Source: http://www.peterbe.com/plog/uniqifiers-benchmark
Suppose you have a list in python that looks like this:
[
'a'
,
'b'
,
'a'
]
# or like this:
[
1
,
2
,
2
,
2
,
3
,
4
,
5
,
6
,
6
,
6
,
6
]
and you want to remove all duplicates so you get this result:
[
'a'
,
'b'
]
# or
[
1
,
2
,
3
,
4
,
5
,
6
]
How do you do that? ...the fastest way? I wrote a couple of
alternative implementations and did a quick benchmark loop on the
various implementations to find out which way was the fastest. (I
haven't looked at memory usage). The slowest function was 78 times
slower than the fastest function.
However, there's one very important difference between the various
functions. Some are order preserving and some are not. For example, in
an order preserving function, apart from the duplicates, the order is
guaranteed to be the same as it was inputted. Eg, uniqify([1,2,2,3])==[1,2,3]
Here are the functions:
def
f1
(
seq
):
# not order preserving
set
=
{}
map
(
set
.
__setitem__
,
seq
,
[])
return
set
.
keys
()
def
f2
(
seq
):
# order preserving
checked
=
[]
for
e
in
seq
:
if
e
not
in
checked
:
checked
.
append
(
e
)
return
checked
def
f3
(
seq
):
# Not order preserving
keys
=
{}
for
e
in
seq
:
keys
[
e
]
=
1
return
keys
.
keys
()
def
f4
(
seq
):
# order preserving
noDupes
=
[]
[
noDupes
.
append
(
i
)
for
i
in
seq
if
not
noDupes
.
count
(
i
)]
return
noDupes
def
f5
(
seq
,
idfun
=
None
):
# order preserving
if
idfun
is
None
:
def
idfun
(
x
):
return
x
seen
=
{}
result
=
[]
for
item
in
seq
:
marker
=
idfun
(
item
)
# in old Python versions:
# if seen.has_key(marker)
# but in new ones:
if
marker
in
seen
:
continue
seen
[
marker
]
=
1
result
.
append
(
item
)
return
result
def
f6
(
seq
):
# Not order preserving
set
=
Set
(
seq
)
return
list
(
set
)
And what you've all been waiting for (if you're still reading). Here are the results:
*
f2
13.24
*
f4
11.73
*
f5
0.37
f1
0.18
f3
0.17
f6
0.19
(
*
order
preserving
)
Clearly f5
is the "best" solution. Not only is it really
really fast; it's also order preserving and supports an optional
transform function which makes it possible to do this:
>>>
a
=
list
(
'ABeeE'
)
>>>
f5
(
a
)
[
'A'
,
'B'
,
'e'
,
'E'
]
>>>
f5
(
a
,
lambda
x
:
x
.
lower
())
[
'A'
,
'B'
,
'e'
]
Download the benchmark script here
UPDATE
From the comments I've now added a couple of more functions to the
benchmark. Some which don't support uniqify a list of objects that can't
be hashed unless passed with a special hashing method. So see all the
functions download the file
Here are the new results:
*
f5
10.1
*
f5b
9.99
*
f8
6.49
*
f10
6.57
*
f11
6.6
f1
4.28
f3
3.55
f6
4.03
f7
2.59
f9
2.58
(f2 and f4) were too slow for this testdata.
Source: http://www.peterbe.com/plog/uniqifiers-benchmark