I have to sort a list with multiple attributes. I can do that in ascending order for ALL attributes easily with
L.sort(key=operator.attrgetter(attribute))....
but the problem is, that I have to use mixed configurations for ascending/descending… I have to “imitate” a bit the SQL Order By where you can do something like name ASC, year DESC.
Is there a way to do this easily in Python without having to implement a custom compare function?
Answers:
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Method 1
If your attributes are numeric, you have this.
def mixed_order( a ):
return ( a.attribute1, -a.attribute2 )
someList.sort( key=mixed_order )
If your attributes includes strings or other more complex objects, you have some choices.
The .sort() method is stable: you can do multiple passes. This is perhaps the simplest. It’s also remarkably fast.
def key1( a ): return a.attribute1 def key2( a ): return a.attribute2 someList.sort( key=key2, reverse=True ) someList.sort( key=key1 )
If this is the only sort, you can define your own special-purpose comparison operators. Minimally, you need __eq__ and __lt__. The other four can be derived from these two by simple logic.
Method 2
A custom function will render your code more readable. If you have many sorting operations and you don’t want to create those functions though, you can use lambda’s:
L.sort(lambda x, y: cmp(x.name, y.name) or -cmp(x.year, y.year))
Method 3
You can’t, but writing the compare function is easy:
def my_cmp(a, b):
return cmp(a.foo, b.foo) or cmp(b.bar, a.bar)
L.sort(my_cmp)
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0