CODEWITHSUNDEEP
python
whoisearth
whoisearth

Reputation: 4160

python - most recent file in list

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

Upvotes: 0

Views: 64

Answers (1)

Jack
Jack

Reputation: 21163

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

Upvotes: 2

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