How can I sort a list in python 3 with priority given to specific values?

Question

I have a list called V. The elements of V will be one of be one of: 0,1,2 or +inf

I would like to sort this list such that 2 is always at the beginning and the list is sorted in ascending order from then on.

For example, if the list is initially V = [0,1,2,inf] It should then become V = [2,0,1,inf]

I have a good idea of how to do this with conditionals and appending into a new list, but if possible I'd like to sort this with a key or a comprehension.

I am writing this in Python3 but I would prefer to not import anything

dawg · Accepted Answer · 2020-05-05 23:38:00Z

2

Use a lambda to create a tuple and test if the first element is one of the preferred values:

>>> V = [0,1,2,float('inf')] 
>>> sorted(V, key=lambda e: (e!=2, e))
[2, 0, 1, inf]

So the tuple will be one of (True, the_value) or (False, the_value). False < True and that trumps whatever the second actual value in the tuple is.

If you have more than one preferred value, you can use in with a tuple of preferred values:

>>> sorted(V, key=lambda e: (e not in (1,2), e))
[1, 2, 0, inf]
>>> sorted(V, key=lambda e: (e not in (2,0), e))
[0, 2, 1, inf]

You could also create a comprehension:

>>> [(e not in (0,2),e) for e in V]
[(False, 0), (True, 1), (False, 2), (True, inf)]

And then sort that:

>>> [t[1] for t in sorted((e not in (0,2),e) for e in V)]
[0, 2, 1, inf]

Which is a form of Decorate, Sort, Undecorate

edited May 5, 2020 at 23:38

answered May 5, 2020 at 23:20

dawg

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