Elegant Python code for Integer Partitioning

While this answer is fine, I'd recommend skovorodkin's answer.

>>> def partition(number):
...     answer = set()
...     answer.add((number, ))
...     for x in range(1, number):
...         for y in partition(number - x):
...             answer.add(tuple(sorted((x, ) + y)))
...     return answer
... 
>>> partition(4)
set([(1, 3), (2, 2), (1, 1, 2), (1, 1, 1, 1), (4,)])

If you want all permutations(ie (1, 3) and (3, 1)) change answer.add(tuple(sorted((x, ) + y)) to answer.add((x, ) + y)

I needed to solve a similar problem, namely the partition of an integer n into d nonnegative parts, with permutations. For this, there's a simple recursive solution (see here):

def partition(n, d, depth=0):
    if d == depth:
        return [[]]
    return [
        item + [i]
        for i in range(n+1)
        for item in partition(n-i, d, depth=depth+1)
        ]


# extend with n-sum(entries)
n = 5
d = 3
lst = [[n-sum(p)] + p for p in partition(n, d-1)]

print(lst)

Output:

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

A smaller and faster than Nolen's function:

def partitions(n, I=1):
    yield (n,)
    for i in range(I, n//2 + 1):
        for p in partitions(n-i, i):
            yield (i,) + p

Let's compare them:

In [10]: %timeit -n 10 r0 = nolen(20)
1.37 s ± 28.7 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [11]: %timeit -n 10 r1 = list(partitions(20))
979 µs ± 82.9 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [13]: sorted(map(sorted, r0)) == sorted(map(sorted, r1))
Out[14]: True

Looks like it's 1370 times faster for n = 20.

Anyway, it's still far from accel_asc:

def accel_asc(n):
    a = [0 for i in range(n + 1)]
    k = 1
    y = n - 1
    while k != 0:
        x = a[k - 1] + 1
        k -= 1
        while 2 * x <= y:
            a[k] = x
            y -= x
            k += 1
        l = k + 1
        while x <= y:
            a[k] = x
            a[l] = y
            yield a[:k + 2]
            x += 1
            y -= 1
        a[k] = x + y
        y = x + y - 1
        yield a[:k + 1]

It's not only slower, but requires much more memory (but apparently is much easier to remember):

In [18]: %timeit -n 5 r2 = list(accel_asc(50))
114 ms ± 1.04 ms per loop (mean ± std. dev. of 7 runs, 5 loops each)

In [19]: %timeit -n 5 r3 = list(partitions(50))
527 ms ± 8.86 ms per loop (mean ± std. dev. of 7 runs, 5 loops each)

In [24]: sorted(map(sorted, r2)) == sorted(map(sorted, r3))
Out[24]: True

---

You can find other versions on ActiveState: Generator For Integer Partitions (Python Recipe).

---

I use Python 3.6.1 and IPython 6.0.0.

I've compared the solution with perfplot (a little project of mine for such purposes) and found that Nolen's top-voted answer is also the slowest.

Both answers supplied by skovorodkin are much faster. (Note the log-scale.)

---

To to generate the plot:

import perfplot
import collections


def nolen(number):
    answer = set()
    answer.add((number,))
    for x in range(1, number):
        for y in nolen(number - x):
            answer.add(tuple(sorted((x,) + y)))
    return answer


def skovorodkin(n):
    return set(skovorodkin_yield(n))


def skovorodkin_yield(n, I=1):
    yield (n,)
    for i in range(I, n // 2 + 1):
        for p in skovorodkin_yield(n - i, i):
            yield (i,) + p


def accel_asc(n):
    return set(accel_asc_yield(n))


def accel_asc_yield(n):
    a = [0 for i in range(n + 1)]
    k = 1
    y = n - 1
    while k != 0:
        x = a[k - 1] + 1
        k -= 1
        while 2 * x <= y:
            a[k] = x
            y -= x
            k += 1
        l = k + 1
        while x <= y:
            a[k] = x
            a[l] = y
            yield tuple(a[: k + 2])
            x += 1
            y -= 1
        a[k] = x + y
        y = x + y - 1
        yield tuple(a[: k + 1])


def mct(n):
    partitions_of = []
    partitions_of.append([()])
    partitions_of.append([(1,)])
    for num in range(2, n + 1):
        ptitions = set()
        for i in range(num):
            for partition in partitions_of[i]:
                ptitions.add(tuple(sorted((num - i,) + partition)))
        partitions_of.append(list(ptitions))
    return partitions_of[n]


perfplot.show(
    setup=lambda n: n,
    kernels=[nolen, mct, skovorodkin, accel_asc],
    n_range=range(1, 17),
    logy=True,
    # https://stackoverflow.com/a/7829388/353337
    equality_check=lambda a, b: collections.Counter(set(a))
    == collections.Counter(set(b)),
    xlabel="n",
)