|
P: n/a
|
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(x) > 1:
mylist.remove(x)
Method 2 (not so traditional)
mylist = set(mylist)
mylist = list(mylist)
Converting to a set drops all the duplicates and converting back to a
list, well, gets it back to a list which is what I want.
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.
What do you think?
Cheers,
Robin
| |
Share this Question
|
P: n/a
|
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)
mylist = list(mylist)
Converting to a set drops all the duplicates and converting back to a
list, well, gets it back to a list which is what I want.
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.
What do you think?
Hi,
I would use "set":
mylist=list(set(mylist))
Thomas
--
Thomas Güttler, http://www.thomas-guettler.de/
E-Mail: guettli (*) thomas-guettler + de
Spam Catcher: ni**************@thomas-guettler.de | |
|
P: n/a
|
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:
if mylist.count(x) > 1:
mylist.remove(x)
Method 2 (not so traditional)
mylist = set(mylist)
mylist = list(mylist)
Converting to a set drops all the duplicates and converting back to a
list, well, gets it back to a list which is what I want.
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. Method 1 uses
'count' which must make a pass through every element of the list, which
would be slower than the efficient hashing that set does. I'm also not
sure about removing an element whilst iterating, I think thats a no-no.
Will McGugan
-- http://www.willmcgugan.com
"".join({'*':'@','^':'.'}.get(c,0) or chr(97+(ord(c)-84)%26) for c in
"jvyy*jvyyzpthtna^pbz") | |
|
P: n/a
|
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:
if*mylist.count(x)*>*1:
mylist.remove(x)
That would be an odd tradition: mylist = [1, 2, 1, 3, 2, 3]
for x in mylist:
.... if mylist.count(x) > 1:
.... mylist.remove(x)
.... mylist
[2, 1, 2, 3] # oops!
See "Unexpected Behavior Iterating over a Mutating Object" http://mail.python.org/pipermail/pyt...er/298993.html
thread for the most recent explanation.
Rather, the traditional approach for an algorithmic problem in Python is to
ask Tim Peters, see his recipe at http://aspn.activestate.com/ASPN/Coo.../Recipe/52560/
(which predates Python's set class).
Peter | |
|
P: n/a
|
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 approach)
for x in mylist:
if mylist.count(x) > 1:
mylist.remove(x)
That would be an odd tradition:
By tradition I wasn't really talking Python tradition; what I meant was
that the above pattern is similar to what would be generated by people
used to traditional programming languages.
mylist = [1, 2, 1, 3, 2, 3]
for x in mylist: ... if mylist.count(x) > 1:
... mylist.remove(x)
... mylist
[2, 1, 2, 3] # oops!
But you're absolutely right, it doesn't work! Oops indeed :)
I've gone with Thomas's suggestion above of: mylist=list(set(mylist))
Thanks,
Robin | |
|
P: n/a
|
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.
| |
|
P: n/a
|
<ma*****@gamecreators.nl> wrote in message
news:11*********************@o13g2000cwo.googlegro ups.com... 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
if all keys are unique.
Conversion to a set just might use a better sorting
algorithm than this (i.e. n*log(n)) and throwing out
duplicates (which, after sorting, are positioned
next to each other) is O(n). If conversion
to a set should turn out to be slower than O(n*log(n))
[depending on the implementation], then you are well
advised to sort the list first (n*log(n)) and then
throw out the duplicate keys with a single walk over
the list. In this case you know at least what to
expect for large n...
Regards,
Christian | |
|
P: n/a
|
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 length of the list, but
also the estimated redundancy. (i.e. a technique that gives good
performance with a list that has only one or two elements duplicated
might be painfully slow when there is 10-100 copies of each element.)
There really is no substitute for profiling with representitive data sets. | |
|
P: n/a
|
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 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.
Method 1 uses
'count' which must make a pass through every element of the list, which
would be slower than the efficient hashing that set does.
But count passes through the list in C and is also very fast. Is that
faster or slower than the hashing code used by sets? I don't know, and
I'll bet you don't either.
--
Steven. | |
|
P: n/a
|
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 I think you do sometimes have to
guess. Otherwise you would write multiple versions of every line of code.
But count passes through the list in C and is also very fast. Is that
faster or slower than the hashing code used by sets? I don't know, and
I'll bet you don't either.
Sure. But if I'm not currently optimizing I would go for the method with
the best behaviour, which usualy means hashing rather than searching.
Since even if it is actualy slower - its not likely to be _very_ slow.
Will McGugan
-- http://www.willmcgugan.com
"".join({'*':'@','^':'.'}.get(c,0) or chr(97+(ord(c)-84)%26) for c in
"jvyy*jvyyzpthtna^pbz") | |
|
P: n/a
|
> 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(x) > 1:
mylist.remove(x)
Method 2 (not so traditional)
mylist = set(mylist)
mylist = list(mylist)
Converting to a set drops all the duplicates and converting back to a
list, well, gets it back to a list which is what I want.
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.
What do you think?
Cheers,
Robin
Hi,
Try this:
def unique(s):
e = {}
for x in s:
if not e.has_key(x):
e[x] = 1
return e.keys()
Regards
Przemek | |
|
P: n/a
|
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] | |
|
P: n/a
|
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 coming in Python
2.5, there's even more of a reason to use them when you can.
STeVe | |
|
P: n/a
|
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 | |
|
P: n/a
|
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, 2, 3, 4, 1, 3, 2]
[0, 1, 2, 3, 4]
[0, 1, 2, 4, 3]
As you can see the end result is not the same.
I must get the end result [0, 1, 2, 4, 3] and not [0, 1, 2, 3, 4].
Thats why I use unique2()
| |
|
P: n/a
|
there wasn't any information about ordering...
maybe i'll find something better which don't destroy original ordering
regards
przemek
| |
|
P: n/a
|
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 | |
|
P: n/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:
if not e.has_key(x):
e[x] = 1
ret.append(x)
return ret
tmp = [0,1,2,4,2,2,3,4,1,3,2]
s = """\
try:
str.__nonzero__
except AttributeError:
pass
"""
t = timeit.Timer(stmt=s)
print "%.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
print tmp
print "%.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
print unique2(tmp)
print "%.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
print unique3(tmp)
print "%.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
---------------------
5.80 usec/pass
[0, 1, 2, 4, 2, 2, 3, 4, 1, 3, 2]
7.51 usec/pass
[0, 1, 2, 4, 3]
6.93 usec/pass
[0, 1, 2, 4, 3]
6.45 usec/pass <--- your code unique2(s):
| |
|
P: n/a
|
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')
t2.timeit(2000)
1.8392596235778456 t.timeit(2000)
51.52945844819817
unique2 has quadratic complexity
unique3 has amortized linear complexity
what it means?
it means that speed of your algorithm strongly depends on
len(unique2(a)). the greater distinct elements in a the greater
difference in execution time of both implementations
regards
przemek | |
|
P: n/a
|
Thanks for all the information.
And now I understand the timeit module ;)
GC-Martijn
| |
|
P: n/a
|
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)
return ret
I'll repeat Peter Otten's link to Tim Peters's recipe here: http://aspn.activestate.com/ASPN/Coo.../Recipe/52560/
Read the comments at the end, they talk about order-preserving lists.
See Raymond Hettinger's response:
def uniq(alist) # Fastest order preserving
set = {}
return [set.setdefault(e,e) for e in alist if e not in set]
STeVe | | This discussion thread is closed Replies have been disabled for this discussion. | |  Question stats - viewed: 2921
- replies: 20
- date asked: Sep 14 '05
|
