我有一个文档名称列表，其中一些具有修订版，例如：
docs = ["ab-14-001", "ab-14-001A", "ab-14-001B", "ab-14-002", "jk-9-12B", "jk-9-12C", "io-34-003"]
我想保留每一个的最新修订，意思是：
docs_final = ["ab-14-001B", "ab-14-002", "jk-9-12C", "io-34-003"]
正如你所看到的，有些文档不会以第一种状态出现(末尾没有字母rev(，它们只是以rev字母出现(如jk-9-12B(。
有没有一种快速的方法可以解析并将这个列表分为docs_final和docs_old？
谢谢！
澄清：结尾的字母是唯一用来复习的东西。例如，这两份文件并不是相互修订的，它们完全不同：
"ab-14-001" and "ab-14-002A"
您可以使用itertools.groupby(我假设您的列表docs已排序(：
from itertools import groupby
docs = ["ab-14-001", "ab-14-001A", "ab-14-001B", "jk-9-12B", "jk-9-12C", "io-34-003"]
docs_old, docs_new = [], []
for _, g in groupby(docs, lambda k: k.rsplit('-', maxsplit=1)[0]):
*a, b = g
docs_old.extend(a)
docs_new.append(b)
print('Old = ', docs_old)
print('New = ', docs_new)
打印：
Old = ['ab-14-001', 'ab-14-001A', 'jk-9-12B']
New = ['ab-14-001B', 'jk-9-12C', 'io-34-003']
编辑：
import re
from itertools import groupby
docs = ["ab-14-001", "ab-14-001A", "ab-14-001B", "ab-14-002", "jk-9-12B", "jk-9-12C", "io-34-003"]
docs_old, docs_new = [], []
for _, g in groupby(docs, lambda k: re.search(r'(.*?)[A-Z]*$', k).group(1)):
*a, b = g
docs_old.extend(a)
docs_new.append(b)
print('Old = ', docs_old)
print('New = ', docs_new)
打印：
Old = ['ab-14-001', 'ab-14-001A', 'jk-9-12B']
New = ['ab-14-001B', 'ab-14-002', 'jk-9-12C', 'io-34-003']
您可以将以下代码示例用于此类文章名称的情况。不过要小心，因为这只适用于不包括破折号("-"(的版本。版本号/修订号是最后一个破折号后面的一组数字。如果修订命名样式发生更改，则必须相应地调整代码。
docs = ["ab-14-001", "ab-14-001A", "ab-14-001B", "jk-9-12B", "jk-9-12C", "io-34-003"]
#sort articles:
docs.sort()
#split identifier into article and revision number (version number) as tuple
splitted = [x.split('-') for x in docs]
revisions = [("-".join(doc[:-1]), doc[-1]) for doc in splitted]
#iterate over same doc and overwriting older versions with the newest found:
result = ["-".join([doc, version]) for (doc, version) in {x[0]: x[1] for x in revisions}.items()]
print(result)
作为替代方案，此oneliner还将返回最新版本的列表
#One Liner:
result = ["-".join([d, v]) for (d, v) in {x[0]: x[1] for x in [("-".join(doc[:-1]), doc[-1]) for doc in [x.split('-') for x in docs]]}.items()]
from string import ascii_uppercase
docs = ["ab-14-001", "ab-14-001A", "ab-14-001B", "ab-14-002", "jk-9-12B", "jk-9-12C", "io-34-003"]
revisions = tuple(ascii_uppercase)
latest_docs = {}
for doc in sorted(docs):
key = doc[:-1] if doc.endswith(revisions) else doc
latest_docs[key] = doc
print(list(latest_docs.values()))
输出
['ab-14-001B', 'ab-14-002', 'io-34-003', 'jk-9-12C']
使用dict理解可以缩短它，但我肯定会选择以上的那个
from string import ascii_uppercase
docs = ["ab-14-001", "ab-14-001A", "ab-14-001B", "ab-14-002", "jk-9-12B", "jk-9-12C", "io-34-003"]
latest_docs = {doc[:-1] if doc.endswith(tuple(ascii_uppercase)) else doc:doc for doc in sorted(docs)}
print(list(latest_docs.values()))