python list controlled increment

Question

I'm trying to increment a python list of numbers like an abacus.

list = [0,0,0,0]
units = 4

def m(list, units):
  for e in range(len(list)):
    if units:
      list[e] = list[e] + 1
      units -= 1

This code works fine in that if I run m(list, units) the list will be [1,1,1,1] -- the problem I am trying to solve is that when the units value is something like units = 2 the list will increment to [2,2,1,1] (which again is fine) the problem is when I run the m()function from an uneven list, the list will increment from list[0] to end up [3,3,1,1] rather than [2,2,2,2].

Is there a pythonic way I can have the function increment the list from the lowest value to achieve an even spread?

Answer 1

In order to begin at the first element with the minimum value, you can set the starting index to starting_index = abacus.index(min(abacus)) as @jonrsharpe suggests.

However, you need to avoid going beyond the end of the list, for example if units = 4 and starting_index = 3, so you should take the remainder of your calculated index after dividing by len(abacus), i.e. calculated_index % len(abacus).

Finally, I think it would be easier just to loop over range(units). That way you don't need to decrement units yourself, and can handle adding more units at once than the length of the abacus.

Here is an example implementation:

def m(abacus, units):
    starting_index = abacus.index(min(abacus))
    for raw_index in range(units):
        index = (raw_index + starting_index) % len(abacus)
        abacus[index] = abacus[index] + 1

And test:

abacus = [0,0,0,0]    
print abacus
m(abacus, 2)
print abacus
m(abacus, 4)
print abacus
m(abacus, 3)
print abacus
m(abacus, 3)
print abacus
m(abacus, 7)
print abacus

With result:

[0, 0, 0, 0]
[1, 1, 0, 0]
[2, 2, 1, 1]
[3, 2, 2, 2]
[3, 3, 3, 3]
[5, 5, 5, 4]