如何打印与原始列表具有相同总和的列表的所有可能组合？

Question

我正在尝试打印列表的所有可能组合，但前提是这些组合加起来是一个数字。

lst = [0, 1, 2] #The goal is print all combinations that sum up to 3

import itertools

def fun(lst, total):
    sum = 0
    for element in lst:
        sum += element
        all_possible_combinations = set(itertools.product(lst, repeat=2)) # This prints all possible combinations of the elements in the list with length of 2
        for items in all_possible_combinations:
            a = 0
            for i in items:
                a += i
            if a == total:
                x = items
                print(x)
    print('These are all combinations:', all_possible_combinations)

fun([0, 1, 2], 2)

此程序打印总和为 3 的列表，但它多次打印这些列表。

>>>>>>>>
(2, 0)
(1, 1)
(0, 2)  #I was expecting the programme to stop here.
(2, 0)
(1, 1)
(0, 2)
(2, 0)
(1, 1)
(0, 2)
These are all combinations: {(0, 1), (1, 2), (0, 0), (2, 1), (2, 0), (1, 1), (2, 2), (1, 0), (0, 2)}
>>>>>>>>

我认为这是因为 for element in lst: 循环，所以我尝试在此循环之外打印

import itertools

def fun(lst, total):
    sum = 0
    for element in lst:
        sum += element
        all_possible_combinations = set(itertools.product(lst, repeat=2)) # This prints all possible combinations of the elements in the list with length of 2
        for items in all_possible_combinations:
            a = 0
            for i in items:
                a += i
            if a == total:
                x = items
    print(x)
    print('These are all combinations:', all_possible_combinations)

fun([0, 1, 2], 2)

这个程序只返回一个列表

>>>>>>>>>
(0, 2)
>>>>>>>>>

我该如何纠正这个问题？我期待最终结果是

>>>>>>>>>
(2, 0)
(1, 1)
(0, 2)
>>>>>>>>>

Answer 1

另一种可能的解决方案：

import itertools

a = [0, 1, 2]

result = [b for b in itertools.product(a, repeat=2) if sum(b)<=sum(a)]

为了得到你想要的，代码如下：

import itertools

a = [0, 1, 2]

result = [b for b in itertools.product(a, repeat=2) if sum(b)<sum(a) and sum(b)>1]
print result

Answer 2

让列表为l = [0, 1, 2] 和sum_list = sum(l) 编辑：抱歉最初读错了

那么你可以这样做：

import itertools as it

for i in range(len(l)):
    ans = list(filter(lambda x: sum(x)==sum_list, list(it.combinations(l, i)))

print ans

Answer 3

您实际上只需要对元素进行循环，因为您的 all_possible_combinations 实际上确实在列表中具有 2 元组的所有组合。当您循环遍历 lst 时，它会重复列表的长度，因此会在输出中重复。

修订后的第一个版本：

import itertools

def fun(lst, total):
        all_possible_combinations = set(itertools.product(lst, repeat=2)) # This prints all possible combinations of the elements in the list with length of 2
        for items in all_possible_combinations:
            a = 0
            for i in items:
                a += i
            if a == total:
                x = items
                print(x)
        print('These are all combinations:', all_possible_combinations)

fun([0, 1, 2], 2)

输出：

>>> fun([0, 1, 2], 2)
(1, 1)
(2, 0)
(0, 2)
('These are all combinations:', set([(0, 1), (1, 2), (0, 0), (2, 1), (1, 1), (2, 0), (2, 2), (1, 0), (0, 2)]))
>>> 

Answer 4

如果我正确理解您的代码，您只需继续覆盖这些值，只留下 x 的最后一个值。创建一个数组并将项目附加到这个数组应该会显示所有结果。

import itertools

def fun(lst, total):
    sum = 0
    for element in lst:
        sum += element
        x = []
        all_possible_combinations = set(itertools.product(lst, repeat=2)) # This prints all possible combinations of the element in the list with length of 2
        for items in all_possible_combinations:
            a = 0
            for i in items:
                a += i
            if a == total:
                x.append(items)
    print(x)
    print('These are all combinations:', all_possible_combinations)

fun([0, 1, 2], 2)