Python：以条件递归创建一个整数列表的列表

Question

I want to create all possible distributions of n items. This refers to the commonly known pigeonhole principle.

The following values are the result of Microsoft Excel:

get_distributions(list, number_of_items_to_distribute)
get_distributions([], 1) = [[1]]
get_distributions([], 2) = [[1, 1], [2]]
get_distributions([], 3) = [[1, 1, 1], [1, 2], [2, 1], [3]]
get_distributions([], 4) = [[1, 1, 1, 1], [1, 1, 2], [1, 2, 1], [1, 3], [2, 1, 1], [2, 2], [3, 1], [4]]

I already have some code, but there are some issues with deleting the temporary lists.

all_distributions = []

def get_distributions(distribution, items):

    print('call with distribution = ' + str(distribution) + ', items = ' + str(items))
    print('---------------')

    # base case
    if items == 0:
        all_distributions.append(distribution)
        print('end: ' + str(distribution))
        distribution.clear()
        return []

    # recursion
    else:
        for i in range(1, items + 1):
            distribution.append(i)
            get_distributions(distribution, items - i)

With this I get good results printed out after "end: ", but some values like [1, 2] (calling with n = 3) are missing. Additionally to this the values are not appended to my all_distributions.

I'm interested in the way I tried to solve this problem. Is this a good approach or am I absolutely wrong?

Answer 1

The main problem with your code is that the list all_distributions ends up containing many references to the same input list distribution . When you call all_distributions.append(distribution) , the list distribution is not copied into the list all_distributions , but merely a reference to the list is appended. You can fix this by explicitly inserting a copy: all_distributions.append(list(distribution))

A minimal fix to your code is to insert copies, remove distribution.clear() in the base case, and adding distribution.pop() after the recursive call:

all_distributions = []

def get_distributions(distribution, items):
    if items == 0:
        all_distributions.append(list(distribution))
    else:
        for i in range(1, items + 1):
            distribution.append(i)
            get_distributions(distribution, items - i)
            distribution.pop()

get_distributions([], 3)
print(all_distributions)

Outputs: [[1, 1, 1], [1, 2], [2, 1], [3]]

---

A better way is to avoid using distribution.append , and instead using the plus operator on lists, like so:

def get_distributions(distribution, items):
    if items == 0:
        all_distributions.append(distribution)
    else:
        for i in range(1, items + 1):
            d = distribution + [i]
            get_distributions(d, items - i)

The plus operator on lists creates a new list by concatenating the two given lists. In this case, we are concatenating a single element i on the right side of distribution to get a new copy containing the elements in distribution followed by i .

---

Another improvement is to avoid the global variable all_distributions , and instead return the list of distributions:

def get_distributions(distribution, items):
    if items == 0:
        return [distribution]
    else:
        all_distributions = []
        for i in range(1, items + 1):
            d = distribution + [i]
            all_distributions += get_distributions(d, items - i)
        return all_distributions

print(get_distributions([], 4))

Outputs: [[1, 1, 1, 1], [1, 1, 2], [1, 2, 1], [1, 3], [2, 1, 1], [2, 2], [3, 1], [4]]