I have a 2 dimensional list. I want to sort this list by multiple criteria in the sublist, but provide a default if one of the chosen index criteria doesn't exist.
my_list = [[5, 4], [1], [6, 8, 1]]
my_list.sort(key=lambda x: (x[0], x[1])
Obviously, this will throw an out of range error. How can I get around this and provide a default value if the index doesn't exist?
CodePudding user response：
You can write it the explicit way, checking length and using a different value if it's too short:
my_list.sort(key=lambda x: (x[0], x[1] if len(x) > 1 else 12345)
or you can append the fallback value unconditionally, then slice down to the number of elements you care about:
my_list.sort(key=lambda x: [*x, 12345][:2])
CodePudding user response：
In your lambda, you could add two default values (in this case I've used None) to the end of the list ensuring it is at least two values long.
>>> my_list = [[5, 4], [1], [6, 8, 1]]
>>> sorted(my_list, key=lambda x: ((x := [*x, None, None])[0], x[1]))
[[1], [5, 4], [6, 8, 1]]
Requires Python 3.8 or later for := operator.
CodePudding user response：
You can take advantage of the fact that indexing a list with an out-of-bound slice returns an empty list, and use the or operator to make the empty list default to a singleton list enclosing a value you want in its place. Since the or operator is widely used as an idiom to provide default values, this approach makes the code arguably more readable:
DEFAULT = [1]
my_list.sort(key=lambda x: (x[0], x[1:2] or DEFAULT))