[Python] Converting a set into list

TheSaint wrote:

I've stumble to find a solution to get a list from a set

<code>
['a', 'b', 'c', 'f']

TheSaint <nobody at nowhere.net.no> writes:

Hello

I've stumble to find a solution to get a list from a set

<code>

Ben Finney wrote:

Another method to do what?

TheSaint <nobody at 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.

On May 15, 4:11?am, Chris Torek wrote:
In article <871v00j2bh.... at benfinney.id.au>
Ben Finney ?wrote:
I have no idea why someone *else* is doing that, but I have used
this very expression to unique-ize a list:

? ? >>> x = [3, 1, 4, 1, 5, 9, 2, 6]
? ? >>> x
? ? [3, 1, 4, 1, 5, 9, 2, 6]
? ? >>> list(set(x))
? ? [1, 2, 3, 4, 5, 6, 9]
? ? >>>

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

Am 15.05.2011 17:56 schrieb TheSaint:
SigmundV wrote:
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.

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.

Thomas Rachel wrote:

Which loops do you mean here?

TheSaint <nobody at nowhere.net.no> writes:

Thomas Rachel wrote:
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]

['three', 'one', 'four', 'one', 'five']
['four', 'five', 'three', 'one']

On Mon, 16 May 2011 00:05:44 +0800, TheSaint wrote:

Chris Torek wrote:
Why one *"one"* has purged out?
Removing double occurences in a list?

set([42])

[3, 1, 4, 1, 5, 9, 2, 6]
[1, 2, 3, 4, 5, 6, 9]
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.

Peter Otten wrote:

mylist = list(myset)
Do you notice the similarity to converting a list to a set?

On Sun, May 15, 2011 at 12:14 AM, TheSaint wrote:
newset= set(myset1) & set(myset2)
list= [newset]

<< [{'bla', 'alb', 'lab'}]

Probably list(set) is not like [set].