I have an array which contains elements that are boolean values (True , False) and strings which represent logical operators ('and', 'or', 'not').

For example:

array = [True, 'and', False]

I'd like to take the elements of this array in order and turn them into an actual expression I can assign a variable to. Meaning the following:

expression = True and False

So that I can then print the boolean result of said expression

print("The result of the expression is = ", expression)

Which should return

The result of the expression is = False

How could I achieve this? My vague idea is to iterate through the array, but I'm having a hard time figuring out how to turn a string ('and') to an operator (and).

TIA.

CodePudding user response：

Here is a simple 3-step iterative algorithm to solve this without relying on eval, and assuming and has higher operator precedence than or. If there is no operator precedence, you could combine steps 2 and 3 into a single iteration.

1. First, iterate through the array, getting rid of all not. For instance, convert all not True to False and vice versa. Now we are left with only and, True, or, and False.
2. Then, iterate through the array evaluating all and expressions.
3. Finally, return True if there are any True left in the array, otherwise return False.

Here's an implementation. It's not the most efficient, since deleting in the middle of a list takes O(n), but it shows how the algorithm works:

from copy import copy

def remove_nots(lst):
    i = 0
    while i < len(lst):
        # if we encounter a `not`, remove it, and negate the following element
        if lst[i] == 'not':
            del lst[i]
            # to negate a `not`, we remove it
            if lst[i] == 'not':
                del lst[i]
            # otherwise, it must be a boolean, so we can use native not
            else:
                lst[i] = not lst[i]
        else:
            i  = 1

def resolve_and(lst):
    i = 0
    while i   2 < len(lst):
        # if we encounter an `and` expression, replace it with the result
        if lst[i   1] == 'and':
            result = lst[i] and lst[i   2]
            del lst[i 2]
            del lst[i 1]
            lst[i] = result
        else:
            i  = 1
    

def resolve(lst):
    # create a copy to work on, since we'll be modifying the list
    lst_copy = copy(lst)
    remove_nots(lst_copy)
    resolve_and(lst_copy)
    return True in lst_copy

print(resolve(lst = [False, 'or', True, 'and', 'not', 'not', 'not', False])) # True

CodePudding user response：

This should work:

array = [True, 'and', False]
expression = eval(' '.join([str(i) for i in array]))
print("The result of the expression is = ", expression)

Output:

The result of the expression is =  False