python - most recent file in list

Question

I have a list being provided like this -

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']]

I want to scan the list and get file associated with the most recent date.

For list my expected output would be -

multinational_intel_2014-11-14T172344Z.zip

Answer 1

One way is to use sorted :

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