如何在 python 中将括号操作拆分为必须首先评估哪些块的顺序？

Question

我正在尝试创建一个布尔逻辑解析器 - 例如 A & B | ！C。 我为此制作了一个解析器，最终以 pygame 显示的维恩图结束。 （代码如下）。 但是，这不支持括号。

我想要做的是按照需要评估的顺序记录每个括号术语的开始和结束，如下所示： ((a & b) | c) | （c 和！b） = [[1, 7], [0, 12], [16, 23]]

我该如何做到这一点（没有模块）？ 我有信心一旦获得这些信息我就可以评估方程。

代码：

def parse(expression, sets, total, classifications):
# First split the expression down into the terms
terms = expression.split(" ")
operators = []
arrays = []
# Replace A, B, C with their corresponding arrays, inverse if necessary.
for index, term in enumerate(terms):
    # Do not change operators but record indices.
    if term == "&" or term == "|":
        arrays.append(term)
        operators.append(index)
    is_not = "!" in term
    term = term.strip("!")
    # If the term contains a '!' character, instead append the complement of that set.
    if term in sets.keys():
        if is_not:
            arrays.append(complement(sets[term], total))
        else:
            arrays.append(sets[term])

# [[1, 2, 3], "&", [3, 4, 5]] for example.
for operator in operators:
    # We know that for each operator, there is a set on either side.
    set_1 = arrays[operator - 1]
    logic = arrays[operator]
    set_2 = arrays[operator + 1]
    # Combine the sets in either way.
    if logic == "&":
        result = intersection(set_1, set_2)
    elif logic == "|":
        result = union(set_1, set_2)
    # Remove the old sets & operator, add the combined set.
    for i in range(3): arrays.pop(operator - 1)
    arrays.insert(operator - 1, result)
    # Account for the size of the array shrinking.
    for index in range(len(operators)):
        operators[index] -= 2

# Parsing done, clean up array.
sections = []
arrays = arrays[0]

# Figure out which sections must be rendered.
for number in arrays:
    sections.append(classifications[number])
return sections

地点：

total = [1, 2, 3, 4, 5, 6, 7, 8]
sets = {
    "a":[1, 2, 3, 5],
    "b":[2, 3, 4, 8],
    "c":[3, 4, 5, 6]
}

# The sets were carefully chosen to ensure each set has 2 numbers corresponding toeach of the 
other sets, 1 number corresponding
#   to all three sets, and 1 number that is unique to that set. There is also 1 number (8) 
that is not in any sets.
# This way, sections can be shader based on whether or not that number is present.
classifications = {
    1:"a",
    2:"ab",
    3:"abc",
    4:"bc",
    5:"ac",
    6:"c",
    7:"none",
    8:"b"
}

Answer 1

您可以实现一个跟踪左括号和右括号位置的函数：

def find_parentheses_spans(expression):
    stack = []
    spans = []
    for index, char in enumerate(expression):
        if char == '(':
            stack.append(index)
        elif char == ')':
            start = stack.pop()  # Get the position of the matching '('
            spans.append([start, index])
    # Optional: sort spans if needed, based on the starting index
    spans.sort(key=lambda x: x[0])
    return spans

expression = "((a & b) | c) | (c & !b)"
spans = find_parentheses_spans(expression)
print(spans)