python - 将列表转换为字典的最佳方法，其中键是每个对象的值？

Question

我正在尝试获取对象列表，并将该列表转换为字典。字典值是列表中的每个对象，字典键是在每个对象中找到的值。

这里是一些代表我在做什么的代码:

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

现在代码可以工作了，但是它有点难看，而且有点慢。谁能举一个更快/“更好”的例子？

编辑:好的，谢谢回复。我必须说，我很惊讶地看到更多 pythonic 方式看起来比手工方式慢。

编辑2:好吧，我更新了测试代码以使其更具可读性，有这么多测试嘿。

这是我们在代码方面所处的位置，我将作者放在代码中，如果我搞砸了，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))

列表中有 5 个对象，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]"

还有 50 个:

    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，有 500 个对象:

    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!我对结果感到非常惊讶。如果有任何其他更快方法的提示，我很乐意听到它们。谢谢大家!

Answer 1

在 python 3.0 中你可以使用字典理解:

{an_object.name : an_object for an_object in object_list}

这在 Python 2 中也是可行的，但有点丑陋:

dict([(an_object.name, an_object) for an_object in object_list])