Python eval表達式用於算術運算符的所有置換

Question

給定string = '1*7/5-3'

我有表達式來評估像eval('1*7/5-3')這樣的字符串

代碼：

import __future__
string = '1*7/5-3'
print eval(compile(string, '<string>', 'eval', __future__.division.compiler_flag))

我想評估所有排列

example 
        eval('((1*7)/5)-3') 
        eval('1*((7/5)-3)')
        and so on

Answer 1

我不應該對其進行編輯以消除“過分的”括號。 實際上，它們是必要的。 我正在恢復原始代碼。

想法是依次將每個運算符視為主要操作- 二進制表達式樹的根。 這將字符串分成兩部分，然后我們遞歸地應用該過程。

 def parenthesize(string):
    '''
    Return a list of all ways to completely parenthesize operator/operand string
    '''
    operators = ['+','-','*','/']
    depth = len([s for s in string if s in operators]) 
    if depth == 0:
        return [string]
    if depth== 1:
        return ['('+ string + ')']
    answer = []
    for index, symbol in enumerate(string):
        if symbol in operators:
            left = string[:index]
            right = string[(index+1):]
            strings = ['(' + lt + ')' + symbol +'(' + rt + ')' 
                           for lt in parenthesize(left) 
                           for rt in parenthesize(right) ]
            answer.extend(strings)
    return answer    

 string='4+7/5-3'
 for t in parenthesize(string):print(t, eval(t))

此打印

(4)+((7)/((5-3))) 7.5
(4)+(((7/5))-(3)) 2.4
((4+7))/((5-3)) 5.5
((4)+((7/5)))-(3) 2.4000000000000004
(((4+7))/(5))-(3) -0.7999999999999998

順便說一句，這是針對Euler Project問題93的嗎？