[ {'time':33}, {'time':11}, {'time':66} ]
How to sort by the "time" element, DESC.
+23
A:
Like this:
from operator import itemgetter
l = sorted(l, key=itemgetter('time'), reverse=True)
Or:
l = sorted(l, key=lambda a: a['time'], reverse=True)
output:
[{'time': 66}, {'time': 33}, {'time': 11}]
If you don't want to keep the original order you can use your_list.sort which modifies the original list instead of creating a copy like sorted(your_list)
l.sort(key=lambda a: a['time'], reverse=True)
Nadia Alramli
2009-10-10 11:41:33
the operator.itemgetter version is preferred. It has one less function call for each element.
nosklo
2009-10-10 13:26:26
@nosklo, itemgetter actually returns a function that works almost the same as lambda a: a['time'] so there isn't really much of a difference from this prospective. Both methods involves a function call for each element.
Nadia Alramli
2009-10-10 17:04:56