查找不固定长度的数字的所有可能排列以达到给定的总和或乘积

Question

使用普通的 Python 或任何 Python 库，您将如何使用 go 查找列表l中等于给定值val的所有可能的元素组合、加法或乘法？ 假设列表的长度并不总是相同，假设列表中的每个元素在每个组合中只能使用一次，并且假设没有使用任何括号。

例如：

• 我们得到一个数字列表： l = [1,2,3,4]
• 我们得到一个等于值组合的值： val = 6
• output 将包括以下内容：
  • [2,4] ，因为2+4=6
  • 和[4,2] ，因为4+2=6
  • 和[1,3,2] ，因为1+3+2=6
  • 和[1,2,4] ，因为1*2+4=6
  • 等等

---

我试过使用itertools.permutations ：

>>> from itertools import permutations
>>> l = [1,2,3,4]
>>> val = 6
>>> correct_combos = []

>>> for i in range(1, len(l)+1):
...   for p in permutations(l, r=i):
...     if sum(p) == val:
...       correct_combos.append(p)

我只能实现代码来测试列表中所有元素组合的总和。

>>> print(correct_combos)
[(2, 4), (4, 2)]

我一直坚持使用加法、减法和乘法的组合来查找列表中元素的排列。

Answer 1

我不知道这个算法是否有效，但它工作正常：

from itertools import permutations, product
l = [1,2,3,4]
val = 6
operator = ['+', '-', '*']
correct_combos=[]
for r in range(1, len(l)+1):
    for item in permutations(l,r):
        for unit in product(operator, repeat=r-1):
            res=""
            for idx in range(0,r-1):
                res+=str(item[idx])+unit[idx]
            res+=str(item[-1])
            if(val==eval(res)):
                if item not in correct_combos:
                    correct_combos.append(item)
print(correct_combos)

Output

[(2, 3), (2, 4), (3, 2), (4, 2), (1, 2, 3), (1, 2, 4), (1, 3, 2), (1, 4, 2), (2, 1, 3), (2, 1, 4), (2, 3, 1), (2, 4, 1), (3, 1, 2), (3, 1, 4), (3, 2, 1), (3, 4, 1), (4, 1, 2), (4, 1, 3), (4, 2, 1), (4, 3, 1), (1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1), (4, 1, 2, 3), (4, 1, 3, 2), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1)]

Answer 2

您可以使用递归生成器 function：

from operator import add, sub, mul
l = [1,2,3,4]
val = 6
def combos(c = [], r_exp = None):
   if r_exp == val:
      yield c
   else:
      for i in filter(lambda x:x not in c, l):
         if not c:
            yield from combos(c=[i], r_exp = i)
         else:
            for s, f in [['+', add], ['-', sub], ['*', mul]]:
               yield from combos(c=c+[s, i], r_exp = f(r_exp, i))

print([''.join(map(str, i)) for i in combos()])

Output：

['1+2+3', '1-2+3+4', '1-2+4+3', '1*2*3', '1*2+4', '1+3+2', '1+3-2+4', '1+3+4-2', '1*3*2', '1+4-2+3', '1+4+3-2', '1*4+2', '1*4-2*3', '2+1+3', '2*1*3', '2*1+4', '2+3+1', '2*3', '2+4', '2*4+1-3', '2*4-3+1', '3+1+2', '3+1-2+4', '3+1+4-2', '3-1+4', '3-1*4-2', '3*1*2', '3+2+1', '3-2+1+4', '3-2+4+1', '3*2', '3+4+1-2', '3+4-1', '3+4-2+1', '4+1-2+3', '4+1+3-2', '4-1*2', '4-1+3', '4*1+2', '4*1-2*3', '4+2', '4-2+1+3', '4-2*1*3', '4-2+3+1', '4-2*3', '4*2+1-3', '4*2-3+1', '4+3+1-2', '4+3-1', '4+3-2+1']

为了得到元组结果，只需要做一个小的改变：

for f in [add, sub, mull]:
   yield from combos(c=c+[i], r_exp = f(r_exp, i))
...
print(list(map(tuple, combos())))

Output：

[(1, 2, 3), (1, 2, 3, 4), (1, 2, 4, 3), (1, 2, 3), (1, 2, 4), (1, 3, 2), (1, 3, 2, 4), (1, 3, 4, 2), (1, 3, 2), (1, 4, 2, 3), (1, 4, 3, 2), (1, 4, 2), (1, 4, 2, 3), (2, 1, 3), (2, 1, 3), (2, 1, 4), (2, 3, 1), (2, 3), (2, 4), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 2), (3, 1, 2, 4), (3, 1, 4, 2), (3, 1, 4), (3, 1, 4, 2), (3, 1, 2), (3, 2, 1), (3, 2, 1, 4), (3, 2, 4, 1), (3, 2), (3, 4, 1, 2), (3, 4, 1), (3, 4, 2, 1), (4, 1, 2, 3), (4, 1, 3, 2), (4, 1, 2), (4, 1, 3), (4, 1, 2), (4, 1, 2, 3), (4, 2), (4, 2, 1, 3), (4, 2, 1, 3), (4, 2, 3, 1), (4, 2, 3), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 1), (4, 3, 2, 1)]

Answer 3

非递归解决方案：

from itertools import permutations, product, accumulate
from collections import deque

from operator import add, sub, mul

l = [1, 2, 3, 4]
val = 6
correct_combos = []


def is_correct(p, val, ops=[add, sub, mul]):
    if len(p) == 1:
        return p[0] == val

    for op in product(ops, repeat=len(p) - 1):
        iop = iter(op)
        l = deque(accumulate(p, lambda a, b: next(iop)(a, b)), maxlen=1)[0]
        if l == val:
            return True

    return False


for i in range(1, len(l) + 1):
    for p in permutations(l, r=i):
        if is_correct(p, val):
            correct_combos.append(p)

print(correct_combos)

印刷：

[(2, 3), (2, 4), (3, 2), (4, 2), (1, 2, 3), (1, 2, 4), (1, 3, 2), (1, 4, 2), (2, 1, 3), (2, 1, 4), (2, 3, 1), (2, 4, 1), (3, 1, 2), (3, 1, 4), (3, 2, 1), (3, 4, 1), (4, 1, 2), (4, 1, 3), (4, 2, 1), (4, 2, 3), (4, 3, 1), (1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1), (4, 1, 2, 3), (4, 1, 3, 2), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1)]

Answer 4

from itertools import chain, combinations


def subSet(l):
    return chain.from_iterable(combinations(l, r) for r in range(len(l) + 1))


def whatWeWant(l, val):
    x = 0
    for i in l:
        x += i
    if x == val:
        return True

    return False


l = [1, 2, 3, 4]
val = 6

allSubSets = list(subSet(l))

ans = []

for subSet in allSubSets:
    if whatWeWant(subSet, val):
        ans.append(subSet)

使用此代码，您可以找到所需的内容，但您必须明确说明可能的组合（在 whatWeWant 方法中）。