Pythonの中で一番速い辞書の並べ替え方法
12948 ワード
コード
def
sbv0(adict,reverse
=
False):
"""
proposed at Digital Sanitation Engineering
http://blog.modp.com/2007/11/sorting-python-dict-by-value.html
"""
return
sorted(adict.iteritems(), key
=
lambda
(k,v): (v,k), reverse
=
reverse)
def
sbv1(d,reverse
=
False):
"""
explicit list expansion
"""
L
=
[(k,v)
for
(k,v)
in
d.iteritems()]
return
sorted(L, key
=
lambda
x: x[
1
] , reverse
=
reverse)
def
sbv2(d,reverse
=
False):
"""
generator
"""
L
=
((k,v)
for
(k,v)
in
d.iteritems())
return
sorted(L, key
=
lambda
x: x[
1
] , reverse
=
reverse)
def
sbv3(d,reverse
=
False):
"""
using a lambda to get the key, rather than “double-assignment
"""
return
sorted(d.iteritems(), key
=
lambda
x: x[
1
] , reverse
=
reverse)
def
sbv4(d,reverse
=
False):
"""
using a formal function to get the sorting key, rather than a lambda
"""
def
sk(x):
return
x[
1
]
return
sorted(d.iteritems(), key
=
sk , reverse
=
reverse)
def
sk(x):
return
x[
1
]
def
sbv5(d,reverse
=
False):
"""
using a formal function, defined in outer scope
to get the sorting key, rather than a lambda
"""
return
sorted(d.iteritems(), key
=
sk , reverse
=
reverse)
from
operator
import
itemgetter
def
sbv6(d,reverse
=
False):
"""
proposed in PEP 265, using the itemgetter
"""
return
sorted(d.iteritems(), key
=
itemgetter(
1
), reverse
=
True)
D
=
dict(zip(range(
100
),range(
100
)))
from
profile
import
run
run(“
for
ii
in
xrange(
10000
): sbv0(D, reverse
=
True)”)
run(“
for
ii
in
xrange(
10000
): sbv1(D, reverse
=
True)”)
run(“
for
ii
in
xrange(
10000
): sbv2(D, reverse
=
True)”)
run(“
for
ii
in
xrange(
10000
): sbv3(D, reverse
=
True)”)
run(“
for
ii
in
xrange(
10000
): sbv4(D, reverse
=
True)”)
run(“
for
ii
in
xrange(
10000
): sbv5(D, reverse
=
True)”)
run(“
for
ii
in
xrange(
10000
): sbv6(D, reverse
=
True)”)
総じて、PEP 265で紹介された方法は、基本的に別の方法の10倍ぐらいの速度である.
回転:http://www.python8.org/a/fenleiwenzhang/yuyanjichu/2009/0913/318.html