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How can I sort this list in descending order?
timestamps = [
"2010-04-20 10:07:30",
"2010-04-20 10:07:38",
"2010-04-20 10:07:52",
"2010-04-20 10:08:22",
"2010-04-20 10:08:22",
"2010-04-20 10:09:46",
"2010-04-20 10:10:37",
"2010-04-20 10:10:58",
"2010-04-20 10:11:50",
"2010-04-20 10:12:13",
"2010-04-20 10:12:13",
"2010-04-20 10:25:38"
]
352
Sort in Descending order The sort() method accepts a reverse parameter as an optional argument. Setting reverse = True sorts the list in the descending order.
The ORDER BY keyword sorts the records in ascending order by default. To sort the records in descending order, use the DESC keyword.
Method #1 : Naive method The simplest way to check this is run a loop for first element and check if we could find any larger element than it after that element, if yes, the list is not reverse sorted. if ( not flag) : print ( "Yes, List is reverse sorted." )
This will give you a sorted version of the array.
sorted(timestamps, reverse=True)
If you want to sort in-place:
timestamps.sort(reverse=True)
Check the docs at Sorting HOW TO
65
In one line, using a lambda:
timestamps.sort(key=lambda x: time.strptime(x, '%Y-%m-%d %H:%M:%S')[0:6], reverse=True)
Passing a function to list.sort:
def foo(x):
return time.strptime(x, '%Y-%m-%d %H:%M:%S')[0:6]
timestamps.sort(key=foo, reverse=True)
You can simply do this:
timestamps.sort(reverse=True)
you simple type:
timestamps.sort()
timestamps=timestamps[::-1]
Since your list is already in ascending order, we can simply reverse the list.
>>> timestamps.reverse()
>>> timestamps
['2010-04-20 10:25:38',
'2010-04-20 10:12:13',
'2010-04-20 10:12:13',
'2010-04-20 10:11:50',
'2010-04-20 10:10:58',
'2010-04-20 10:10:37',
'2010-04-20 10:09:46',
'2010-04-20 10:08:22',
'2010-04-20 10:08:22',
'2010-04-20 10:07:52',
'2010-04-20 10:07:38',
'2010-04-20 10:07:30']
Here is another way
timestamps.sort()
timestamps.reverse()
print(timestamps)
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