delete duplicates in list

Hi,

this must have come up before, so i am already sorry for asking but a
quick googling did not give me any answer.

i have a list from which i want a simpler list without the duplicates
an easy but somehow contrived solution would be

In python 2.3, this should work:

import sets
l = [1,2,2,3]
s = sets.Set(l)

A set is a container for which the invariant that no object is in it twice
holds. So it suits your needs.

Regards,

Diez

> >>> a = [1, 2, 2, 3]

>>> d = {}.fromkeys(a)
>>> b = d.keys()
>>> print b

[1, 2, 3]

That, or using a Set (python 2.3+). Actually I seem to recall that
"The Python CookBook" still advises building a dict as the fastest
solution - if your elements can be hashed. See the details at

http://aspn.activestate.com/ASPN/Coo...n/Recipe/52560

Cheers,

Bernard.

christof hoeke wrote:
...

i have a list from which i want a simpler list without the duplicates
Canonical is:

import sets
simplerlist = list(sets.Set(t helist))

if you're allright with destroying order, as your example solution suggests.
But dict.fromkeys(a ).keys() is probably faster. Your assertion:
there should be an easier or more intuitive solution, maybe with a list
comprehension=

doesn't seem self-evident to me. A list-comprehension might be, e.g:

[ x for i, x in enumerate(a) if i==a.index(x) ]

and it does have the advantages of (a) keeping order AND (b) not
requiring hashable (nor even inequality-comparable!) elements -- BUT
it has the non-indifferent cost of being O(N*N) while the others
are about O(N). If you really want something similar to your approach:

>>> b = [x for x in a if x not in b]

you'll have, o horrors!-), to do a loop, so name b is always bound to
"the result list so far" (in the LC, name b is only bound at the end):

b = []
for x in a:
if x not in b:
b.append(x)

However, this is O(N*N) too. In terms of "easier or more intuitive",
I suspect only this latter solution might qualify.
Alex

Bernard Delmée wrote:

>>> a = [1, 2, 2, 3]
>>> d = {}.fromkeys(a)
>>> b = d.keys()
>>> print b

[1, 2, 3]

That, or using a Set (python 2.3+). Actually I seem to recall that
"The Python CookBook" still advises building a dict as the fastest
solution - if your elements can be hashed. See the details at

http://aspn.activestate.com/ASPN/Coo...n/Recipe/52560

Yep, but a Set is roughly as fast as a dict -- it has a small
penalty wrt a dict, but, emphasis on small. Still, if you're
trying to squeeze every last cycle out of a bottleneck, it's
worth measuring, and perhaps moving from Set to dict.
Alex

Diez B. Roggisch wrote:

this must have come up before, so i am already sorry for asking but a
quick googling did not give me any answer.

i have a list from which i want a simpler list without the duplicates
an easy but somehow contrived solution would be

In python 2.3, this should work:

import sets
l = [1,2,2,3]
s = sets.Set(l)

A set is a container for which the invariant that no object is in it twice
holds. So it suits your needs.

....except it's not a LIST, which was part of the specifications given
by the original poster. It may, of course, be that you've read his
mind correctly and that, despite his words, he doesn't really care
whether he gets a list or a very different container:-).
Alex

Hi,

...except it's not a LIST, which was part of the specifications given
by the original poster. It may, of course, be that you've read his
mind correctly and that, despite his words, he doesn't really care
whether he gets a list or a very different container:-).

You're right - mathematically. However, there is no such thing like a set in
computers - you always end up with some sort of list :)

So - he'll have a list anyway. But if it respects the order the list
parameter... <just checking, standby>

.... nope:

l = [1,2,3,2]
import sets
sets.Set(l) Set([1, 2, 3]) l = [2,1,2,3,2]
sets.Set(l)

Set([1, 2, 3])

Which is of course reasonable, as the check for existence in the set might
be performed in O(ln n) instead of O(n).

Regards,

Diez

"Diez B. Roggisch" <de************ @web.de> wrote in message news:<bn******* ******@news.t-online.com>...

Hi,

...except it's not a LIST, which was part of the specifications given
by the original poster. It may, of course, be that you've read his
mind correctly and that, despite his words, he doesn't really care
whether he gets a list or a very different container:-).
You're right - mathematically. However, there is no such thing like a set in
computers - you always end up with some sort of list :)

Those are just implementation details. There could be a group of monkeys
emulating Python under the hood and their implementation of a set
would be a neural network instead of any kind of sequence, but you
still wouldn't care as a programmer. The only thing that matters is,
if the interface stays same. Nope, the items in a set are no longer
accessible by their index (among other differences).
So - he'll have a list anyway.
So I can't agree with this. You don't know if his Python virtual machine
is a group of monkeys. Python is supposed to be a high level language.
But if it respects the order the list
parameter... <just checking, standby>

... nope:

l = [1,2,3,2]
import sets
sets.Set(l) Set([1, 2, 3]) l = [2,1,2,3,2]
sets.Set(l)

Set([1, 2, 3])

Which is of course reasonable, as the check for existence in the set might
be performed in O(ln n) instead of O(n).

Actually the Set in sets module has an average lookup of O(1), worst
case O(n) (not 100% sure of worst case, but 99% sure). It's been
implemented with dictionaries, which in turn are hash tables.

Alex Martelli <al***@aleax.it > wrote in
news:L7******** ************@ne ws1.tin.it:

Your assertion:

there should be an easier or more intuitive solution, maybe with a list
comprehension=

doesn't seem self-evident to me. A list-comprehension might be, e.g:

[ x for i, x in enumerate(a) if i==a.index(x) ]

and it does have the advantages of (a) keeping order AND (b) not
requiring hashable (nor even inequality-comparable!) elements -- BUT
it has the non-indifferent cost of being O(N*N) while the others
are about O(N). If you really want something similar to your approach:

>>> b = [x for x in a if x not in b]
you'll have, o horrors!-), to do a loop, so name b is always bound to
"the result list so far" (in the LC, name b is only bound at the end):

b = []
for x in a:
if x not in b:
b.append(x)

However, this is O(N*N) too. In terms of "easier or more intuitive",
I suspect only this latter solution might qualify.

I dunno about more intuitive, but here's a fairly simple list comprehension
solution which is O(N) and preserves the order. Of course its back to
requiring hashable elements again:

startList = [5,1,2,1,3,4,2,5 ,3,4]
d = {}
[ d.setdefault(x, x) for x in startList if x not in d ] [5, 1, 2, 3, 4]

And for the 'I must do it on one line' freaks, here's the single expression
variant of the above: :^)
[ d.setdefault(x, x) for d in [{}] for x in startList if x not in d ]

[5, 1, 2, 3, 4]

--
Duncan Booth du****@rcp.co.u k
int month(char *p){return(1248 64/((p[0]+p[1]-p[2]&0x1f)+1)%12 )["\5\x8\3"
"\6\7\xb\1\x9\x a\2\0\4"];} // Who said my code was obscure?

Hannu Kankaanp?? wrote:
...

So - he'll have a list anyway.
So I can't agree with this. You don't know if his Python virtual machine
is a group of monkeys. Python is supposed to be a high level language.

So, in that case those monkeys would surely be perched up on tall trees.

Which is of course reasonable, as the check for existence in the set
might be performed in O(ln n) instead of O(n).

Actually the Set in sets module has an average lookup of O(1), worst
case O(n) (not 100% sure of worst case, but 99% sure). It's been

Hmmm -- could you give an example of that worstcase...? a _full_
hashtable would give such behavior, but Python's dicts always ensure
the underlying hashtables aren't too full...
implemented with dictionaries, which in turn are hash tables.

Yep - check it out...:

[alex@lancelot bo]$ timeit.py -c -s'import sets' -s'x=sets.Set(xr ange(100))'
'7 in x'
100000 loops, best of 3: 2.3 usec per loop
[alex@lancelot bo]$ timeit.py -c -s'import sets'
-s'x=sets.Set(xr ange(1000))' '7 in x'
100000 loops, best of 3: 2.3 usec per loop
[alex@lancelot bo]$ timeit.py -c -s'import sets'
-s'x=sets.Set(xr ange(10000))' '7 in x'
100000 loops, best of 3: 2.2 usec per loop
[alex@lancelot bo]$ timeit.py -c -s'import sets'
-s'x=sets.Set(xr ange(100000))' '7 in x'
100000 loops, best of 3: 2.2 usec per loop
[alex@lancelot bo]$ timeit.py -c -s'import sets'
-s'x=sets.Set(xr ange(1000000))' '7 in x'
100000 loops, best of 3: 2.3 usec per loop

see the pattern...?-)
Same when something is NOT there, BTW:

[alex@lancelot bo]$ timeit.py -c -s'import sets'
-s'x=sets.Set(xr ange(1000000))' 'None in x'
100000 loops, best of 3: 2.3 usec per loop

etc, etc.