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

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]

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]