Find third latest date in a list

Question

I have a situation where I need to get the third latest date, i.e

INPUT :

['14-04-2001', '29-12-2061', '21-10-2019',
 '07-01-1973', '19-07-2014','11-03-1992','21-10-2019']

Also , INPUT

OUTPUT : 19-07-2014

import datetime
datelist = ['14-04-2001', '29-12-2061', '21-10-2019', '07-01-1973', '19-07-2014','11-03-1992','21-10-2019' ]
for d in datelist:
    x = datetime.datetime.strptime(d,'%d-%m-%Y')
    print x

How can i achieve this?

Please don't call your list `list`, You'll find a lot of trouble that way. As for the problem: 1) build a new list with the datetime objects, 2) sort that list 3) get the third last element from that list. — dhke, Aug 19 '17 at 09:05
@dhke edited , i tried using sorted , but not sure how to implement — Kittystone, Aug 19 '17 at 09:08
`sorted(list, key=lambda d: datetime.datetime.strptime(d,'%d-%m-%Y'), reverse=True)`. Note that sorted returns a generator, so you can't just `[2]` on it. — dhke, Aug 19 '17 at 09:10
@Bijoy Not today, sort in increasing to decresing and get the 3rd one, but get rid of duplicate — Kittystone, Aug 19 '17 at 09:15
@dhke no - `sorted` returns a `list` - so indexing into it is fine... — Jon Clements, Aug 19 '17 at 09:15

shad0w_wa1k3r · Accepted Answer · 2017-08-19T09:39:19.287

2

You can sort the list and take the 3rd element from it.

my_list = [datetime.datetime.strptime(d,'%d-%m-%Y') for d in list]
# [datetime.datetime(2001, 4, 14, 0, 0), datetime.datetime(2061, 12, 29, 0, 0), datetime.datetime(2019, 10, 21, 0, 0), datetime.datetime(1973, 1, 7, 0, 0), datetime.datetime(2014, 7, 19, 0, 0), datetime.datetime(1992, 3, 11, 0, 0), datetime.datetime(2019, 10, 21, 0, 0)]
my_list.sort(reverse=True)
my_list[2]
# datetime.datetime(2019, 10, 21, 0, 0)

Also, as per Kerorin's suggestion, if you don't need to sort in-place and just need the 3rd element always, you can simply do

sorted(my_list, reverse=True)[2]

Update

To remove the duplicates, taking inspiration from this answer, you can do the following -

import datetime

datelist = ['14-04-2001', '29-12-2061', '21-10-2019', '07-01-1973', '19-07-2014', '11-03-1992', '21-10-2019']
seen = set()
my_list = [datetime.datetime.strptime(d,'%d-%m-%Y')
           for d in datelist
           if d not in seen and not seen.add(d)]
my_list.sort(reverse=True)

edited Aug 19 '17 at 09:39

answered Aug 19 '17 at 09:05

shad0w_wa1k3r

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2

Additional: `sorted(my_list, reverse=True)[2]` is also accepted. – harukaeru Aug 19 '17 at 09:11
Thanks, added to the answer. – shad0w_wa1k3r Aug 19 '17 at 09:13
@AshishNitinPatil Does this gets rid of duplicate 21-10-2019' – Kittystone Aug 19 '17 at 09:14
@AshishNitinPatil Can you make it more of a code rather like input on console, also getting rid of duplicate – Kittystone Aug 19 '17 at 09:17
@Kittystone Updated the answer for deduping, had already edit it for non-console code style. – shad0w_wa1k3r Aug 19 '17 at 09:24
@AshishNitinPatil Not able to execute properly, can you please create a proper def ? , also i update the input – Kittystone Aug 19 '17 at 09:29
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/152322/discussion-between-kittystone-and-ashish-nitin-patil). – Kittystone Aug 19 '17 at 09:33
To get rid of duplicating, juse use `sorted(set([datetime.datetime.strptime(d, '%d-%m-%Y') for d in datelist]), reverse=True)[2]`. – harukaeru Aug 19 '17 at 09:33

score 0 · Answer 2 · answered Aug 19 '17 at 09:52

You can use heapq.nlargest to do this.

import heapq
from datetime import datetime


datelist = [
    '14-04-2001',
    '29-12-2061', 
    '21-10-2019',
    '07-01-1973', 
    '19-07-2014', 
    '11-03-1992', 
    '21-10-2019'
]

heapq.nlargest(3, {datetime.strptime(d, "%d-%m-%Y") for d in datelist})[-1]

This return datetime.datetime(2014, 7, 19, 0, 0)

Find third latest date in a list

2 Answers2