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 setset solution, in contrary, might require one loop
while transforming to a set and another one for the
Chris Torek nos...@torek.net 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
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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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
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 Xns9EE772D313153duncanbooth@127.0.0.1,
Duncan Booth
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:
In article 871v00j2bh@benfinney.id.au
Ben Finney
I'm sorry I top posted. I'll remember not to top post next time.
Sigmund
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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
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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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
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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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
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
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 setset solution, in contrary, might require one loop while transforming
to a set and another one for the operation.
python -m timeit
Hello
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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TheSaint wrote:
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
TheSaint nob...@nowhere.net.no writes:
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',
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]
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
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
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
In article 871v00j2bh@benfinney.id.au
Ben Finney ben+pyt...@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
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