I am trying to figure an elegant way of implementing the distribution of an amount into a given set of slots in python.

For example:

7 oranges distributed onto 4 plates would return:

[2, 2, 2, 1]

10 oranges across 4 plates would be:

[3, 3, 2, 2]

Conceptually, what you want to do is compute 7 // 4 = 1 and 7 % 4 = 3. This means that all the plates get 1 whole orange. The remainder of 3 tells you that three of the plates get an extra orange.

The divmod builtin is a shortcut for getting both quantities simultaneously:

def distribute(oranges, plates):
    base, extra = divmod(oranges, plates)
    return [base + (i < extra) for i in range(plates)]

With your example:

>>> distribute(oranges=7, plates=4)
[2, 2, 2, 1]

For completeness, you’d probably want to check that oranges is non-negative and plates is positive. Given those conditions, here are some additional test cases:

>>> distribute(oranges=7, plates=1)
[7]

>>> distribute(oranges=0, plates=4)
[0, 0, 0, 0]

>>> distribute(oranges=20, plates=2)
[10, 10]

>>> distribute(oranges=19, plates=4)
[5, 5, 5, 4]

>>> distribute(oranges=10, plates=4)
[3, 3, 2, 2]

You want to look at Bresenham’s algorithm for drawing lines (i.e. distributing X pixels on a Y range as “straightly” as possible; the application of this to the distribution problem is straightforward).

This is an implementation I found here:

def get_line(start, end):
    """Bresenham's Line Algorithm
    Produces a list of tuples from start and end

    >>> points1 = get_line((0, 0), (3, 4))
    >>> points2 = get_line((3, 4), (0, 0))
    >>> assert(set(points1) == set(points2))
    >>> print points1
    [(0, 0), (1, 1), (1, 2), (2, 3), (3, 4)]
    >>> print points2
    [(3, 4), (2, 3), (1, 2), (1, 1), (0, 0)]
    """
    # Setup initial conditions
    x1, y1 = start
    x2, y2 = end
    dx = x2 - x1
    dy = y2 - y1

    # Determine how steep the line is
    is_steep = abs(dy) > abs(dx)

    # Rotate line
    if is_steep:
        x1, y1 = y1, x1
        x2, y2 = y2, x2

    # Swap start and end points if necessary and store swap state
    swapped = False
    if x1 > x2:
        x1, x2 = x2, x1
        y1, y2 = y2, y1
        swapped = True

    # Recalculate differentials
    dx = x2 - x1
    dy = y2 - y1

    # Calculate error
    error = int(dx / 2.0)
    ystep = 1 if y1 < y2 else -1

    # Iterate over bounding box generating points between start and end
    y = y1
    points = []
    for x in range(x1, x2 + 1):
        coord = (y, x) if is_steep else (x, y)
        points.append(coord)
        error -= abs(dy)
        if error < 0:
            y += ystep
            error += dx

    # Reverse the list if the coordinates were swapped
    if swapped:
        points.reverse()
    return points

Mad Physicist answer is perfect. But if you want to distribute the oranges uniformley on the plates (eg. 2 3 2 3 vs 2 2 3 3 in the 7 oranges and 4 plates example), here’s an simple idea.

Easy case

Take an example with 31 oranges and 7 plates for example.

Step 1: You begin like Mad Physicist with an euclidian division: 31 = 4*7 + 3. Put 4 oranges in each plate and keep the remaining 3.

[4, 4, 4, 4, 4, 4, 4]

Step 2: Now, you have more plates than oranges, and that’s quite different: you have to distribute plates among oranges. You have 7 plates and 3 oranges left: 7 = 2*3 + 1. You will have 2 plates per orange (you have a plate left, but it doesn’t matter). Let’s call this 2 the leap. Start at leap/2 will be pretty :

[4, 5, 4, 5, 4, 5, 4]

Not so easy case

That was the easy case. What happens with 34 oranges and 7 plates?

Step 1: You still begin like Mad Physicist with an euclidian division: 34 = 4*7 + 6. Put 4 oranges in each plate and keep the remaining 6.

[4, 4, 4, 4, 4, 4, 4]

Step 2: Now, you have 7 plates and 6 oranges left: 7 = 1*6 + 1. You will have one plate per orange. But wait.. I don’t have 7 oranges! Don’t be afraid, I lend you an apple:

[5, 5, 5, 5, 5, 5, 4+apple]

But if you want some uniformity, you have to place that apple elsewhere! Why not try distribute apples like oranges in the first case? 7 plates, 1 apple : 7 = 1*7 + 0. The leap is 7, start at leap/2, that is 3:

[5, 5, 5, 4+apple, 5, 5, 5]

Step 3. You owe me an apple. Please give me back my apple :

[5, 5, 5, 4, 5, 5, 5]

To summarize : if you have few oranges left, you distribute the peaks, else you distribute the valleys. (Disclaimer: I’m the author of this “algorithm” and I hope it is correct, but please correct me if I’m wrong !)

The code

Enough talk, the code:

def distribute(oranges, plates):
    base, extra = divmod(oranges, plates) # extra < plates
    if extra == 0:
        L = [base for _ in range(plates)]
    elif extra <= plates//2:
        leap = plates // extra
        L = [base + (i%leap == leap//2) for i in range(plates)]
    else: # plates/2 < extra < plates
        leap = plates // (plates-extra) # plates - extra is the number of apples I lent you
        L = [base + (1 - (i%leap == leap//2)) for i in range(plates)]
    return L

Some tests:

>>> distribute(oranges=28, plates=7)
[4, 4, 4, 4, 4, 4, 4]
>>> distribute(oranges=29, plates=7)
[4, 4, 4, 5, 4, 4, 4]
>>> distribute(oranges=30, plates=7)
[4, 5, 4, 4, 5, 4, 4]
>>> distribute(oranges=31, plates=7)
[4, 5, 4, 5, 4, 5, 4]
>>> distribute(oranges=32, plates=7)
[5, 4, 5, 4, 5, 4, 5]
>>> distribute(oranges=33, plates=7)
[5, 4, 5, 5, 4, 5, 5]
>>> distribute(oranges=34, plates=7)
[5, 5, 5, 4, 5, 5, 5]
>>> distribute(oranges=35, plates=7)
[5, 5, 5, 5, 5, 5, 5]

See also more_itertools.distribute, a third-party tool and its source code.

Code

Here we distributes m items into n bins, one-by-one, and count each bin.

import more_itertools as mit


def sum_distrib(m, n):
    """Return an iterable of m items distributed across n spaces."""
    return [sum(x) for x in mit.distribute(n, [1]*m)]

Demo

sum_distrib(10, 4)
# [3, 3, 2, 2]

sum_distrib(7, 4)
# [2, 2, 2, 1]

sum_distrib(23, 17)
# [2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

Example

This idea is akin to distributing a deck of cards among players. Here is an initial game of Slapjack

import random
import itertools as it


players = 8
suits = list("????")
ranks = list(range(2, 11)) + list("JQKA")
deck = list(it.product(ranks, suits))
random.shuffle(deck)

hands = [list(hand) for hand in mit.distribute(players, deck)]
hands
# [[('A', '?'), (9, '?'), ('K', '?'), (7, '?'), ('A', '?'), (5, '?'), (2, '?')],
#  [(6, '?'), ('Q', '?'), (5, '?'), (5, '?'), (3, '?'), (8, '?'), (7, '?')],
#  [(7, '?'), (9, '?'), (2, '?'), (9, '?'), (7, '?'), ('K', '?')],
#   ...]

where the cards are distributed “as equally as possible between all [8] players”:

[len(hand) for hand in hands]
# [7, 7, 7, 7, 6, 6, 6, 6]

Not sure how this works. But it returns the exact same results

a = 23
b = 17
s = pd.Series(pd.cut(mylist, b), index=mylist)
s.groupby(s).size().values