如何使 Python 代码更高效？

Question

Newbie here.I have this code, that displays all possible sums to reach a certain number. But its complexity is too high and it takes too longe when numbers are too high. How can I refactor this to something more simple?

import itertools

def combos(n):

    result = []
    for i in range(n,0,-1):
        for seq in itertools.combinations_with_replacement(range(1,n+1), i):
            if sum(seq) == n:
                seq = list(seq)
                result.append(seq)
    return(result)

combos(4)

output:

[[1,1,1,1],[1,1,2],[1,3],[2,2],[4]]

Answer 1

The recursive version could be like this:

def combinations_max_sum(sum_max, i_max=None):

    if sum_max == 0:
        return [()]

    if not i_max or i_max > sum_max:
        i_max = sum_max

    combinations = [(i, *right_part)
                    for i in range(i_max, 0, -1)
                    for right_part in combinations_max_sum(sum_max-i, i_max=i)]

    return combinations

Test:

print(combinations_max_sum(4)) # [(4,), (3, 1), (2, 2), (2, 1, 1), (1, 1, 1, 1)]
print(combinations_max_sum(4, i_max=1)) # [(1, 1, 1, 1)]
print(combinations_max_sum(5))
# [(5,), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1)]

The idea to decompose the problem: a set of combinations can be written as a number concatenated with all the combinations for which the sum is n minus that first number.

A simpler code without dealing with repetition could be like this:

def combinations_with_repetition(n):

    if n == 0:
        return [()]

    combinations = [(i, *right_part)  # concatenation
                    for i in range(1, n+1)  # for all possible first number 
                    for right_part in combinations_with_repetition(n-i)]
                    # ^ all possible combinations
                    #   given the first number i

    return combinations

which gives:

combinations_with_repetition(3)
# [(1, 2), (1, 1, 1), (2, 1), (3,)]

(1, 2) and (2, 1) are similar, to prevent this a i_max argument is added (see the first function). The idea here is to go always in descending order. A number on the right is always equal or smaller than a number on the left. This maximum number is passed as an argument, and the loop start with it instead of the total sum asked.