Re: Converting a set into list

2011-05-16 Thread Chris Torek
>Chris Torek  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 ,
Duncan Booth   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". :-)
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Re: Converting a set into list

2011-05-16 Thread Ben Finney
TheSaint  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.

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Re: Converting a set into list

2011-05-16 Thread TheSaint
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]

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Re: Converting a set into list

2011-05-16 Thread Duncan Booth
Chris Torek  wrote:

> >>> 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]


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Re: Converting a set into list

2011-05-15 Thread Peter Otten
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, 1); 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, 1); 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, 10); l2 = range(5, 15)"
> "l3 = list(set(l1) & set(l2))"
> 10 loops, best of 3: 28 msec per loop
> 
> python -m timeit -s "l1 = range(1, 10); l2 = range(5, 15)"
> "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, 1); 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, 1); l2 = range(5000, 15000)" "s = 
set(l1); s.intersection_update(l2); list(s)"
100 loops, best of 3: 1.99 msec per loop


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Re: Converting a set into list

2011-05-15 Thread Daniel Kluev
> 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, 1); 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, 1); 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, 10); l2 = range(5, 15)"
"l3 = list(set(l1) & set(l2))"
10 loops, best of 3: 28 msec per loop

python -m timeit -s "l1 = range(1, 10); l2 = range(5, 15)"
"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.

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Re: Converting a set into list

2011-05-15 Thread Thomas Rachel

Am 15.05.2011 17:56 schrieb TheSaint:

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.


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
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Re: Converting a set into list

2011-05-15 Thread Roy Smith
In article 
<34fc571c-f382-405d-94b1-0a673da5f...@t16g2000vbi.googlegroups.com>,
 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.

Both ways work, but the first is O(n) and the second is O(n^2).

import time
n = 1
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 10
list(set) method took 0.004791 seconds
loop method took 1.437322 seconds
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Re: Converting a set into list

2011-05-15 Thread TheSaint
Steven D'Aprano wrote:

 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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Re: Converting a set into list

2011-05-15 Thread Steven D'Aprano
On Mon, 16 May 2011 00:05:44 +0800, TheSaint wrote:

> Chris Torek wrote:
> 
>> >>> 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?

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])




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Re: Converting a set into list

2011-05-15 Thread TheSaint
Chris Torek wrote:

> >>> 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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Re: Converting a set into list

2011-05-15 Thread TheSaint
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.
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Re: Converting a set into list

2011-05-15 Thread SigmundV
I'm sorry I top posted. I'll remember not to top post next time.

Sigmund
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Re: Converting a set into list

2011-05-15 Thread SigmundV
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  wrote:
> In article <871v00j2bh@benfinney.id.au>
> Ben Finney   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']
>     >>>
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> 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

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Re: Converting a set into list

2011-05-14 Thread Chris Torek
In article <871v00j2bh@benfinney.id.au>
Ben Finney   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']
>>> 
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Re: Converting a set into list

2011-05-14 Thread Ben Finney
TheSaint  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?

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Re: Converting a set into list

2011-05-14 Thread Chris Angelico
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].

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.

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Re: Converting a set into list

2011-05-14 Thread TheSaint
Ben Finney wrote:

> 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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Re: Converting a set into list

2011-05-14 Thread TheSaint
Peter Otten wrote:

> 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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Re: Converting a set into list

2011-05-14 Thread Ben Finney
TheSaint  writes:

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

> 
> 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?

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Re: Converting a set into list

2011-05-14 Thread Peter Otten
TheSaint wrote:

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

> 
> 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?

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