如何比较 Python 中的版本号？

Question

I am walking a directory that contains eggs to add those eggs to the sys.path . If there are two versions of the same.egg in the directory, I want to add only the latest one.

I have a regular expression r"^(?P<eggName>\w+)-(?P<eggVersion>[\d\.]+)-.+\.egg$ to extract the name and version from the filename. The problem is comparing the version number, which is a string like 2.3.1 .

Since I'm comparing strings, 2 sorts above 10, but that's not correct for versions.

>>> "2.3.1" > "10.1.1"
True

I could do some splitting, parsing, casting to int, etc., and I would eventually get a workaround. But this is Python, not Java . Is there an elegant way to compare version strings?

Answer 1

Use packaging.version.parse .

>>> from packaging import version
>>> version.parse("2.3.1") < version.parse("10.1.2")
True
>>> version.parse("1.3.a4") < version.parse("10.1.2")
True
>>> isinstance(version.parse("1.3.a4"), version.Version)
True
>>> isinstance(version.parse("1.3.xy123"), version.LegacyVersion)
True
>>> version.Version("1.3.xy123")
Traceback (most recent call last):
...
packaging.version.InvalidVersion: Invalid version: '1.3.xy123'

packaging.version.parse is a third-party utility but is used by setuptools (so you probably already have it installed) and is conformant to the current PEP 440 ; it will return a packaging.version.Version if the version is compliant and a packaging.version.LegacyVersion if not. The latter will always sort before valid versions.

Note : packaging has recently been vendored into setuptools .

---

An ancient alternative still used by a lot of software is distutils.version , built in but undocumented and conformant only to the superseded PEP 386 ;

>>> from distutils.version import LooseVersion, StrictVersion
>>> LooseVersion("2.3.1") < LooseVersion("10.1.2")
True
>>> StrictVersion("2.3.1") < StrictVersion("10.1.2")
True
>>> StrictVersion("1.3.a4")
Traceback (most recent call last):
...
ValueError: invalid version number '1.3.a4'

As you can see it sees valid PEP 440 versions as “not strict” and therefore doesn't match modern Python's notion of what a valid version is.

As distutils.version is undocumented, here 's the relevant docstrings.

Answer 2

The packaging library contains utilities for working with versions and other packaging-related functionality. This implements PEP 0440 -- Version Identification and is also able to parse versions that don't follow the PEP. It is used by pip, and other common Python tools to provide version parsing and comparison.

$ pip install packaging

from packaging.version import parse as parse_version
version = parse_version('1.0.3.dev')

This was split off from the original code in setuptools and pkg_resources to provide a more lightweight and faster package.

---

Before the packaging library existed, this functionality was (and can still be) found in pkg_resources, a package provided by setuptools. However, this is no longer preferred as setuptools is no longer guaranteed to be installed (other packaging tools exist), and pkg_resources ironically uses quite a lot of resources when imported. However, all the docs and discussion are still relevant.

From the parse_version() docs :

Parsed a project's version string as defined by PEP 440. The returned value will be an object that represents the version. These objects may be compared to each other and sorted. The sorting algorithm is as defined by PEP 440 with the addition that any version which is not a valid PEP 440 version will be considered less than any valid PEP 440 version and the invalid versions will continue sorting using the original algorithm.

The "original algorithm" referenced was defined in older versions of the docs, before PEP 440 existed.

Semantically, the format is a rough cross between distutils' StrictVersion and LooseVersion classes; if you give it versions that would work with StrictVersion , then they will compare the same way. Otherwise, comparisons are more like a "smarter" form of LooseVersion . It is possible to create pathological version coding schemes that will fool this parser, but they should be very rare in practice.

The documentation provides some examples:

If you want to be certain that your chosen numbering scheme works the way you think it will, you can use the pkg_resources.parse_version() function to compare different version numbers:

 >>> from pkg_resources import parse_version >>> parse_version('1.9.a.dev') == parse_version('1.9a0dev') True >>> parse_version('2.1-rc2') < parse_version('2.1') True >>> parse_version('0.6a9dev-r41475') < parse_version('0.6a9') True

Answer 3

def versiontuple(v):
    return tuple(map(int, (v.split("."))))

>>> versiontuple("2.3.1") > versiontuple("10.1.1")
False

Answer 4

What's wrong with transforming the version string into a tuple and going from there? Seems elegant enough for me

>>> (2,3,1) < (10,1,1)
True
>>> (2,3,1) < (10,1,1,1)
True
>>> (2,3,1,10) < (10,1,1,1)
True
>>> (10,3,1,10) < (10,1,1,1)
False
>>> (10,3,1,10) < (10,4,1,1)
True

@kindall's solution is a quick example of how good the code would look.

Answer 5

There is packaging package available, which will allow you to compare versions as per PEP-440 , as well as legacy versions.

>>> from packaging.version import Version, LegacyVersion
>>> Version('1.1') < Version('1.2')
True
>>> Version('1.2.dev4+deadbeef') < Version('1.2')
True
>>> Version('1.2.8.5') <= Version('1.2')
False
>>> Version('1.2.8.5') <= Version('1.2.8.6')
True

Legacy version support:

>>> LegacyVersion('1.2.8.5-5-gdeadbeef')
<LegacyVersion('1.2.8.5-5-gdeadbeef')>

Comparing legacy version with PEP-440 version.

>>> LegacyVersion('1.2.8.5-5-gdeadbeef') < Version('1.2.8.6')
True

Answer 6

The way that setuptools does it, it uses the pkg_resources.parse_version function. It should be PEP440 compliant.

Example:

#! /usr/bin/python
# -*- coding: utf-8 -*-
"""Example comparing two PEP440 formatted versions
"""
import pkg_resources

VERSION_A = pkg_resources.parse_version("1.0.1-beta.1")
VERSION_B = pkg_resources.parse_version("v2.67-rc")
VERSION_C = pkg_resources.parse_version("2.67rc")
VERSION_D = pkg_resources.parse_version("2.67rc1")
VERSION_E = pkg_resources.parse_version("1.0.0")

print(VERSION_A)
print(VERSION_B)
print(VERSION_C)
print(VERSION_D)

print(VERSION_A==VERSION_B) #FALSE
print(VERSION_B==VERSION_C) #TRUE
print(VERSION_C==VERSION_D) #FALSE
print(VERSION_A==VERSION_E) #FALSE

Answer 7

You can use the semver package to determine if a version satisfies a semantic version requirement. This is not the same as comparing two actual versions, but is a type of comparison.

For example, version 3.6.0+1234 should be the same as 3.6.0.

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

Answer 8

Posting my full function based on Kindall's solution. I was able to support any alphanumeric characters mixed in with the numbers by padding each version section with leading zeros.

While certainly not as pretty as his one-liner function, it seems to work well with alpha-numeric version numbers. (Just be sure to set the zfill(#) value appropriately if you have long strings in your versioning system.)

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

Answer 9

I was looking for a solution which wouldn't add any new dependencies. Check out the following (Python 3) solution:

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()

EDIT: added variant with tuple comparison. Of course the variant with tuple comparison is nicer, but I was looking for the variant with integer comparison

Answer 10

... and getting back to easy ... for simple scripts you can use:

import sys
needs = (3, 9) # or whatever
pvi = sys.version_info.major, sys.version_info.minor    

later in your code

try:
    assert pvi >= needs
except:
    print("will fail!")
    # etc.

Answer 11

similar to standard strverscmp and similar to this solution by Mark Byers but using findall instead of split to avoid empty case.

import re
num_split_re = re.compile(r'([0-9]+|[^0-9]+)')

def try_int(i, fallback=None):
    try:
        return int(i)
    except ValueError:
        pass
    except TypeError:
        pass
    return fallback

def ver_as_list(a):
    return [try_int(i, i) for i in num_split_re.findall(a)]

def strverscmp_lt(a, b):
    a_ls = ver_as_list(a)
    b_ls = ver_as_list(b)
    return a_ls < b_ls

Answer 12

Here is something that will work assuming your semantic versions are "clean" (eg xxx ) and you have a list of versions you need to sort.

# 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']

To get the latest version you could just select the last element in the list versions[-1] or reverse sort by using the reverse attribute of sorted() , setting it to True , and getting the [0] element.

You could of course then wrap all this up in a convenient function for reuse.

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"]))

Answer 13

To increment version using python

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))

Answer 14

This is a compact code for comparing three version numbers. Note that the string comparison fails for all pairs here.

from itertools import permutations

for v1, v2 in permutations(["3.10.21", "3.10.3", "3.9.9"], 2):
    print(f"\nv1 = {v1}, v2 = {v2}")
    print(f"v1 < v2      version.parse(v1) < version.parse(v2)")
    print(f"{v1 < v2}         {version.parse(v1) < version.parse(v2)}")

That gives us:

v1='3.10.21', v2='3.10.3'
v1 < v2      version.parse(v1) < version.parse(v2)
True         False

v1='3.10.21', v2='3.9.9'
v1 < v2      version.parse(v1) < version.parse(v2)
True         False

v1='3.10.3', v2='3.10.21'
v1 < v2      version.parse(v1) < version.parse(v2)
False         True

v1='3.10.3', v2='3.9.9'
v1 < v2      version.parse(v1) < version.parse(v2)
True         False

v1='3.9.9', v2='3.10.21'
v1 < v2      version.parse(v1) < version.parse(v2)
False         True

v1='3.9.9', v2='3.10.3'
v1 < v2      version.parse(v1) < version.parse(v2)
False         True

permutations(iterable, 2) gives us all the 2-length permutations of an iterable. So for example

list(permutations('ABC', 2))

gives us [('A', 'B'), ('A', 'C'), ('B', 'A'), ('B', 'C'), ('C', 'A'), ('C', 'B')] .

Answer 15

If you want to create a filter on a library version, you may use the __version__ attribute (here an example with the jwt library):

from packaging import version
import jwt

if version.parse(jwt.__version__) < version.parse('2.0.0'):
    # TODO: your code

Answer 16

simple few-liner:

import sys
if (sys.version_info.major, sys.version_info.minor) >= (3, 9):
    ...
else:
    ...

Answer 17

i will go more for the touple option, doing a test, using LooseVersion, i do get in my test the second biggest one, (might be doing something wront since is my first time using that library)

import itertools
from distutils.version import LooseVersion, StrictVersion

lista_de_frameworks = ["1.1.1", "1.2.5", "10.5.2", "3.4.5"]

for a, b in itertools.combinations(lista_de_frameworks, 2):
    if LooseVersion(a) < LooseVersion(b):
        big = b
print big

list_test = []
for a in lista_de_frameworks:
    list_test.append( tuple(map(int, (a.split(".")))))

print max(list_test)

and this is what i got:

3.4.5 with Loose

(10, 5, 2) and with the touples