将列表转为字典的最佳方法，键为每个对象的值？

Question

我正在尝试把一个对象列表转换成一个字典。字典的值就是列表中的每个对象，而字典的键则是每个对象中找到的一个值。

下面是我正在使用的代码：

class SomeClass(object):

    def __init__(self, name):
        self.name = name

object_list = [
    SomeClass(name='a'),
    SomeClass(name='b'),
    SomeClass(name='c'),
    SomeClass(name='d'),
    SomeClass(name='e'),
]

object_dict = {}
for an_object in object_list:
    object_dict[an_object.name] = an_object

这段代码可以运行，但看起来有点丑，而且速度也有点慢。有没有人能给我一个更快或者更好的例子呢？

编辑：好的，感谢大家的回复。我必须说，看到一些更符合Python风格的方法看起来比我手动写的方式还要慢，我感到很惊讶。

编辑2：好的，我更新了测试代码，让它看起来更易读，因为测试的数量太多了，哈哈。

这是我们目前的代码，我在代码中加了作者的名字，如果我搞错了请告诉我。

from itertools import izip
import timeit

class SomeClass(object):

    def __init__(self, name):
        self.name = name

object_list = []

for i in range(5):
    object_list.append(SomeClass(name=i))

def example_1():
    'Original Code'
    object_dict = {}
    for an_object in object_list:
        object_dict[an_object.name] = an_object

def example_2():
    'Provided by hyperboreean'
    d = dict(zip([o.name for o in object_list], object_list))

def example_3():
    'Provided by Jason Baker'
    d = dict([(an_object.name, an_object) for an_object in object_list])

def example_4():
    "Added izip to hyperboreean's code, suggested by Chris Cameron"
    d = dict(izip([o.name for o in object_list], object_list))

def example_5():
    'zip, improved by John Fouhy'
    d = dict(zip((o.name for o in object_list), object_list))

def example_6():
    'izip, improved by John Fouhy'
    d = dict(izip((o.name for o in object_list), object_list))

def example_7():
    'Provided by Jason Baker, removed brackets by John Fouhy'
    d = dict((an_object.name, an_object) for an_object in object_list)

timeits = []
for example_index in range(1, 8):
    timeits.append(
        timeit.Timer(
            'example_%s()' % example_index,
            'from __main__ import example_%s' % example_index)
    )

for i in range(7):
    timeit_object = timeits[i]
    print 'Example #%s Result: "%s"' % (i+1, timeit_object.repeat(2))

当列表中有5个对象时，我得到的结果是：

    Example #1 Result: "[1.2428441047668457, 1.2431108951568604]"
    Example #2 Result: "[3.3567759990692139, 3.3188660144805908]"
    Example #3 Result: "[2.8346641063690186, 2.8344728946685791]"
    Example #4 Result: "[3.0710639953613281, 3.0573830604553223]"
    Example #5 Result: "[5.2079918384552002, 5.2170760631561279]"
    Example #6 Result: "[3.240635871887207, 3.2402129173278809]"
    Example #7 Result: "[3.0856869220733643, 3.0688989162445068]"

而当有50个对象时：

    Example #1 Result: "[9.8108220100402832, 9.9066231250762939]"
    Example #2 Result: "[16.365023136138916, 16.213981151580811]"
    Example #3 Result: "[15.77024507522583, 15.771029949188232]"
    Example #4 Result: "[14.598290920257568, 14.591825008392334]"
    Example #5 Result: "[20.644147872924805, 20.64064884185791]"
    Example #6 Result: "[15.210831165313721, 15.212569952011108]"
    Example #7 Result: "[17.317100048065186, 17.359367847442627]"

最后，当有500个对象时：

    Example #1 Result: "[96.682723999023438, 96.678673028945923]"
    Example #2 Result: "[137.49416589736938, 137.48705387115479]"
    Example #3 Result: "[136.58069896697998, 136.5823769569397]"
    Example #4 Result: "[115.0344090461731, 115.1088011264801]"
    Example #5 Result: "[165.08325910568237, 165.06769108772278]"
    Example #6 Result: "[128.95187497138977, 128.96077489852905]"
    Example #7 Result: "[155.70515990257263, 155.74126601219177]"

感谢所有回复的人！我对这个结果感到非常惊讶。如果还有其他更快的方法，我很想听听。谢谢大家！

性能优化数据结构代码可读性编程技巧代码效率列表转字典对象处理字典键值

将列表转为字典的最佳方法，键为每个对象的值？

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