13

我有一个对象列表：

[Object_1, Object_2, Object_3]

每个对象都有一个属性：时间：

Object_1.time = 20
Object_2.time = 30
Object_3.time = 40

我想创建一个时间属性列表：

[20, 30, 40]

获得此输出的最有效方法是什么？不可能遍历对象列表，对吗？：

items = []
for item in objects:
    items.append(item.time)
4

4 回答 4

34

列表理解是您所追求的：

list_of_objects = [Object_1, Object_2, Object_3]
[x.time for x in list_of_objects]
于 2012-06-19T01:59:14.637 回答
3 
from operator import attrgetter
items = map(attrgetter('time'), objects)
于 2012-06-19T02:06:59.083 回答
2

最快（也是最容易理解）的是列表理解。

看时间：

import timeit
import random
c=10000

class SomeObj:
    def __init__(self, i):
        self.attr=i

def loopCR():
    l=[]
    for i in range(c):
        l.append(SomeObj(random.random()))

    return l 

def compCR():
    return [SomeObj(random.random()) for i in range(c)]   

def loopAc():
    lAttr=[]
    for e in l:
        lAttr.append(e.attr)

    return lAttr

def compAc():
    return [e.attr for e in l]             

t1=timeit.Timer(loopCR).timeit(10)
t2=timeit.Timer(compCR).timeit(10)
print "loop create:", t1,"secs"   
print "comprehension create:", t2,"secs"   
print 'Faster of those is', 100.0*abs(t1-t2) / max(t1,t2), '% faster'
print 

l=compCR()

t1=timeit.Timer(loopAc).timeit(10)
t2=timeit.Timer(compAc).timeit(10)
print "loop access:", t1,"secs"   
print "comprehension access:", t2,"secs"   
print 'Faster of those is', 100.0*abs(t1-t2) / max(t1,t2), '% faster'

印刷：

loop create: 0.103852987289 secs
comprehension create: 0.0848100185394 secs
Faster of those is 18.3364670069 % faster

loop access: 0.0206878185272 secs
comprehension access: 0.00913000106812 secs
Faster of those is 55.8677438315 % faster

所以列表推导式写起来更快，执行起来也更快。

于 2012-06-19T05:51:05.573 回答
2

怎么样：

items=[item.time for item in objects]
于 2012-06-19T01:59:47.423 回答