List sort based on another shorter list

Question

I need to sort a list based on the order of the elements in another list which is shorter ie, doesn't have all the elements compared to the list I'm sorting. I run into this error when using the sort(key=short_list):

long_list = ['y', 'z', 'x', 'c', 'a', 'b']
short_list = ['b', 'c', 'a']
long_list.sort(key=short_list.index)

ValueError: 'x' is not in list

Is there another way to sort the long_list to result in a list that maintains the order of short_list followed by the order of the elements in the long_list?

['b', 'c', 'a', 'y', 'z', 'x']

Answer 1

Something like this should work:

def position(value):
    try:
        return short_list.index(value)
    except ValueError:
        return len(short_list)

long_list.sort(key=position)

Sorting is guaranteed to be stable, so using len(short_list) ensures that the unknown values sort last.

Answer 2

You can use in to detect if the element is in the short_list and a ternary to return a tuple based on that.:

>>> long_list = ['y', 'z', 'x', 'c', 'a', 'b']
>>> short_list = ['b', 'c', 'a']
>>> sorted(long_list, key=lambda e: (short_list.index(e),e) if e in short_list  else (len(short_list),e))
['b', 'c', 'a', 'x', 'y', 'z']

Since Python sorts are stable, the order will only change based on a change of the elements themselves. To make that change, we can use a tuple with either a (index_of_the_element, element) of (len(short_list), element) to effect that change.

If you want the elements to not change order if the element is not in short list, just return an empty tuple:

>>> sorted(long_list, key=lambda e: (short_list.index(e),e) if e in short_list  else (len(short_list),))
['b', 'c', 'a', 'y', 'z', 'x']