我试图让 1.7.0 在 1.7.0.rc0 之后但在 1.8.0 之前得到它，如果你正在对版本进行排序，它应该是这样。我认为 LooseVersion 的全部意义在于它正确地处理了这类事情的排序和比较。
>>> from distutils.version import LooseVersion
>>> versions = ["1.7.0", "1.7.0.rc0", "1.8.0"]
>>> lv = [LooseVersion(v) for v in versions]
>>> sorted(lv, reverse=True)
[LooseVersion ('1.8.0'), LooseVersion ('1.7.0.rc0'), LooseVersion ('1.7.0')]
>>> from distutils.version import LooseVersion
>>> versions = ["1.7.0", "1.7.0rc0", "1.11.0"]
>>> sorted(versions, key=LooseVersion)
['1.7.0', '1.7.0rc0', '1.11.0']
从文档
无政府主义者和软件现实主义者的版本编号。实现上述版本号类的标准接口。版本号由一系列数字组成，由句点或字母字符串分隔。比较版本号时，数字组件将按数字进行比较，字母组件将按词汇进行比较。
……
其实这个方案下不存在版本号无效的问题；比较规则简单且可预测，但可能并不总是给出您想要的结果（对于“想要”的某些定义）。
所以你会看到特别对待“rc”没有什么聪明之处
你可以看到版本号是如何分解成这样的
>>> LooseVersion('1.7.0rc0').version
[1, 7, 0, 'rc', 0]
主要编辑：旧答案太不合情理了。这里有两个更漂亮的解决方案。
所以，我目前看到了三种实现预期排序的方法，在实际发布之前发布候选“rc”。
StrictVersion来自同一个包的 ,Version类以添加对任意标签和标签排序的支持from distutils.version import LooseVersion
versions = ["1.7.0", "1.7.0.rc0", "1.8.0"]
lv = [LooseVersion(v) for v in versions]
lv.sort()
sorted_rc = [v.vstring for v in lv]
import re
p = re.compile('rc\\d+$')
i = 0
# skip the first RCs
while i + 1 < len(sorted_rc):
m = p.search(sorted_rc[i])
if m:
i += 1
else:
break
while i + 1 < len(sorted_rc):
tmp = sorted_rc[i]
m = p.search(sorted_rc[i+1])
if m and sorted_rc[i+1].startswith(tmp):
sorted_rc[i] = sorted_rc[i+1]
sorted_rc[i+1] = tmp
i += 1
有了这个我得到：
['1.7.0rc0', '1.7.0', '1.11.0']
如果您被允许编写为或注明 alpha 或 beta 版本，则该包distutils.version还有另一个类StrictVersion可以完成这项工作。1.7.0.rc01.7.0a01.7.0b0
那是：
from distutils.version import StrictVersion
versions = ["1.7.0", "1.7.0b0", "1.11.0"]
sorted(versions, key=StrictVersion)
这给出了：
['1.7.0b0', '1.7.0', '1.11.0']
可以使用re模块完成从一种形式到另一种形式的翻译。
上一个解决方案的明显问题是缺乏灵活性StrictVersion。更改version_re类属性以使用rc而不是aor b，即使它接受1.7.1rc0，仍将其打印为1.7.1r0（从 python 2.7.3 开始）。
我们可以通过实现我们自己的自定义版本类来做到这一点。这可以像这样完成，至少在某些情况下通过一些单元测试来确保正确性：
#!/usr/bin/python
# file: version2.py
from distutils import version
import re
import functools
@functools.total_ordering
class NumberedVersion(version.Version):
"""
A more flexible implementation of distutils.version.StrictVersion
This implementation allows to specify:
- an arbitrary number of version numbers:
not only '1.2.3' , but also '1.2.3.4.5'
- the separator between version numbers:
'1-2-3' is allowed when '-' is specified as separator
- an arbitrary ordering of pre-release tags:
1.1alpha3 < 1.1beta2 < 1.1rc1 < 1.1
when ["alpha", "beta", "rc"] is specified as pre-release tag list
"""
def __init__(self, vstring=None, sep='.', prerel_tags=('a', 'b')):
version.Version.__init__(self)
# super() is better here, but Version is an old-style class
self.sep = sep
self.prerel_tags = dict(zip(prerel_tags, xrange(len(prerel_tags))))
self.version_re = self._compile_pattern(sep, self.prerel_tags.keys())
self.sep_re = re.compile(re.escape(sep))
if vstring:
self.parse(vstring)
_re_prerel_tag = 'rel_tag'
_re_prerel_num = 'tag_num'
def _compile_pattern(self, sep, prerel_tags):
sep = re.escape(sep)
tags = '|'.join(re.escape(tag) for tag in prerel_tags)
if tags:
release_re = '(?:(?P<{tn}>{tags})(?P<{nn}>\d+))?'\
.format(tags=tags, tn=self._re_prerel_tag, nn=self._re_prerel_num)
else:
release_re = ''
return re.compile(r'^(\d+)(?:{sep}(\d+))*{rel}$'\
.format(sep=sep, rel=release_re))
def parse(self, vstring):
m = self.version_re.match(vstring)
if not m:
raise ValueError("invalid version number '{}'".format(vstring))
tag = m.group(self._re_prerel_tag)
tag_num = m.group(self._re_prerel_num)
if tag is not None and tag_num is not None:
self.prerelease = (tag, int(tag_num))
vnum_string = vstring[:-(len(tag) + len(tag_num))]
else:
self.prerelease = None
vnum_string = vstring
self.version = tuple(map(int, self.sep_re.split(vnum_string)))
def __repr__(self):
return "{cls} ('{vstring}', '{sep}', {prerel_tags})"\
.format(cls=self.__class__.__name__, vstring=str(self),
sep=self.sep, prerel_tags = list(self.prerel_tags.keys()))
def __str__(self):
s = self.sep.join(map(str,self.version))
if self.prerelease:
return s + "{}{}".format(*self.prerelease)
else:
return s
def __lt__(self, other):
"""
Fails when the separator is not the same or when the pre-release tags
are not the same or do not respect the same order.
"""
# TODO deal with trailing zeroes: e.g. "1.2.0" == "1.2"
if self.prerel_tags != other.prerel_tags or self.sep != other.sep:
raise ValueError("Unable to compare: instances have different"
" structures")
if self.version == other.version and self.prerelease is not None and\
other.prerelease is not None:
tag_index = self.prerel_tags[self.prerelease[0]]
other_index = self.prerel_tags[other.prerelease[0]]
if tag_index == other_index:
return self.prerelease[1] < other.prerelease[1]
return tag_index < other_index
elif self.version == other.version:
return self.prerelease is not None and other.prerelease is None
return self.version < other.version
def __eq__(self, other):
tag_index = self.prerel_tags[self.prerelease[0]]
other_index = other.prerel_tags[other.prerelease[0]]
return self.prerel_tags == other.prerel_tags and self.sep == other.sep\
and self.version == other.version and tag_index == other_index and\
self.prerelease[1] == other.prerelease[1]
import unittest
class TestNumberedVersion(unittest.TestCase):
def setUp(self):
self.v = NumberedVersion()
def test_compile_pattern(self):
p = self.v._compile_pattern('.', ['a', 'b'])
tests = {'1.2.3': True, '1a0': True, '1': True, '1.2.3.4a5': True,
'b': False, '1c0': False, ' 1': False, '': False}
for test, result in tests.iteritems():
self.assertEqual(result, p.match(test) is not None, \
"test: {} result: {}".format(test, result))
def test_parse(self):
tests = {"1.2.3.4a5": ((1, 2, 3, 4), ('a', 5))}
for test, result in tests.iteritems():
self.v.parse(test)
self.assertEqual(result, (self.v.version, self.v.prerelease))
def test_str(self):
tests = (('1.2.3',), ('10-2-42rc12', '-', ['rc']))
for t in tests:
self.assertEqual(t[0], str(NumberedVersion(*t)))
def test_repr(self):
v = NumberedVersion('1,2,3rc4', ',', ['lol', 'rc'])
expected = "NumberedVersion ('1,2,3rc4', ',', ['lol', 'rc'])"
self.assertEqual(expected, repr(v))
def test_order(self):
test = ["1.7.0", "1.7.0rc0", "1.11.0"]
expected = ['1.7.0rc0', '1.7.0', '1.11.0']
versions = [NumberedVersion(v, '.', ['rc']) for v in test]
self.assertEqual(expected, list(map(str,sorted(versions))))
if __name__ == '__main__':
unittest.main()
所以，它可以像这样使用：
import version2
versions = ["1.7.0", "1.7.0rc2", "1.7.0rc1", "1.7.1", "1.11.0"]
sorted(versions, key=lambda v: version2.NumberedVersion(v, '.', ['rc']))
输出：
['1.7.0rc1', '1.7.0rc2', '1.7.0', '1.7.1', '1.11.0']
所以，总而言之，使用 python 随附的电池或推出自己的电池。
关于这个实现：可以通过处理版本中的尾随零来改进它，并记住正则表达式的编译。
我像这样使用pkg_resources模块：
from pkg_resources import parse_version
def test_version_sorting():
expected = ['1.0.0dev0',
'1.0.0dev1',
'1.0.0dev2',
'1.0.0dev10',
'1.0.0rc0',
'1.0.0rc2',
'1.0.0rc5',
'1.0.0rc21',
'1.0.0',
'1.1.0',
'1.1.1',
'1.1.11',
'1.2.0',
'1.3.0',
'1.23.0',
'2.0.0', ]
alphabetical = sorted(expected)
shuffled = sorted(expected, key=lambda x: random())
assert expected == sorted(alphabetical, key=parse_version)
assert expected == sorted(shuffled, key=parse_version)
请注意，从预期版本列表中创建随机排序会使这成为潜在的不稳定单元测试，因为两次运行不会有相同的数据。不过，在这种情况下，这应该没关系……希望如此。
我发现这很有帮助，而且更简单：
from packaging import version
vers = ["1.7.0", "1.7.0rc2", "1.7.0rc1", "1.7.1", "1.11.0"]
sorted(vers, key=lambda x: version.Version(x))
结果是：
['1.7.0rc1', '1.7.0rc2', '1.7.0', '1.7.1', '1.11.0']
添加reverse=True使它们按“降序”顺序排列，我觉得这很有帮助。
['1.11.0', '1.7.1', '1.7.0', '1.7.0rc2', '1.7.0rc1']
它可以对各种版本风格的数字进行排序（我的测试平台是 Linux 版本 v4.11.16 等）
我用这个：
#!/usr/bin/python
import re
def sort_software_versions(versions = [], reverse = False):
def split_version(version):
def toint(x):
try:
return int(x)
except:
return x
return map(toint, re.sub(r'([a-z])([0-9])', r'\1.\2', re.sub(r'([0-9])([a-z])', r'\1.\2', version.lower().replace('-', '.'))).split('.'))
def compare_version_list(l1, l2):
def compare_version(v1, v2):
if isinstance(v1, int):
if isinstance(v2, int):
return v1 - v2
else:
return 1
else:
if isinstance(v2, int):
return -1
else:
return cmp(v1, v2)
ret = 0
n1 = len(l1)
n2 = len(l2)
if n1 < n2:
l1.extend([0]*(n2 - n1))
if n2 < n1:
l2.extend([0]*(n1 - n2))
n = max(n1, n2)
i = 0
while not ret and i < n:
ret = compare_version(l1[i], l2[i])
i += 1
return ret
return sorted(versions, cmp = compare_version_list, key = split_version, reverse = reverse)
print(sort_software_versions(['1.7.0', '1.7.0.rc0', '1.8.0']))
['1.7.0.rc0', '1.7.0', '1.8.0']
这样它就可以正确处理 alpha、beta、rc。它可以处理包含连字符的版本，或者当人们将“rc”粘贴到版本时。re.sub 可以使用已编译的正则表达式，但这已经足够了。
在我的情况下，我想使用“.devX”作为“预发布”标识符，所以这是另一个实现，主要基于distutils.version.StrictVersion
class ArtefactVersion(Version):
"""
Based on distutils/version.py:StrictVersion
"""
version_re = re.compile(r'^(\d+) \. (\d+) \. (\d+) (\.dev\d+)?$', re.VERBOSE | re.ASCII)
def parse(self, vstring):
match = self.version_re.match(vstring)
if not match:
raise ValueError("invalid version number '%s'" % vstring)
(major, minor, patch, prerelease) = match.group(1, 2, 3, 4)
self.version = tuple(map(int, [major, minor, patch]))
if prerelease:
self.prerelease = prerelease[4:]
else:
self.prerelease = None
def __str__(self):
vstring = '.'.join(map(str, self.version))
if self.prerelease:
vstring = vstring + f".dev{str(self.prerelease)}"
return vstring
def _cmp(self, other):
if isinstance(other, str):
other = ArtefactVersion(other)
if self.version != other.version:
# numeric versions don't match
# prerelease stuff doesn't matter
if self.version < other.version:
return -1
else:
return 1
# have to compare prerelease
# case 1: neither has prerelease; they're equal
# case 2: self has prerelease, other doesn't; other is greater
# case 3: self doesn't have prerelease, other does: self is greater
# case 4: both have prerelease: must compare them!
if (not self.prerelease and not other.prerelease):
return 0
elif (self.prerelease and not other.prerelease):
return -1
elif (not self.prerelease and other.prerelease):
return 1
elif (self.prerelease and other.prerelease):
if self.prerelease == other.prerelease:
return 0
elif self.prerelease < other.prerelease:
return -1
else:
return 1
else:
assert False, "never get here"