Sort Dates in Python Array

sort dates in python array

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

Numpy sorting by increasing datetime

You can also use numpy for this. Assuming you have:

dates = numpy.array([datetime(2012,02,03,12,00,00), datetime(2012,02,03,15,00,00), datetime(2012,02,03,13,00,00)])
values = numpy.array([[1, 1], [3, 3], [2, 2]])

You can do at first:

unified = numpy.concatenate((dates.reshape(len(dates), 1), values), axis=1)

This will merge your two list, which considering what you want should be a better suited way to hold your data anyway. So now you have:

unified = array([[2012-02-03 12:00:00, 1, 1],
                 [2012-02-03 15:00:00, 3, 3],
                 [2012-02-03 13:00:00, 2, 2]], dtype=object)

Now you can do:

unified = numpy.sort(unified, axis=0)

This will sort after date. Now if you still want just the separate lists you can get them with:

unified[:, 0]

>>> array([2012-02-03 12:00:00, 2012-02-03 13:00:00, 2012-02-03 15:00:00], dtype=object)

unified[:, 1:]

>>> array([[1, 1],
           [2, 2],
           [3, 3]], dtype=object)

EDIT after your comment
Ok now that I fully understand what you want you can achieve that by replacing

   unified = numpy.sort(unified, axis=0)    

with:

   unified = numpy.array(sorted(unified, key= lambda x: x[0]))

EDIT

have you tried what I just suggested? In my terminal:

    unified = numpy.array([[datetime(2012,02,03,12,00,00), 4, 1],[datetime(2012,02,03,15,00,00), 5, 2],[datetime(2012,02,03,13,00,00), 2, 1]], dtype=object)
    >>> unified
        array([[2012-02-03 12:00:00, 4, 1],
               [2012-02-03 15:00:00, 5, 2],
               [2012-02-03 13:00:00, 2, 1]], dtype=object)

    >>> unified = numpy.array(sorted(unified, key=lambda x: x[0]))
    >>> unified
        array([[2012-02-03 12:00:00, 4, 1],
               [2012-02-03 13:00:00, 2, 1],
               [2012-02-03 15:00:00, 5, 2]], dtype=object)

How to sort a list by datetime in python?

from datetime import datetime
list = 
[
('2017/09/10 13:19:38', 'employee_id', 'enrolled'),
('2017/09/10 12:15:21', 'employee_id', 'deleted'),
('2017/09/10 21:19:34', 'employee_id', 'enrolled'),
('2017/09/10 22:42:50', 'employee_id', 'deleted'),
('2017/09/10 16:53:03', 'employee_id', 'enrolled')
]
sorted_list = sorted(list, key=lambda t: datetime.strptime(t[0], '%Y/%m/%d %H:%M:%S'))

Use the key parameter of the sorted function, in this case it tells the function to parse the first element of each tuple as a datetime string with the format '%Y/%m/%d %H:%M:%S' and use that value for sorting.

How to sort an array by date in python?

I am not sure if you really tried every other solution on this website, but here's how you can do in a fool-proof way:

from datetime import datetime

lst=[['08/12/2020,gggggh,medium'], 
     ['17/12/2020,hhhhhhh,medium'], 
     ['09/12/2020,bbbbbb,low'], 
     ['30/12/2020,bbbbbb,low']]

lst.sort(key = lambda x: datetime.strptime(x[0].split(',')[0], "%d/%m/%Y"))
# Your list `lst` is sorted now.
print(lst)

Outputs

[['08/12/2020,gggggh,medium'],
['09/12/2020,bbbbbb,low'],
['17/12/2020,hhhhhhh,medium'],
['30/12/2020,bbbbbb,low']]

Update: It might be helpful for you to also check this official documentation out regarding the date-time formatting that I used in the above code.

sort dates in python according to hour, day, month

import datetime
dates = [datetime.datetime.strptime(ts, "%Y-%m-%d %H:%M:%S %z") for ts in timestamps]
dates.sort()
sorteddates = [datetime.datetime.strftime(ts, "%Y-%m-%d %H:%M:%S %z") for ts in dates]

Do double confirm if the DateTime format corresponds. https://www.w3schools.com/python/python_datetime.asp

Sort python array field as date string by date

Just remove the array access in the lambda expression

key=lambda x['status_changed_at']: ...

and do

key=lambda x: ...

The array access is done later:

datetime.strptime(x['status_changed_at'] ...

For descending sorting, you also need

..., reverse=True)

Other than that, you need to rework your expression in strftime, because you have the month first and it's locale dependent. %Y%m%d%H%M probably sorts better.

Format in sorting an array of dates in Python 3

sales['date'] = sales['date'].dt.strftime('%Y/%d/%m')
dates = sales.date.sort_values()
dates.unique()

Out:
array(['2013/01/01', '2013/01/02', '2013/01/03', ..., '2015/31/07','2015/31/08', '2015/31/10'], dtype=object)

thank you for the comments above!

How to sort an array of objects by datetime in Python?

You can use the built-in sorted function, along with datetime.strptime

from datetime import datetime

sortedArray = sorted(
    unsortedArray,
    key=lambda x: datetime.strptime(x['Created'], '%m/%d/%y %H:%M'), reverse=True
)

Python - How to sort date string in a Data Frame?

Convert the dates to timestamps first:

import pandas as pd
df["date"] = pd.to_datetime(df["date"], format="%m/%d/%y")
>>> df["date"].max()

How do I sort a list of datetime or date objects?

You're getting None because list.sort() it operates in-place, meaning that it doesn't return anything, but modifies the list itself. You only need to call a.sort() without assigning it to a again.

There is a built in function sorted(), which returns a sorted version of the list - a = sorted(a) will do what you want as well.

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