python-列表中最新的文件_Python

python-列表中最新的文件

python

python-列表中最新的文件,python,Python,我有一份这样的清单- list = [['multinational_intel_2014-10-22T110406Z.zip', '2014 10 22 11:05:19'], ['multinational_intel_2014-10-24T140006Z.zip', '2014 10 24 14:02:51'], ['multinational_intel_2014-11-12T104622Z.zip', '2014 11 12 10:47:49'],

我有一份这样的清单-

list = [['multinational_intel_2014-10-22T110406Z.zip', '2014 10 22 11:05:19'],
        ['multinational_intel_2014-10-24T140006Z.zip', '2014 10 24 14:02:51'],
        ['multinational_intel_2014-11-12T104622Z.zip', '2014 11 12 10:47:49'],
        ['multinational_intel_2014-11-10T131155Z.zip', '2014 11 10 13:13:57'],
        ['multinational_intel_2014-11-14T172344Z.zip', '2014 11 14 17:25:17'],
        ['multinational_intel_2014-11-11T103518Z.zip', '2014 11 11 10:36:47']]

我想扫描列表并获取与最近日期关联的文件

对于

列表

我的预期输出是-

多国英特尔2014-11-14T172344Z.zip

一种方法是使用：

list = [['multinational_intel_2014-10-22T110406Z.zip', '2014 10 22 11:05:19'],
        ['multinational_intel_2014-10-24T140006Z.zip', '2014 10 24 14:02:51'],
        ['multinational_intel_2014-11-12T104622Z.zip', '2014 11 12 10:47:49'],
        ['multinational_intel_2014-11-10T131155Z.zip', '2014 11 10 13:13:57'],
        ['multinational_intel_2014-11-14T172344Z.zip', '2014 11 14 17:25:17'],
        ['multinational_intel_2014-11-11T103518Z.zip', '2014 11 11 10:36:47']]

sorted_list = sorted(list, key=lambda x: x[1], reverse=True)
most_recent = sorted_list[0][0]
print most_recent