sort dates in python array

Question

How to sort the below array of dates on python 2.4

 timestamps = ['2011-06-2', '2011-08-05', '2011-02-04', '2010-1-14', '2010-12-13', '2010-1-12', '2010-2-11', '2010-2-07', '2010-12-02', '2011-11-30', '2010-11-26', '2010-11-23', '2010-11-22', '2010-11-16']

Why are you using Python 2.4?

Tim Pietzcker
– Tim Pietzcker

2011-03-02 11:34:04 +00:00
Commented Mar 2, 2011 at 11:34 — Tim Pietzcker
– Tim Pietzcker, Commented Mar 2, 2011 at 11:34

Tim Pietzcker · Accepted Answer · 2017-01-10 10:09:28Z

74

>>> import datetime
>>> dates = [datetime.datetime.strptime(ts, "%Y-%m-%d") for ts in timestamps]
>>> dates.sort()
>>> sorteddates = [datetime.datetime.strftime(ts, "%Y-%m-%d") for ts in dates]
>>> sorteddates
['2010-01-12', '2010-01-14', '2010-02-07', '2010-02-11', '2010-11-16', '2010-11-
22', '2010-11-23', '2010-11-26', '2010-12-02', '2010-12-13', '2011-02-04', '2011
-06-02', '2011-08-05', '2011-11-30']

edited Jan 10, 2017 at 10:09

answered Mar 2, 2011 at 11:33

Tim Pietzcker

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4 Comments

CornSmith Over a year ago

This is the most pythonic way I think. It scales to different formats of time, unlike the accepted answer.

Antoine Over a year ago

best answer (+1) but you probably should not call the last object sorted as it is also the name of a function

Frankie Drake Over a year ago

How to disaccept another answer and accept this one? :)

Debashis Sahoo Over a year ago

this is most logical approach of sorting dates, if they are in string format.

jd. · Accepted Answer · 2011-03-02 11:36:57Z

37

sorted(timestamps, key=lambda d: map(int, d.split('-')))

answered Mar 2, 2011 at 11:36

jd.

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3 Comments

Tim Pietzcker Over a year ago

Cool. Raises a TypeError in Python 3, though.

jd. Over a year ago

I don't have Python 3 installed, but would that be because map returns an iterable? In that case, the key function should be lambda d: tuple(map(int, d.split('-'))).

SeanFromIT Over a year ago

If you need the latest date in the array, you can add the reverse=True argument.

eyquem · Accepted Answer · 2011-03-02 11:40:31Z

13

Just doing that:

timestamps.sort()

result:

['2010-1-12',
 '2010-1-14',
 '2010-11-16',
 '2010-11-22',
 '2010-11-23',
 '2010-11-26',
 '2010-12-02',
 '2010-12-13',
 '2010-2-07',
 '2010-2-11',
 '2011-02-04',
 '2011-06-2',
 '2011-08-05',
 '2011-11-30']

The order year-month-day allow such a sorting because a day changes before a month and a month changes before a year when time is passing.

It's like for a number: the unity digit (the rightmost digit) changes before the ten digit, and this latter changes before the hundred digit, when 1 is progressively added.

And there is the fact that sort() processes from left to right : if the characters at one precise position are the same in two strings to sort, it will examine the two characters in the two string at the following position to decide which one is logically preceding.

Plus the fact that '0' < '1' is True, '1' < '2' is True etc

answered Mar 2, 2011 at 11:40

eyquem

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3 Comments

Tim Pietzcker Over a year ago

That way, 2010-10-1 will be sorted before 2010-2-1.

eyquem Over a year ago

@Tim Pietzcker Oh yes. It only works if the numbers are zfilled(2) which is more frequently the case

Mohsen_Fatemi Over a year ago

and your result is wrong, because '2010-2-07' comes after '2010-12-13' , this is radix sort

Senthil Kumaran · Accepted Answer · 2011-03-02 11:41:02Z

7

>>> import time
>>> timestamps = ['2011-06-2', '2011-08-05', '2011-02-04', '2010-1-14', '2010-12-13', '2010-1-12', '2010-2-11', '2010-2-07', '2010-12-02', '2011-11-30', '2010-11-26', '2010-11-23', '2010-11-22', '2010-11-16']
>>> timestamps.sort(key=lambda x: time.mktime(time.strptime(x,"%Y-%m-%d")))
>>> timestamps
['2010-1-12', '2010-1-14', '2010-2-07', '2010-2-11', '2010-11-16', '2010-11-22', '2010-11-23', '2010-11-26', '2010-12-02', '2010-12-13', '2011-02-04', '2011-06-2', '2011-08-05', '2011-11-30']

answered Mar 2, 2011 at 11:41

Senthil Kumaran

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Comments

Jakob Bowyer · Accepted Answer · 2011-03-02 11:34:07Z

2

If you sort them into the same format you can just call timestamps.sort()

answered Mar 2, 2011 at 11:34

Jakob Bowyer

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Comments

pyanon · Accepted Answer · 2011-03-14 21:50:31Z

1

map(lambda x:x[1], sorted(map(lambda a:[map(int,a.split('-')),a], timestamps)))

['2010-1-12',
 '2010-1-14',
 '2010-2-07',
 '2010-2-11',
 '2010-11-16',
 '2010-11-22',
 '2010-11-23',
 '2010-11-26',
 '2010-12-02',
 '2010-12-13',
 '2011-02-04',
 '2011-06-2',
 '2011-08-05',
 '2011-11-30']

edited Mar 14, 2011 at 21:50

answered Mar 11, 2011 at 22:27

pyanon

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Γιώργος Δήμας · Accepted Answer · 2018-04-25 06:51:00Z

1

This sort ascending dates in dd/mm/yyyy format

from datetime import datetime 
c_array=['07/12/2017', '30/01/2018', '31/05/2016', '30/09/2016', '30/01/2017', '31/05/2017']

sorted(c_array, key=lambda x: datetime.strptime(x, "%d/%m/%Y").strftime("%Y-%m-%d"))
#out: ['31/05/2016', '30/09/2016', '30/01/2017', '31/05/2017', '07/12/2017', '30/01/2018']

answered Apr 25, 2018 at 6:51

Γιώργος Δήμας

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Comments

S.S. Anne · Accepted Answer · 2020-02-26 17:43:35Z

1

print(sorted(list_of_strings,key=lambda x :(int(x.split('-')[0]),int(x.split('-')[1]),int(x.split('-')[2])))

edited Feb 26, 2020 at 17:43

S.S. Anne

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answered Feb 26, 2020 at 17:23

bhanuprakash bhimarati

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Comments

M.Abdullah Iqbal · Accepted Answer · 2022-12-06 15:04:44Z

1

In Python 3 and using (my personal favorite) comprehensions. For dates, I would always use Python's datetime lib:

from datetime import date
timestamps = ['2011-06-2', '2011-08-05', '2011-02-04', '2010-1-14', '2010-12-13', '2010-1-12', '2010-2-11', '2010-2-07', '2010-12-02', '2011-11-30', '2010-11-26', '2010-11-23', '2010-11-22', '2010-11-14']
timestamps = [date(*[int(y) for y in x.split("-")]) for x in timestamps]
sorted_timestamps = sorted(timestamps)

edited Dec 6, 2022 at 15:04

M.Abdullah Iqbal

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answered Nov 11, 2018 at 7:51

peterb

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