how to sort the date list in python
Question:
I have a list in python as follows..
[Timestamp('2016-01-03 10:38:52'),
Timestamp('2016-01-18 09:37:29'),
Timestamp('2016-02-06 09:44:44'),
Timestamp('2016-02-07 11:11:28'),
Timestamp('2016-02-15 11:24:41'),
Timestamp('2016-02-20 12:46:07'),
Timestamp('2016-02-21 11:07:11')]
I want to sort this with ascending order
I tried with temp_list.sort() but it does not display any output
Answers:
temp_list.sort() will sort the list in place. That means that it will not return anything. You can say x = sorted(y) to assign x to a sorted version of y, but you could also say y.sort() to define y as the sorted version.
Example with sort:
>>> xx = [2, 1]
>>> xx.sort()
>>> print(xx)
[1, 2] # xx got sorted
and with sorted:
>>> xx = [2, 1]
>>> sorted_xx = sorted(my_list)
>>> print(sorted_xx)
[1, 2]
>>> print(xx)
[2, 1] # xx is still unsorted
Python sorts lists in place, meaning there is no return. If you were to look at the value of your list again you will see it sorted in ascending order.
I have a list in python as follows..
[Timestamp('2016-01-03 10:38:52'),
Timestamp('2016-01-18 09:37:29'),
Timestamp('2016-02-06 09:44:44'),
Timestamp('2016-02-07 11:11:28'),
Timestamp('2016-02-15 11:24:41'),
Timestamp('2016-02-20 12:46:07'),
Timestamp('2016-02-21 11:07:11')]
I want to sort this with ascending order
I tried with temp_list.sort() but it does not display any output
temp_list.sort() will sort the list in place. That means that it will not return anything. You can say x = sorted(y) to assign x to a sorted version of y, but you could also say y.sort() to define y as the sorted version.
Example with sort:
>>> xx = [2, 1]
>>> xx.sort()
>>> print(xx)
[1, 2] # xx got sorted
and with sorted:
>>> xx = [2, 1]
>>> sorted_xx = sorted(my_list)
>>> print(sorted_xx)
[1, 2]
>>> print(xx)
[2, 1] # xx is still unsorted
Python sorts lists in place, meaning there is no return. If you were to look at the value of your list again you will see it sorted in ascending order.