Question

如何根据与字符串

组合的最大日期对python列表进行排序

['', 'q//Attachments/Swoop_coverletter_311386_20120103.doc', 'q//Attachments/Swoop_RESUME_311386_20091012.doc', 'q//Attachments/Swoop_Resume_311386_20100901.doc', 'q//Attachments/Swoop_reSume_311386_20120103.doc', 'q//Attachments/Swoop_coverletter_311386_20100901.doc', 'q//Attachments/Swoop_coverletter_311386_20091012.doc']

上面是列表，预期结果是这个

['q//Attachments/Swoop_coverletter_311386_20120103.doc','q//Attachments/Swoop_reSume_311386_20120103.doc','q//Attachments/Swoop_Resume_311386_20100901.doc','q//Attachments/Swoop_coverletter_311386_20100901.doc','q//Attachments/Swoop_RESUME_311386_20091012.doc','q//Attachments/Swoop_coverletter_311386_20091012.doc','']

我编写了一个脚本，该脚本没有排序，但只在结尾处打印一个值

a = ['q//Attachments/Swoop_coverletter_311386_20120103.doc','q//Attachments/Swoop_reSume_311386_20120103.doc','q//Attachments/Swoop_Resume_311386_20100901.doc','q//Attachments/Swoop_coverletter_311386_20100901.doc','q//Attachments/Swoop_RESUME_311386_20091012.doc','q//Attachments/Swoop_coverletter_311386_20091012.doc','']
print max(a)

结果：

q//Attachments/Swoop_reSume_311386_20120103.doc

我如何获得这样的预期输出

预期产出：

['q//Attachments/Swoop_coverletter_311386_20120103.doc','q//Attachments/Swoop_reSume_311386_20120103.doc','q//Attachments/Swoop_Resume_311386_20100901.doc','q//Attachments/Swoop_coverletter_311386_20100901.doc','q//Attachments/Swoop_RESUME_311386_20091012.doc','q//Attachments/Swoop_coverletter_311386_20091012.doc','']

Answer 1

编写一个函数，用正则表达式从字符串中提取日期，并将其用作sorted的键：

import re

l = ['',
     'q//Attachments/Swoop_coverletter_311386_20120103.doc',
     'q//Attachments/Swoop_RESUME_311386_20091012.doc',
     'q//Attachments/Swoop_Resume_311386_20100901.doc',
     'q//Attachments/Swoop_reSume_311386_20120103.doc',
     'q//Attachments/Swoop_coverletter_311386_20100901.doc',
     'q//Attachments/Swoop_coverletter_311386_20091012.doc']

def get_date(line):
    pattern = '.*_(\d{8}).doc'
    m = re.match(pattern, line)
    if m:
        return int(m.group(1))
    else:
        return -1 # or do something else with lines that contain no date


print sorted(l, key=get_date, reverse=True)

打印：

['q//Attachments/Swoop_coverletter_311386_20120103.doc', 
 'q//Attachments/Swoop_reSume_311386_20120103.doc', 
 'q//Attachments/Swoop_Resume_311386_20100901.doc', 
 'q//Attachments/Swoop_coverletter_311386_20100901.doc', 
 'q//Attachments/Swoop_RESUME_311386_20091012.doc', 
 'q//Attachments/Swoop_coverletter_311386_20091012.doc', 
 '']

Answer 2

我认为使用内置函数str.rpartition('_')（https://docs.python.org/3/library/stdtypes.html#str.rpartition）可以更容易解决问题。

我假设您的所有文件当然具有相同的格式，在这种情况下，上述函数将始终返回<date>.doc。然后，您只需删除.doc。

Answer 3

您可以尝试使用备用的单行解决方案（种类）。您必须先删除空元素来清理列表。

given_list = filter(None, given_list)
sorted(given_list, key=lambda x: datetime.strptime(x.split(".")[0][-8:], "%Y%m%d"), reverse=True)

或者像在BioGeek's answer中一样简化它，而不是使用datetime只需转换为int并对其进行排序。

given_list = filter(None, given_list)
sorted(a, key=lambda x: int(x.split(".")[0][-8:]), reverse=True)

输出：

['q//Attachments/Swoop_coverletter_311386_20120103.doc', 
 'q//Attachments/Swoop_reSume_311386_20120103.doc',
 'q//Attachments/Swoop_Resume_311386_20100901.doc',
 'q//Attachments/Swoop_coverletter_311386_20100901.doc', 
 'q//Attachments/Swoop_RESUME_311386_20091012.doc',
 'q//Attachments/Swoop_coverletter_311386_20091012.doc']

Python根据日期排序数组

3 个答案: