Question

输入：

foo = {
    'testing-1.30.5': ['The', 'quick'],
    'testing-1.30.12': ['fox', 'jumped', 'over'],
    'testing-1.30.13': ['the'],
    'testing-1.30.4': ['lazy', 'dog'],
    'testing-1.30.1': ['brown'],
    'testing-1.30.3': ['the'],
    'testing-1.30.6': ['brown'],
    'testing-1.30.2': ['fox', 'jumped', 'over'],
    'testing-1.30.14': ['lazy', 'dog'],
    'testing-1.30.8': ['the'],
    'testing-1.30.0': ['The', 'quick'],
    'testing-1.30.10': ['The', 'quick'],
    'testing-1.30.11': ['brown'],
    'testing-1.30.7': ['fox', 'jumped', 'over'],
    'testing-1.30.9': ['lazy', 'dog']
}

做一些排序

bar = sortfoo(foo)

期望的输出：

for item in bar:
    print '{}: {}'.format(item, bar[item])



testing-1.30.0: ['The', 'quick']
testing-1.30.1: ['brown']
testing-1.30.2: ['fox', 'jumped', 'over']
testing-1.30.3: ['the']
testing-1.30.4: ['lazy', 'dog']
testing-1.30.5: ['The', 'quick']
testing-1.30.6: ['brown']
testing-1.30.7: ['fox', 'jumped', 'over']
testing-1.30.8: ['the']
testing-1.30.9: ['lazy', 'dog']
testing-1.30.10: ['The', 'quick']
testing-1.30.11: ['brown']
testing-1.30.12: ['fox', 'jumped', 'over']
testing-1.30.13: ['the']
testing-1.30.14: ['lazy', 'dog']

理想情况下，我希望这是pythonic，因为我不会做一些疯狂的事情，比如将密钥分成组件并根据它构建新的字典;

我尝试了什么：

from collections import OrderedDict
od = OrderedDict(sorted(foo.items()))
    print '{}: {}'.format(item, od[item])

谢谢

Answer 1

知道了！没关系，找到了LooseVersion / strict版本

from distutils.version import LooseVersion
from collections import OrderedDict

orderedKeys = sorted(foo, key=LooseVersion)

odict = OrderedDict((key, foo[key]) for key in orderedKeys)

for item in odict:
     print '{}: {}'.format(item, odict[item])

Answer 2

使用适当的排序键对foo进行排序。你必须切断＆＃34;测试 - ＆＃34;部分，然后将其余部分拆分为句点，然后将其中的每一个转换为整数。此外，结果将是一个键列表，因此请在原始字典中查找这些项目。

>>> bar = sorted(foo, key=lambda x: map(int, x.split('-')[1].split('.')))
>>> for item in bar:
...     print '{}: {}'.format(item, foo[item])
...
testing-1.30.0: ['The', 'quick']
testing-1.30.1: ['brown']
testing-1.30.2: ['fox', 'jumped', 'over']
testing-1.30.3: ['the']
testing-1.30.4: ['lazy', 'dog']
testing-1.30.5: ['The', 'quick']
testing-1.30.6: ['brown']
testing-1.30.7: ['fox', 'jumped', 'over']
testing-1.30.8: ['the']
testing-1.30.9: ['lazy', 'dog']
testing-1.30.10: ['The', 'quick']
testing-1.30.11: ['brown']
testing-1.30.12: ['fox', 'jumped', 'over']
testing-1.30.13: ['the']
testing-1.30.14: ['lazy', 'dog']

请注意，对于Python 3，您必须将map()包裹在list()或tuple()中以使用惰性map对象。

Answer 3

在sort函数中，拆分'.'上的键并将 splitted 列表中的最后一项转换为整数：

for k in sorted(foo, key=lambda x: int(x.rsplit('.')[-1])):
    print('{}: {}'.format(k, foo[k]))

输出：

testing-1.30.0: ['The', 'quick']
testing-1.30.1: ['brown']
testing-1.30.2: ['fox', 'jumped', 'over']
testing-1.30.3: ['the']
testing-1.30.4: ['lazy', 'dog']
testing-1.30.5: ['The', 'quick']
testing-1.30.6: ['brown']
testing-1.30.7: ['fox', 'jumped', 'over']
testing-1.30.8: ['the']
testing-1.30.9: ['lazy', 'dog']
testing-1.30.10: ['The', 'quick']
testing-1.30.11: ['brown']
testing-1.30.12: ['fox', 'jumped', 'over']
testing-1.30.13: ['the']
testing-1.30.14: ['lazy', 'dog']

如何按版本键对字典进行排序

3 个答案: