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"
]
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)
Here is another way
timestamps.sort()
timestamps.reverse()
print(timestamps)
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']
you simple type:
timestamps.sort()
timestamps=timestamps[::-1]
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)
You can simply do this:
timestamps.sort(reverse=True)
Source: Stackoverflow.com