Sort list of list with dates strings python

Question

I'm trying to sort my list of list with a.sort(key=lambda x: x[0]) or a.sort(). The first element of my date is the day and I need to sort by it.

a = [['1/8/2018', 71], ['10/8/2018', 76], ['11/8/2018', 6], ['12/8/2018', 60], ['13/8/2018', 81], ['14/8/2018', 74], ['15/8/2018', 7], ['16/8/2018', 77], ['17/8/2018', 81], ['18/8/2018', 69], ['19/8/2018', 66], ['2/8/2018', 81], ['20/8/2018', 74], ['21/8/2018', 88], ['22/8/2018', 92], ['23/8/2018', 90], ['24/8/2018', 84], ['25/8/2018', 66], ['26/8/2018', 64], ['27/8/2018', 82], ['28/8/2018', 82], ['29/8/2018', 79], ['3/8/2018', 8], ['4/8/2018', 64], ['5/8/2018', 64], ['6/8/2018', 12], ['7/8/2018', 8], ['8/8/2018', 83], ['9/8/2018', 77]]

the results I need:

[['1/8/2018', 71],['2/8/2018', 81], ['3/8/2018', 8], ['4/8/2018', 64], ['5/8/2018', 64], ['6/8/2018', 12],['7/8/2018', 8], ['8/8/2018', 83], ['9/8/2018', 77], ['10/8/2018', 76], ['11/8/2018', 6], ['12/8/2018', 60], ['13/8/2018', 81], ['14/8/2018', 74], ['15/8/2018', 7], ['16/8/2018', 77], ['17/8/2018', 81], ['18/8/2018', 69], ['19/8/2018', 66], ['20/8/2018', 74], ['21/8/2018', 88], ['22/8/2018', 92], ['23/8/2018', 90], ['24/8/2018', 84], ['25/8/2018', 66], ['26/8/2018', 64], ['27/8/2018', 82], ['28/8/2018', 82], ['29/8/2018', 79]]

Looks like your current code is comparing those dates as a string instead of numbers or dates. Try to convert that into date than do your stuff. — Kamaldeep singh Bhatia
– Kamaldeep singh Bhatia, Commented Aug 30, 2018 at 13:39

Rakesh · Accepted Answer · 2018-08-30 13:41:17Z

3

Use datetime.datetime.strptime

Ex:

import datetime
a = [['1/8/2018', 71], ['10/8/2018', 76], ['11/8/2018', 6], ['12/8/2018', 60], ['13/8/2018', 81], ['14/8/2018', 74], ['15/8/2018', 7], ['16/8/2018', 77], ['17/8/2018', 81], ['18/8/2018', 69], ['19/8/2018', 66], ['2/8/2018', 81], ['20/8/2018', 74], ['21/8/2018', 88], ['22/8/2018', 92], ['23/8/2018', 90], ['24/8/2018', 84], ['25/8/2018', 66], ['26/8/2018', 64], ['27/8/2018', 82], ['28/8/2018', 82], ['29/8/2018', 79], ['3/8/2018', 8], ['4/8/2018', 64], ['5/8/2018', 64], ['6/8/2018', 12], ['7/8/2018', 8], ['8/8/2018', 83], ['9/8/2018', 77]]
a.sort(key=lambda x: datetime.datetime.strptime(x[0], "%d/%m/%Y"))
print(a)

Output:

[['1/8/2018', 71], ['2/8/2018', 81], ['3/8/2018', 8], ['4/8/2018', 64], ['5/8/2018', 64], ['6/8/2018', 12], ['7/8/2018', 8], ['8/8/2018', 83], ['9/8/2018', 77], ['10/8/2018', 76], ['11/8/2018', 6], ['12/8/2018', 60], ['13/8/2018', 81], ['14/8/2018', 74], ['15/8/2018', 7], ['16/8/2018', 77], ['17/8/2018', 81], ['18/8/2018', 69], ['19/8/2018', 66], ['20/8/2018', 74], ['21/8/2018', 88], ['22/8/2018', 92], ['23/8/2018', 90], ['24/8/2018', 84], ['25/8/2018', 66], ['26/8/2018', 64], ['27/8/2018', 82], ['28/8/2018', 82], ['29/8/2018', 79]]

answered Aug 30, 2018 at 13:41

Rakesh

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1 Comment

Nisarg Over a year ago

See for more reference: tutorialspoint.com/How-to-sort-a-Python-date-string-list

Ajax1234 · Accepted Answer · 2018-08-30 13:46:51Z

2

You can use str.split:

a = [['1/8/2018', 71], ['10/8/2018', 76], ['11/8/2018', 6], ['12/8/2018', 60], ['13/8/2018', 81], ['14/8/2018', 74], ['15/8/2018', 7], ['16/8/2018', 77], ['17/8/2018', 81], ['18/8/2018', 69], ['19/8/2018', 66], ['2/8/2018', 81], ['20/8/2018', 74], ['21/8/2018', 88], ['22/8/2018', 92], ['23/8/2018', 90], ['24/8/2018', 84], ['25/8/2018', 66], ['26/8/2018', 64], ['27/8/2018', 82], ['28/8/2018', 82], ['29/8/2018', 79], ['3/8/2018', 8], ['4/8/2018', 64], ['5/8/2018', 64], ['6/8/2018', 12], ['7/8/2018', 8], ['8/8/2018', 83], ['9/8/2018', 77]]
new_a = sorted(a, key=lambda x:[int(i) for i in x[0].split('/')])

Output:

[['1/8/2018', 71], ['2/8/2018', 81], ['3/8/2018', 8], ['4/8/2018', 64], ['5/8/2018', 64], ['6/8/2018', 12], ['7/8/2018', 8], ['8/8/2018', 83], ['9/8/2018', 77], ['10/8/2018', 76], ['11/8/2018', 6], ['12/8/2018', 60], ['13/8/2018', 81], ['14/8/2018', 74], ['15/8/2018', 7], ['16/8/2018', 77], ['17/8/2018', 81], ['18/8/2018', 69], ['19/8/2018', 66], ['20/8/2018', 74], ['21/8/2018', 88], ['22/8/2018', 92], ['23/8/2018', 90], ['24/8/2018', 84], ['25/8/2018', 66], ['26/8/2018', 64], ['27/8/2018', 82], ['28/8/2018', 82], ['29/8/2018', 79]]

edited Aug 30, 2018 at 13:46

answered Aug 30, 2018 at 13:39

Ajax1234

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2 Comments

DeepSpace Over a year ago

Or actually convert the strings to dates to make the code less cryptic.

bereal Over a year ago

This will sort by day then by month then by year.

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