name ‘aa’.
> >>> set(aa) > {'a', 'c', 'b', 'f'}
Creates a new set object, populating it with the contents from the list object referenced by ‘aa’. Doesn't do anything with the new set object, which will soon be garbage-collected.
> >>> [k for k in aa] > ['a', 'b', 'c', 'f']
Creates a new list object by iterating each of the items from the list referenced by ‘aa’. Does nothing with the new list object, which will soon be garbage-collected.
> </code> > I repute the comprehension list too expensive, is there another method?
Another method to do what?
If you want to bind ‘aa’ to a new object, do so with an assignment statement. (Your example has exactly one assignment statement; all the other statements create objects which are never bound to anything.)
But what is it you actually want to do?
-- \ “The fact that I have no remedy for all the sorrows of the | `\ world is no reason for my accepting yours. It simply supports | _o__) the strong probability that yours is a fake.” —Henry L. Mencken | Ben Finney
Sorry, some time we expect to have said it as we thought it.
The example was to show that after having made a set
set(aa)
the need to get that set converted into a list. My knowledge drove me to use a comprehension list as a converter. In another post I got to know the simplest way to state
On Sun, May 15, 2011 at 12:14 AM, TheSaint <nob...@nowhere.net.no> wrote: > newset= set(myset1) & set(myset2) > list= [newset]
> << [{'bla', 'alb', 'lab'}]
> Probably list(set) is not like [set].
list(set) creates a list out of the set. [set] creates a list with one element, the set itself. It's not a copy of the set, it's another reference to the same set; change one and you'll see the change in the other.
TheSaint <nob...@nowhere.net.no> writes: > The example was to show that after having made a set
> set(aa)
> the need to get that set converted into a list.
As pointed out: you already know how to create a set from an object; creating a list from an object is very similar:
list(set(aa))
But why are you doing that? What are you trying to achieve?
-- \ “We are all agreed that your theory is crazy. The question that | `\ divides us is whether it is crazy enough to have a chance of | _o__) being correct.” —Niels Bohr (to Wolfgang Pauli), 1958 | Ben Finney
Of course, this trick only works if all the list elements are hashable.
This might not be the best example since the result is sorted "by accident", while other list(set(...)) results are not. Add sorted() or .sort() if needed:
>>> x = ['three', 'one', 'four', 'one', 'five'] >>> x ['three', 'one', 'four', 'one', 'five'] >>> list(set(x)) ['four', 'five', 'three', 'one'] >>> sorted(list(set(x))) ['five', 'four', 'one', 'three'] >>> -- In-Real-Life: Chris Torek, Wind River Systems Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603 email: gmail (figure it out) http://web.torek.net/torek/index.html
I think the OP wants to find the intersection of two lists. list(set(list1) & set(list2)) is indeed one way to achieve this. [i for i in list1 if i in list2] is another one.
Sigmund
On May 15, 4:11 am, Chris Torek <nos...@torek.net> wrote:
> Of course, this trick only works if all the list elements are > hashable.
> This might not be the best example since the result is sorted > "by accident", while other list(set(...)) results are not. Add > sorted() or .sort() if needed:
> >>> x = ['three', 'one', 'four', 'one', 'five'] > >>> x > ['three', 'one', 'four', 'one', 'five'] > >>> list(set(x)) > ['four', 'five', 'three', 'one'] > >>> sorted(list(set(x))) > ['five', 'four', 'one', 'three'] > >>> > -- > In-Real-Life: Chris Torek, Wind River Systems > Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603 > email: gmail (figure it out) http://web.torek.net/torek/index.html
SigmundV wrote: > I think the OP wants to find the intersection of two lists. > list(set(list1) & set(list2)) is indeed one way to achieve this. [i > for i in list1 if i in list2] is another one
Exactly. I was confused on that I wasn't able to have a list in return. The set intersection is the smartest result better than a "for" loop or a comprehension list. Infact the operatin loops are compiled into python, therfore they are the fastest. -- goto /dev/null
In article <34fc571c-f382-405d-94b1-0a673da5f...@t16g2000vbi.googlegroups.com>,
SigmundV <sigmu...@gmail.com> wrote: > I think the OP wants to find the intersection of two lists. > list(set(list1) & set(list2)) is indeed one way to achieve this. [i > for i in list1 if i in list2] is another one.
Both ways work, but the first is O(n) and the second is O(n^2).
import time n = 10000 list1 = range(n) list2 = range(n) t0 = time.time() list(set(list1) & set(list2)) t1 = time.time() print "list(set) method took %f seconds" % (t1 - t0) t0 = time.time() [i for i in list1 if i in list2] t1 = time.time() print "loop method took %f seconds" % (t1 - t0)
./intersect.py 100000 list(set) method took 0.004791 seconds loop method took 1.437322 seconds
>> I think the OP wants to find the intersection of two lists. >> list(set(list1)& set(list2)) is indeed one way to achieve this. [i >> for i in list1 if i in list2] is another one
> Exactly. I was confused on that I wasn't able to have a list in return. > The set intersection is the smartest result better than a "for" loop or a > comprehension list.
I'm not sure about if it is really the smartest way.
s=set(list2); [i for i in list1 if i in s] is in the same order of magnitude as the set operation. Both solutions seem to be equivalent in that concerns the number of needed loop runs, but this two-step operation might require one less loop over list1.
The set&set solution, in contrary, might require one loop while transforming to a set and another one for the & operation.
> Infact the operatin loops are compiled into python, therfore they are the > fastest.
> Both solutions seem to be equivalent in that concerns the number of needed loop runs, but this two-step operation might require one less loop over list1. > The set&set solution, in contrary, might require one loop while transforming to a set and another one for the & operation.
python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)" "l3 = list(set(l1) & set(l2))" 100 loops, best of 3: 2.19 msec per loop
python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)" "s=set(l2); l3 = [i for i in l1 if i in s]" 100 loops, best of 3: 2.45 msec per loop
python -m timeit -s "l1 = range(1, 100000); l2 = range(50000, 150000)" "l3 = list(set(l1) & set(l2))" 10 loops, best of 3: 28 msec per loop
python -m timeit -s "l1 = range(1, 100000); l2 = range(50000, 150000)" "s=set(l2); l3 = [i for i in l1 if i in s]" 10 loops, best of 3: 28.1 msec per loop
So even with conversion back into list set&set is still marginally faster.
Daniel Kluev wrote: >> Both solutions seem to be equivalent in that concerns the number of >> needed loop runs, but this two-step operation might require one less loop >> over list1. The set&set solution, in contrary, might require one loop >> while transforming to a set and another one for the & operation.
> python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)" > "l3 = list(set(l1) & set(l2))" > 100 loops, best of 3: 2.19 msec per loop
> python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)" > "s=set(l2); l3 = [i for i in l1 if i in s]" > 100 loops, best of 3: 2.45 msec per loop
> python -m timeit -s "l1 = range(1, 100000); l2 = range(50000, 150000)" > "l3 = list(set(l1) & set(l2))" > 10 loops, best of 3: 28 msec per loop
> python -m timeit -s "l1 = range(1, 100000); l2 = range(50000, 150000)" > "s=set(l2); l3 = [i for i in l1 if i in s]" > 10 loops, best of 3: 28.1 msec per loop
> So even with conversion back into list set&set is still marginally faster.
If you are looking for speed, consider
s = set(l1) s.intersection_update(l2) l3 = list(s)
$ python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)" "list(set(l1) & set(l2))" 100 loops, best of 3: 4 msec per loop
$ python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)" "s = set(l1); s.intersection_update(l2); list(s)" 100 loops, best of 3: 1.99 msec per loop
TheSaint <nob...@nowhere.net.no> writes: > Thomas Rachel wrote:
> > Which loops do you mean here?
> list(set) has been proved to largely win against > list = [] > for item in set: > list.append(item) > or [list.append(item) for item in set]
Remember that the criterion of speed is a matter of the implementation, and what's fast on one won't necessarily be fast on others. Which implementations did you try?
Where I do agree is that ‘list(foo)’ wins over the other examples you show on the important criteria of concision and readability.
-- \ “A thing moderately good is not so good as it ought to be. | `\ Moderation in temper is always a virtue; but moderation in | _o__) principle is always a vice.” —Thomas Paine | Ben Finney
>Chris Torek <nos...@torek.net> wrote: >> >>> x = [3, 1, 4, 1, 5, 9, 2, 6] >> >>> list(set(x)) >> This might not be the best example since the result is sorted >> "by accident", while other list(set(...)) results are not.
In article <Xns9EE772D313153duncanbo...@127.0.0.1>, Duncan Booth <duncan.bo...@suttoncourtenay.org.uk> wrote:
>A minor change to your example makes it out of order even for integers:
Yes, but then it is no longer "as easy as pi". :-) -- In-Real-Life: Chris Torek, Wind River Systems Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603 email: gmail (figure it out) http://web.torek.net/torek/index.html
