我正在浏览一个包含卵的目录，以便将这些卵添加到sys.path。如果目录中有相同的.egg的两个版本，我只想添加最新的版本。
我有一个正则表达式r ^ (? P < eggName > \ w +) - (? P < eggVersion > (\ d \.]+)-.+\.Egg $从文件名中提取名称和版本。问题是比较版本号，它是一个像2.3.1这样的字符串。
因为我比较字符串，2排序超过10，但这是不正确的版本。
>>> "2.3.1" > "10.1.1"
True
我可以做一些拆分、解析、转换为int等，最终我将得到一个变通方法。但这是Python，不是Java。是否有一种优雅的方法来比较版本字符串?
当前回答
def versiontuple(v):
return tuple(map(int, (v.split("."))))
>>> versiontuple("2.3.1") > versiontuple("10.1.1")
False
其他回答
基于Kindall的解决方案发布我的完整功能。通过用前导零填充每个版本部分，我能够支持混合在数字中的任何字母数字字符。
虽然肯定不如他的一行函数漂亮，但它似乎可以很好地处理字母数字版本号。(如果您的版本控制系统中有较长的字符串，请确保适当地设置zfill(#)值。)
def versiontuple(v):
filled = []
for point in v.split("."):
filled.append(point.zfill(8))
return tuple(filled)
.
>>> versiontuple("10a.4.5.23-alpha") > versiontuple("2a.4.5.23-alpha")
True
>>> "10a.4.5.23-alpha" > "2a.4.5.23-alpha"
False
我正在寻找一个解决方案，不会增加任何新的依赖。检查以下(Python 3)解决方案:
class VersionManager:
@staticmethod
def compare_version_tuples(
major_a, minor_a, bugfix_a,
major_b, minor_b, bugfix_b,
):
"""
Compare two versions a and b, each consisting of 3 integers
(compare these as tuples)
version_a: major_a, minor_a, bugfix_a
version_b: major_b, minor_b, bugfix_b
:param major_a: first part of a
:param minor_a: second part of a
:param bugfix_a: third part of a
:param major_b: first part of b
:param minor_b: second part of b
:param bugfix_b: third part of b
:return: 1 if a > b
0 if a == b
-1 if a < b
"""
tuple_a = major_a, minor_a, bugfix_a
tuple_b = major_b, minor_b, bugfix_b
if tuple_a > tuple_b:
return 1
if tuple_b > tuple_a:
return -1
return 0
@staticmethod
def compare_version_integers(
major_a, minor_a, bugfix_a,
major_b, minor_b, bugfix_b,
):
"""
Compare two versions a and b, each consisting of 3 integers
(compare these as integers)
version_a: major_a, minor_a, bugfix_a
version_b: major_b, minor_b, bugfix_b
:param major_a: first part of a
:param minor_a: second part of a
:param bugfix_a: third part of a
:param major_b: first part of b
:param minor_b: second part of b
:param bugfix_b: third part of b
:return: 1 if a > b
0 if a == b
-1 if a < b
"""
# --
if major_a > major_b:
return 1
if major_b > major_a:
return -1
# --
if minor_a > minor_b:
return 1
if minor_b > minor_a:
return -1
# --
if bugfix_a > bugfix_b:
return 1
if bugfix_b > bugfix_a:
return -1
# --
return 0
@staticmethod
def test_compare_versions():
functions = [
(VersionManager.compare_version_tuples, "VersionManager.compare_version_tuples"),
(VersionManager.compare_version_integers, "VersionManager.compare_version_integers"),
]
data = [
# expected result, version a, version b
(1, 1, 0, 0, 0, 0, 1),
(1, 1, 5, 5, 0, 5, 5),
(1, 1, 0, 5, 0, 0, 5),
(1, 0, 2, 0, 0, 1, 1),
(1, 2, 0, 0, 1, 1, 0),
(0, 0, 0, 0, 0, 0, 0),
(0, -1, -1, -1, -1, -1, -1), # works even with negative version numbers :)
(0, 2, 2, 2, 2, 2, 2),
(-1, 5, 5, 0, 6, 5, 0),
(-1, 5, 5, 0, 5, 9, 0),
(-1, 5, 5, 5, 5, 5, 6),
(-1, 2, 5, 7, 2, 5, 8),
]
count = len(data)
index = 1
for expected_result, major_a, minor_a, bugfix_a, major_b, minor_b, bugfix_b in data:
for function_callback, function_name in functions:
actual_result = function_callback(
major_a=major_a, minor_a=minor_a, bugfix_a=bugfix_a,
major_b=major_b, minor_b=minor_b, bugfix_b=bugfix_b,
)
outcome = expected_result == actual_result
message = "{}/{}: {}: {}: a={}.{}.{} b={}.{}.{} expected={} actual={}".format(
index, count,
"ok" if outcome is True else "fail",
function_name,
major_a, minor_a, bugfix_a,
major_b, minor_b, bugfix_b,
expected_result, actual_result
)
print(message)
assert outcome is True
index += 1
# test passed!
if __name__ == '__main__':
VersionManager.test_compare_versions()
编辑:添加变量与元组比较。当然，具有元组比较的变体更好，但我正在寻找具有整数比较的变体
使用python增加版本
def increment_version(version):
version = version.split('.')
if int(version[len(version) - 1]) >= 99:
version[len(version) - 1] = '0'
version[len(version) - 2] = str(int(version[len(version) - 2]) + 1)
else:
version[len(version) - 1] = str(int(version[len(version) - 1]) + 1)
return '.'.join(version)
version = "1.0.0"
version_type_2 = "1.0"
print("old version",version ,"new version",increment_version(version))
print("old version",version_type_2 ,"new version",increment_version(version_type_2))
假设你的语义版本是“干净的”(例如，x.x.x)，并且你有一个需要排序的版本列表，这里有一些东西可以工作。
# Here are some versions
versions = ["1.0.0", "1.10.0", "1.9.0"]
# This does not work
versions.sort() # Result: ['1.0.0', '1.10.0', '1.9.0']
# So make a list of tuple versions
tuple_versions = [tuple(map(int, (version.split(".")))) for version in versions]
# And sort the string list based on the tuple list
versions = [x for _, x in sorted(zip(tuple_versions, versions))] # Result: ['1.0.0', '1.9.0', '1.10.0']
要获得最新版本，您只需选择列表版本[-1]中的最后一个元素，或者使用sorted()的reverse属性进行反向排序，将其设置为True，并获得[0]元素。
当然，您可以将所有这些打包到一个方便的函数中以供重用。
def get_latest_version(versions):
"""
Get the latest version from a list of versions.
"""
try:
tuple_versions = [tuple(map(int, (version.split(".")))) for version in versions]
versions = [x for _, x in sorted(zip(tuple_versions, versions), reverse=True)]
latest_version = versions[0]
except Exception as e:
print(e)
latest_version = None
return latest_version
print(get_latest_version(["1.0.0", "1.10.0", "1.9.0"]))
您可以使用semver包来确定一个版本是否满足语义版本需求。这与比较两个实际版本不同，但这是一种比较。
例如，3.6.0+1234版本应与3.6.0相同。
import semver
semver.match('3.6.0+1234', '==3.6.0')
# True
from packaging import version
version.parse('3.6.0+1234') == version.parse('3.6.0')
# False
from distutils.version import LooseVersion
LooseVersion('3.6.0+1234') == LooseVersion('3.6.0')
# False
