I tried to write code to solve the standard Integer Partition problem (Wikipedia). The code I wrote was a mess. I need an elegant solution to solve the problem, because I want to improve my coding style. This is not a homework question.

• python
• algorithm

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

>>> 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)

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'
    )

Much quicker than the accepted response and not bad looking, either. The accepted response does lots of the same work multiple times because it calculates the partitions for lower integers multiple times. For example, when n=22 the difference is 12.7 seconds against 0.0467 seconds.

def partitions_dp(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]

The code is essentially the same except we save the partitions of smaller integers so we don't have to calculate them again and again.

In case anyone is interested: 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]
]

I'm a bit late to the game, but I can offer a contribution which might qualify as more elegant in a few senses:

def partitions(n, m = None):
  """Partition n with a maximum part size of m. Yield non-increasing
  lists in decreasing lexicographic order. The default for m is
  effectively n, so the second argument is not needed to create the
  generator unless you do want to limit part sizes.
  """
  if m is None or m >= n: yield [n]
  for f in range(n-1 if (m is None or m >= n) else m, 0, -1):
    for p in partitions(n-f, f): yield [f] + p

Only 3 lines of code. Yields them in lexicographic order. Optionally allows imposition of a maximum part size.

I also have a variation on the above for partitions with a given number of parts:

def sized_partitions(n, k, m = None):
  """Partition n into k parts with a max part of m.
  Yield non-increasing lists.  m not needed to create generator.
  """
  if k == 1:
    yield [n]
    return
  for f in range(n-k+1 if (m is None or m > n-k+1) else m, (n-1)//k, -1): 
    for p in sized_partitions(n-f, k-1, f): yield [f] + p

After composing the above, I ran across a solution I had created almost 5 years ago, but which I had forgotten about. Besides a maximum part size, this one offers the additional feature that you can impose a maximum length (as opposed to a specific length). FWIW:

def partitions(sum, max_val=100000, max_len=100000):
    """ generator of partitions of sum with limits on values and length """
    # Yields lists in decreasing lexicographical order. 
    # To get any length, omit 3rd arg.
    # To get all partitions, omit 2nd and 3rd args. 

    if sum <= max_val:       # Can start with a singleton.
        yield [sum]

    # Must have first*max_len >= sum; i.e. first >= sum/max_len.
    for first in range(min(sum-1, max_val), max(0, (sum-1)//max_len), -1):
        for p in partitions(sum-first, first, max_len-1):
            yield [first]+p

# -*- coding: utf-8 -*-
import timeit

ncache = 0
cache = {}


def partition(number):
    global cache, ncache
    answer = {(number,), }
    if number in cache:
        ncache += 1
        return cache[number]
    if number == 1:
        cache[number] = answer
        return answer
    for x in range(1, number):
        for y in partition(number - x):
            answer.add(tuple(sorted((x, ) + y)))
    cache[number] = answer
    return answer


print('To 5:')
for r in sorted(partition(5))[::-1]:
    print('\t' + ' + '.join(str(i) for i in r))

print(
    'Time: {}\nCache used:{}'.format(
        timeit.timeit(
            "print('To 30: {} possibilities'.format(len(partition(30))))",
            setup="from __main__ import partition",
            number=1
        ), ncache
    )
)

or https://gist.github.com/sxslex/dd15b13b28c40e695f1e227a200d1646

I don't know if my code is the most elegant, but I've had to solve this many times for research purposes. If you modify the

sub_nums

variable you can restrict what numbers are used in the partition.

def make_partitions(number):
    out = []
    tmp = []
    sub_nums = range(1,number+1)
    for num in sub_nums:
        if num<=number:
            tmp.append([num])
        for elm in tmp:
            sum_elm = sum(elm)
            if sum_elm == number:
                out.append(elm)
            else:
                for num in sub_nums:
                    if sum_elm + num <= number:
                         L = [i for i in elm]
                         L.append(num)
                         tmp.append(L)
    return out

F(x,n) = \union_(i>=n) { {i}U g| g in F(x-i,i) }

Just implement this recursion. F(x,n) is the set of all sets that sum to x and their elements are greater than or equal to n.