Algorithm to sort object by attribute value without allowing gaps or duplicates

Question

I have a agenda with multiple dates, each date can contain 0 > ... items. Items can be sorted by position, positions should be Integer values without gaps and duplicates.

class Item(models.Model):
    date = models.DateField()
    position = models.IntegerField()

    def move_to(position):
        qs = self.__class__.objects.filter(date=self.date)

        # if the position is taken, move up all items gte position
        # 1 spot to free the wanted position

        if position in qs.values_list('position', flat=True):
            qs.filter(position__gte=position).update(position=F('position') + 1)
        self.position = position
        self.save()

This kinda works but if I move items back and forth between dates I am left with position gaps, e.g.

"1, 4, 13"

So it doesn't truncate the gaps, I have tried to look for algorithms but MPTT and stuff like that seems an overkill I don't have any requiement for parent hierarchies

update

I came up with an algorithm that seems to do what I want, but I'm not sure how to implement it

l = [0, 2, 13]

def is_compressed(l):
    return len(l) == sum(l)

while not is_compressed(l):
    m = 0
    for i in l[m:]:
        while i - 1 >= 0 and i - 1 not in l:
            m += 1
            index = l.index(i)
            l.remove(i)
            l.insert(index, i - 1)

>>> print l
[0, 1, 2]

Answer 1

The above algorithm would not work , because assume you have the following list -

[0,2,5,9]

You should ideally get -

[0,1,2,3]

The sum of that list is 6 , but the length of the list is 4 , this does not meet your condition in is_compressed() .

The algorithm should be something like -

l = [0, 2, 13, 15]

next_i = 0
for k,j in enumerate(l):
    if j != next_i:
        l[k] = next_i
    next_i = next_i + 1

print(l)
>> [0, 1, 2, 3]

To implement in your program , you can do similar tests with position and change the position inside the object , when not the next expected position.