python - 基于满足条件比较版本字符串的最 Pythonic 和有效的方法

Question

我有一个列表列表，其中包含两个元素 ["Name", "Version"]，所有列表的名称都相同。

[[N1, V1] , [N1, V2], [N1, V3], [N1,V4], [N1,V5] .....[N1,Vn] ]

我想要满足以下条件的两个版本 'Vx' 和 'Vy' 之间的所有 [N1,Vi] 对：

仅在以下情况下检索 Vx 和 Vy 之间的 [N1,Vi] 对： Vy > Max(Vi)

（即版本上限（Vy）大于列表中版本的最大值时）

我试过使用：

from distutils.version import LooseVersion, StrictVersion

但我只能找到布尔结果。

[["pshop","4.6.23.1"], ["pshop","4.6.10"], ["pshop","4.0.1"],

 ["pshop","6.8.1"], ["pshop","5.6.23.1"], ["pshop","7.6.23.1"]]

1. If Vx = (5.5.7) Vy = (9.34.1)

In this case it will return lists which have version numbers between Vx and Vy

[["pshop","6.8.1"], ["pshop","5.6.23.1"], ["pshop","7.6.23.1"]]


2. If Vx = (2.5.7) Vy = (6.0.0)

In this case it should return [] as Vy < max(Vi) (6.0.0 < 7.6.23.1)

标签： pythonpython-3.xpython-2.7versionstring-comparison

Answer 1

用于version.parse解析和比较版本，并使用列表推导过滤所需的版本

>>> from packaging import version
>>> lst = [["pshop","4.6.23.1"], ["pshop","4.6.10"], ["pshop","4.0.1"], ["pshop","6.8.1"], ["pshop","5.6.23.1"], ["pshop","7.6.23.1"]]
>>> compare_ver = lambda x,y: version.parse(x) < version.parse(y)
>>> max_v = max(v for _,v in lst)
>>>
>>> Vx = "2.5.7"; Vy = "9.34.1"
>>> [[n,v] for n,v in lst if compare_ver(Vx, v)] if compare_ver(max_v, Vy) else []
[['pshop', '4.6.23.1'], ['pshop', '4.6.10'], ['pshop', '4.0.1'], ['pshop', '6.8.1'], ['pshop', '5.6.23.1'], ['pshop', '7.6.23.1']]
>>> 
>>> Vx = "2.5.7"; Vy = "6.0.0"
>>> [[n,v] for n,v in lst if compare_ver(Vx, v)] if compare_ver(max_v, Vy) else []
[]