Converting a set into list
TheSaint
I've stumble to find a solution to get a list from a set
<code>
>>> aa= ['a','b','c','f']
>>> aa
['a', 'b', 'c', 'f']
>>> set(aa)
{'a', 'c', 'b', 'f'}
>>> [k for k in aa]
['a', 'b', 'c', 'f']
</code>
I repute the comprehension list too expensive, is there another method?
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Peter Otten
> I've stumble to find a solution to get a list from a set
>
> <code>
>
>>>> aa= ['a','b','c','f']
>>>> aa
> ['a', 'b', 'c', 'f']
>>>> set(aa)
To clarify: this creates a new object, so aa is still a list.
> {'a', 'c', 'b', 'f'}
>>>> [k for k in aa]
> ['a', 'b', 'c', 'f']
So you are actually converting a list to a (new) list here. Of course it
would have worked with a set or an arbitrary iterable, too.
> </code>
> I repute the comprehension list too expensive, is there another method?
mylist = list(myset)
Do you notice the similarity to converting a list to a set?
Ben Finney
> Hello
>
> 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
TheSaint
> mylist = list(myset)
> Do you notice the similarity to converting a list to a set?
>
There was something confusing me yesterday in doing that, but (for me
strangely) I got cleared out.
The point was that after a result from:
newset= set(myset1) & set(myset2)
list= [newset]
<< [{'bla', 'alb', 'lab'}]
Probably list(set) is not like [set].
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TheSaint
> Another method to do what?
>
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
list(aa)
Where aa is a set.
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Chris Angelico
> 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.
Chris Angelico
Ben Finney
> 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
Chris Torek
Ben Finney <ben+p...@benfinney.id.au> wrote:
>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?
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
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:
> In article <871v00j2bh....@benfinney.id.au>
SigmundV
Sigmund
TheSaint
> 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.
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TheSaint
> >>> x = ['three', 'one', 'four', 'one', 'five']
> >>> x
> ['three', 'one', 'four', 'one', 'five']
> >>> list(set(x))
> ['four', 'five', 'three', 'one']
Why one *"one"* has purged out?
Removing double occurences in a list?
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goto /dev/null
Steven D'Aprano
Break the operation up into two steps instead of one:
>>> x = ['three', 'one', 'four', 'one', 'five']
>>> s = set(x)
>>> print s
set(['four', 'five', 'three', 'one'])
>>> list(s)
['four', 'five', 'three', 'one']
Once an element is already in a set, adding it again is a null-op:
>>> s = set()
>>> s.add(42)
>>> s.add(42)
>>> s.add(42)
>>> print s
set([42])
--
Steven
TheSaint
>>>> s = set()
>>>> s.add(42)
>>>> s.add(42)
>>>> s.add(42)
>>>> print s
> set([42])
Good to know. I'll remember it
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goto /dev/null
Roy Smith
<34fc571c-f382-405d...@t16g2000vbi.googlegroups.com>,
SigmundV <sigm...@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
Thomas Rachel
> SigmundV wrote:
>
>> I think the OP wants to find the intersection of two lists.
>> 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.
Which loops do you mean here?
Thomas
Daniel Kluev
> 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.
--
With best regards,
Daniel Kluev
Peter Otten
>> 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
Duncan Booth
> >>> 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.
A minor change to your example makes it out of order even for integers:
>>> x = [7, 8, 9, 1, 4, 1]
>>> list(set(x))
[8, 9, 1, 4, 7]
or for that mattter:
>>> list(set([3, 32, 4, 32, 5, 9, 2, 6]))
[32, 2, 3, 4, 5, 6, 9]
--
Duncan Booth http://kupuguy.blogspot.com
TheSaint
> 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]
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goto /dev/null
Ben Finney
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
>> This might not be the best example since the result is sorted
>> "by accident", while other list(set(...)) results are not.
In article <Xns9EE772D313...@127.0.0.1>,
Duncan Booth <duncan...@suttoncourtenay.org.uk> wrote:
>A minor change to your example makes it out of order even for integers:
>
>>>> x = [7, 8, 9, 1, 4, 1]
>>>> list(set(x))
>[8, 9, 1, 4, 7]
>
>or for that mattter:
>
>>>> list(set([3, 32, 4, 32, 5, 9, 2, 6]))
>[32, 2, 3, 4, 5, 6, 9]
Yes, but then it is no longer "as easy as pi". :-)