python - 如何按括号顺序将字符串分成 block ？

Question

程序描述:我正在尝试创建一个函数 breakInChunks()，它接受一个参数 temp_s: string，其中 temp_s 是一个数学表达式，例如1+(3-e^(x-6))-8+(99-4)*10。该函数然后搜索左括号和右括号，并用以下格式的“ block ”替换其中的表达式:[i$j](其中 i 是左括号的索引，j -最后一个)。如果在一整 block [m$n] 中有多个 block ，程序应该只将字符串中从 m 到 n 的字符替换为[m$n]。最后函数返回 keypairs 字典，其中键应该是 block ，值应该是从初始字符串中剪切出来的实际字符串，例如{'23$28': '23 和 28 个字符以内的字符串'}。所有剩余的符号(括号外的)应该以相同的 chunk: string 方式最后附加到字典中。

breakInChunks() 输入: (7+x+8*(9+10(11+12)+14))-(2*(34))

breakInChunks() 输出: {'12$18': '11+12', '7$22': '9+10[12$18]+14', '0$23 ': '7+x+8*[7$22]', '28$31': '34', '25$32': '2*[28$31]', '24:25': '-'}

问题:当尝试读取更复杂的字符串时，我开始得到非常奇怪的结果。例如:

Input: (7+y+(66+7)+(32+(78*19-(32-0)))+(32-9))+8+9+(9-10)-(9/7)-10
Output: {'5$10': '66+7', '23$28': '32-0', '16$29': '78*19-[23$28', '12$30': 
'32+[16$29]))+8+9+', '32$37': '32-9', '0$38': '7+y+(66+7)+(32+[16$29][23$28]]37]])+8',
'44$49': '9-10', '51$55': '9/7', '11:0': '', '31:0': '', '39:44': '+8+9+', '50:51': '-'}

基本上，当一个 block 中有不同的独立 block 时，程序会开始组合它们，而不是只留下一个外部 block 。我一直试图了解其背后的原因，但每次我尝试更改程序时，问题总是保持不变。我将不胜感激任何帮助，在此先感谢。

代码:

def findall(sstr, substr):
    gen = sstr.find(substr)
    while gen != -1:
        yield gen
        gen = sstr.find(substr, gen + 1)


def findclosest(l: list, el: list):  # find closest string from L to string from EL
    j = el[ 1 ]
    minimum = j
    min_index = 0
    for i in range(len(l)):
        if l[ i ][ 0 ] - j < minimum:
            minimum = l[ i ][ 0 ] - j
            min_index = l[ i ][ 0 ]
    return min_index


def breakInChunks(temp_s):  # main
    list_of_additions = [ ]
    list_of_opened = list(findall(temp_s, '('))
    list_of_closed = list(findall(temp_s, ')'))
    if sum(list_of_opened) < sum(list_of_closed) and len(list_of_opened) == len(
            list_of_closed):
        n = 0
        # <WHILE>
        while len(
                list_of_closed) != 0:  # read strings-expressions from the most inner ones to the most outer ones
            minimum = list_of_closed[ len(list_of_closed) - 1 ]
            j = list_of_closed.pop(0)
            for i in range(len(list_of_opened)):  # find the closest opening bracket to the most inner closing one
                diff = j - list_of_opened[ i ]
                if diff > 0:
                    if diff <= minimum:
                        pop_index = i
                        minimum = j - list_of_opened[ i ]
                else:
                    break
            starting_index = list_of_opened.pop(pop_index)
            # start filling KEYPAIRS
            if len(keypairs) == 0:  # if KEYPAIRS is empty
                keypairs[ f'{starting_index}${j}' ] = temp_s[ starting_index + 1:j ]
            else:  # if KEYPAIRS has at least one key-value pair
                keys = [ key.split('$') for key in
                         keypairs.keys() ]  # reading and unpacking key-value pairs (reading indecies)
                innerSeq = temp_s
                min_index_i = None
                min_index_j = None
                prevExtracted_i = 0
                prevExtracted_j = 0
                for p in range(len(keys) - 1, -1, -1):
                    k = keys[ p ]
                    extracted_i, extracted_j = int(k[ 0 ]), int(k[ 1 ])
                    if starting_index < extracted_i:  # if the chunk we are checking contains another one, we are checking if it's in fact the closest one to the chunk we are checking
                        if (
                                extracted_i < prevExtracted_i and prevExtracted_j < extracted_j) or prevExtracted_i == 0:
                            min_index_i = extracted_i
                            min_index_j = extracted_j
                            if prevExtracted_i == 0:
                                if extracted_i > int(keys[ p - 1 ][ 0 ]) and extracted_j < int(keys[ p - 1 ][ 1 ]):
                                    pass
                                else:
                                    innerSeq = innerSeq[
                                               :extracted_i ] + f'[{extracted_i}${extracted_j}]' + innerSeq[
                                                                                                   extracted_j + 1: ]
                        else:
                            if min_index_i is not None:
                                innerSeq = innerSeq[ :min_index_i ] + f'[{min_index_i}${min_index_j}]' + innerSeq[
                                                                                                         min_index_j + 1: ]
                                min_index_i = None
                                min_index_j = None
                            else:
                                innerSeq = innerSeq[
                                           :prevExtracted_i ] + f'[{prevExtracted_i}${prevExtracted_j}]' + innerSeq[
                                                                                                           prevExtracted_j + 1: ]
    
                        prevExtracted_i = extracted_i
                        prevExtracted_j = extracted_j
                        n += 1
    
                keypairs[ f'{starting_index}${j}' ] = innerSeq[ starting_index + 1:j ]
    
        # </WHILE>
    
        # checking if there are any strings outside parentheses left
        temp = [ [ int(key.split('$')[ 0 ]), int(key.split('$')[ 1 ]) ] for key in sorted(keypairs.keys(),
                                                                                          key=lambda el: int(
                                                                                              el.split('$')[
                                                                                                  1 ])) ]  # sort from the most inner to the most outer
        for i in range(len(temp) - 1):
            if temp[ i ][ 1 ] < temp[ i + 1 ][
                0 ]:  # if there is a gap between parentheses
                # find the closest difference in order to find actual string outside chunks with the help of findclosest()
                # add new chunk to LIST_OF_ADDITIONS
                list_of_additions.append([ temp[ i ][ 1 ] + 1, findclosest(temp[ i + 1: ], temp[ i ]) ])
    
        if len(list_of_additions) > 0:  # if something is inside LIST_OF_ADDITIONS
            # add remaining strings to KEYPAIRS
            for addition in list_of_additions:
                keypairs[ f'{addition[ 0 ]}:{addition[ 1 ]}' ] = self.s[ addition[ 0 ]:addition[ 1 ] ]
    
        return keypairs  # return KEYPAIRS
    else:
        raise RuntimeError(f'Amount of closing and opening brackets does not match')

Answer 1

使用堆栈在嵌套括号的每一层累积子表达式是解决此问题的常用方法。存储左括号的位置和每一层累积的表达式字符串。遇到左括号时添加一个级别。遇到右括号时，弹出当前级别并将其添加到结果中。在这一点上，替换标记被添加到上一级别的表达式(成为当前级别)。

def parGroups(S):
    result = dict()
    stack  = [[0,""]]*2           # parenthesis position, expression
    for i,c in enumerate(S+")"):  # extra ")" to force out main expression
        if c=="(":                
            stack.append([i,""])  # stack up new group
            continue
        if c==")":
            start,expr      = stack.pop(-1)    # unstack current group
            c               = f"[{start}${i}]" # token
            result[c[1:-1]] = expr             # build result
        stack[-1][-1] += c        # accumulate expression in current group
    return result

输出:

S = "(7+y+(66+7)+(32+(78*19-(32-0)))+(32-9))+8+9+(9-10)-(9/7)-10"
print(parGroups(S))

{'5$10' : '66+7',
 '23$28': '32-0',
 '16$29': '78*19-[23$28]',
 '12$30': '32+[16$29]',
 '32$37': '32-9',
 '0$38' : '7+y+[5$10]+[12$30]+[32$37]',
 '44$49': '9-10',
 '51$55': '9/7',
 '0$59' : '[0$38]+8+9+[44$49]-[51$55]-10'}