在数字列表上应用算术运算而无需在python中重复

问题描述

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).对于2/1*3=6.0，2*3/1=6.0，3/1*2=6.0，3*2/1=6.0来说，都被认为是等效的，因为我们使用相同的数字而不管操作如何，并且没有使用10.我希望该功能具有足够的通用性，以便可以使用它从那里开始1到9的数字.感谢您的帮助.我尝试使用itertools和permutation获取所有可能组合的列表，但这似乎不必要，并产生了问题2/1*3=6.0，2*3/1=6.0，3/1*2=6.0，3*2/1=6.0包含在列表中，很难过滤掉

We've got the following python list: [1,2,3,10] I would like to accomplish the following: Create a function that takes in the list and figures out from the list of arithmetic operations: ['+', '-', '/','*'] which combinations give us 6 as the answer. We don't want repetition so we don't want 2*3 and 3*2 in our solution. We do want to list the numbers we haven't used so that's (1 and 10 here). Same for 2/1*3=6.0, 2*3/1=6.0, 3/1*2=6.0, 3*2/1=6.0 are all considered equivalent since we use the same numbers regardless of operations and have not used 10. I want the function to be general enough so that I can use it from there for numbers from 1 to 9.Your help is appreciated. I tried using itertools and permutation to get a list of all possible combinations but this seems unnecessary and produces the problem that 2/1*3=6.0, 2*3/1=6.0, 3/1*2=6.0, 3*2/1=6.0 are included in the list and this is difficult to filter out.

使用itertools的示例:

Example where I got using itertools:

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)

推荐答案

这里是可能的解决方案.我们需要保留已用数字的排序元组，以避免2 * 3和3 * 2之类的重复项.

Here is possible solution. We need to keep a sorted tuple of used numbers to avoid duplicates like 2*3 and 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')

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