drochom wrote:
> i suppose this one is faster (but in most cases efficiency doesn't
> matter)
>
def stable_unique(s):
>
> e = {}
> ret = []
> for x in s:
> if not e.has_key(x):
> e[x] = 1
> ret.append(x)
> retur
Thanks for all the information.
And now I understand the timeit module ;)
GC-Martijn
--
http://mail.python.org/mailman/listinfo/python-list
thanks, nice job. but this benchmark is pretty deceptive:
try this:
(definition of unique2 and unique3 as above)
>>> import timeit
>>> a = range(1000)
>>> t = timeit.Timer('unique2(a)','from __main__ import unique2,a')
>>> t2 = timeit.Timer('stable_unique(a)','from __main__ import stable_unique,a
Ow thanks , i'm I newbie and I did this test. (don't know if this is
the best way to do a small speed test)
import timeit
def unique2(keys):
unique = []
for i in keys:
if i not in unique:unique.append(i)
return unique
def unique3(s):
e = {}
ret = []
for x in s:
i suppose this one is faster (but in most cases efficiency doesn't
matter)
>>> def stable_unique(s):
e = {}
ret = []
for x in s:
if not e.has_key(x):
e[x] = 1
ret.append(x)
return ret
cheers,
przemek
there wasn't any information about ordering...
maybe i'll find something better which don't destroy original ordering
regards
przemek
--
http://mail.python.org/mailman/listinfo/python-list
Look at the code below
def unique(s):
return list(set(s))
def unique2(keys):
unique = []
for i in keys:
if i not in unique:unique.append(i)
return unique
tmp = [0,1,2,4,2,2,3,4,1,3,2]
print tmp
print unique(tmp)
print unique2(tmp)
--
[0, 1, 2, 4, 2
Rubinho napisal(a):
> I've a list with duplicate members and I need to make each entry
> unique.
>
hi,
other possibility (my newest discovery:) )
>>> a = [1,2,2,4,2,1,3,4]
>>> unique = d.fromkeys(a).keys()
>>> unique
[1, 2, 3, 4]
regards
przemek
--
http://mail.python.org/mailman/listinfo/pyth
przemek drochomirecki wrote:
> def unique(s):
> e = {}
> for x in s:
> if not e.has_key(x):
>e[x] = 1
> return e.keys()
This is basically identical in functionality to the code:
def unique(s):
return list(set(s))
And with the new-and-improved C implementation of sets comin
This works too, if speed isn't your thing..
>> a = [ 1,2,3,2,6,1,3,4,1,7,5,6,7]
>> a = dict( ( (i,None) for i in a)).keys()
a
[1, 2, 3, 4, 5, 6, 7]
--
http://mail.python.org/mailman/listinfo/python-list
> I've a list with duplicate members and I need to make each entry
> unique.
>
> I've come up with two ways of doing it and I'd like some input on what
> would be considered more pythonic (or at least best practice).
>
> Method 1 (the traditional approach)
>
> for x in mylist:
> if mylist.count
Steven D'Aprano wrote:
>
>
> Don't imagine, measure.
>
> Resist the temptation to guess. Write some test functions and time the two
> different methods. But first test that the functions do what you expect:
> there is no point having a blindingly fast bug.
Thats is absolutely correct. Although
On Wed, 14 Sep 2005 13:28:58 +0100, Will McGugan wrote:
> Rubinho wrote:
>> I can't imagine one being much faster than the other except in the case
>> of a huge list and mine's going to typically have less than 1000
>> elements.
>
> I would imagine that 2 would be significantly faster.
Don't
Rubinho wrote:
> I can't imagine one being much faster than the other except in the case
> of a huge list and mine's going to typically have less than 1000
> elements.
To add to what others said, I'd imagine that the technique that's going
to be fastest is going to depend not only on the lengt
I do this:
def unique(keys):
unique = []
for i in keys:
if i not in unique:unique.append(i)
return unique
I don't know what is faster at the moment.
--
http://mail.python.org/mailman/listinfo/python-list
<[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
>I do this:
>
> def unique(keys):
>unique = []
>for i in keys:
>if i not in unique:unique.append(i)
>return unique
>
> I don't know what is faster at the moment.
This is quadratic, O(n^2), in the length n of the list
Peter Otten wrote:
> Rubinho wrote:
>
> > I've a list with duplicate members and I need to make each entry
> > unique.
> >
> > I've come up with two ways of doing it and I'd like some input on what
> > would be considered more pythonic (or at least best practice).
> >
> > Method 1 (the traditional
Rubinho wrote:
> I've a list with duplicate members and I need to make each entry
> unique.
>
> I've come up with two ways of doing it and I'd like some input on what
> would be considered more pythonic (or at least best practice).
>
> Method 1 (the traditional approach)
>
> for x in mylist:
>
Rubinho wrote:
> I've a list with duplicate members and I need to make each entry
> unique.
>
> I've come up with two ways of doing it and I'd like some input on what
> would be considered more pythonic (or at least best practice).
>
> Method 1 (the traditional approach)
>
> for x in mylist:
>
Am Wed, 14 Sep 2005 04:38:35 -0700 schrieb Rubinho:
> I've a list with duplicate members and I need to make each entry
> unique.
>
> I've come up with two ways of doing it and I'd like some input on what
> would be considered more pythonic (or at least best practice).
> mylist = set(mylist)
> my
20 matches
Mail list logo
