Question

我们有以下python列表：[1,2,3,10]我要完成以下操作：创建一个接受列表并从算术运算列表中找出数字的函数：['+', '-', '/','*']哪些组合给我们6作为答案。我们不想重复，所以我们不想在解决方案中使用2*3和3*2。我们确实要列出未使用的数字，所以是（这里是1和10）。 2/1*3=6.0，2*3/1=6.0，3/1*2=6.0，3*2/1=6.0都被认为是等效的，因为我们使用相同的数字而不管操作是什么，也没有使用10。我希望函数成为足够通用，因此我可以从那里使用它来表示1到9之间的数字。感谢您的帮助。我尝试使用itertools和permutation来获取所有可能组合的列表，但这似乎不必要，并且产生了2/1*3=6.0，2*3/1=6.0，3/1*2=6.0，3*2/1=6.0包含在内的问题。列表，很难过滤掉。

使用itertools的示例：


from itertools import chain, permutations

def powerset(iterable):
  xs = list(iterable)
  return chain.from_iterable(permutations(xs,n) for n in range(len(xs)+1) )

lst_expr = []
for operands in map(list, powerset(['1','2','3','10'])):
    n = len(operands)
    #print operands
    if n > 1:
        all_operators = map(list, permutations(['+','-','*','/'],n-1))
        #print all_operators, operands
        for operators in all_operators:
            exp = operands[0]
            i = 1
            for operator in operators:
                exp += operator + operands[i]
                i += 1

            lst_expr += [exp]

lst_stages=[]

for equation in lst_expr:
    if eval(equation) == 6:
        lst_stages.append(equation)
        eq = str(equation) + '=' + str(eval(equation))
        print(eq)

Answer 1

我喜欢sanyash的解决方案，但我认为通过使用操作数的组合而不是进行置换，可以更优雅地完成，直到找到具有正确值的第一个为止：

from itertools import chain, permutations, combinations


def powerset(iterable):
    xs = list(iterable)
    return chain.from_iterable(combinations(xs, n) for n in range(len(xs) + 1))

def get_first_perm_with_value(operands, operators, expected_value):
    if len(operands) == 0 or len(operands) == 0:
        return []

    all_operators = list(map(list, permutations(operators, len(operands) - 1)))
    all_operands = list(map(list, permutations(operands, len(operands))))

    for operator in all_operators:
        for operand in all_operands:
            result = [""] * (len(operand) + len(operator))
            result[::2] = operand
            result[1::2] = operator
            eq = ''.join(result)
            if int(eval(eq)) == expected_value:
                return [(f'{eq}={expected_value}', operands)]
    return []


lst_expr = []
for operands in map(list, powerset(['1', '2', '3', '10'])):
    lst_expr += get_first_perm_with_value(operands, ['+','-','*','/'], 6)

print(lst_expr)

返回：

[('2*3=6', ['2', '3']), ('2*3/1=6', ['1', '2', '3']), ('1+10/2=6', ['1', '2', '10']), ('2*10/3=6', ['2', '3', '10']), ('2*3+1/10=6', ['1', '2', '3', '10'])]

Answer 2

这是可能的解决方案。我们需要保留已用数字的排序元组，以避免重复出现2 * 3和3 * 2。

from itertools import chain, permutations

def powerset(iterable):
  xs = list(iterable)
  return chain.from_iterable(permutations(xs,n) for n in range(len(xs)+1) )

lst_expr = []
for operands in map(list, powerset(['1','2','3','10'])):
    n = len(operands)
    #print operands
    if n > 1:
        all_operators = map(list, permutations(['+','-','*','/'],n-1))
        #print all_operators, operands
        for operators in all_operators:
            exp = operands[0]
            numbers = (operands[0],)
            i = 1
            for operator in operators:
                exp += operator + operands[i]
                numbers += (operands[i],)
                i += 1

            lst_expr += [{'exp': exp, 'numbers': tuple(sorted(numbers))}]

lst_stages=[]
numbers_sets = set()

for item in lst_expr:
    equation = item['exp']
    numbers = item['numbers']
    if numbers not in numbers_sets and eval(equation) == 6:
        lst_stages.append(equation)
        eq = str(equation) + '=' + str(eval(equation))
        print(eq, numbers)
        numbers_sets.add(numbers)

输出：

2*3=6 ('2', '3')
1+10/2=6.0 ('1', '10', '2')
2/1*3=6.0 ('1', '2', '3')

在数字列表上应用算术运算而无需在python中重复

2 个答案: