python - 如何按括号顺序将字符串分成块？
问题描述
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。然后该函数搜索左括号和右括号，并用以下格式的“块”替换其中的表达式：（[i$j]其中 i 是左括号的索引，j - 右括号的索引）。如果一个完整的块中有多个块[m$n]，则程序应该只替换字符串 fromm到nwith中的字符[m$n]。最后，该函数返回keypairs字典，其中键应该是块，值应该是从初始字符串中剪切的实际字符串，例如{'23$28': 'string within 23 and 28 characters'}. 所有剩余的符号（括号外的）都应以相同的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$55': '9/7', '11:0': '', '31:0': '', '39:44': '+8+9+', '50:51': '-'}
基本上，当一个块中有不同的独立块时，程序开始梳理它们，而不是只留下一个外部块。我一直试图了解其背后的原因，但每次我尝试更改程序时，问题总是保持不变。我将不胜感激，在此先感谢。
代码：
def findall(sstr, substr):
gen = sstr.find(substr)
while gen != -1:
yield gen
gen = sstr.find(substr, gen + 1)
def findclosest(l: list, el: list): # find closest string from L to string from EL
j = el[ 1 ]
minimum = j
min_index = 0
for i in range(len(l)):
if l[ i ][ 0 ] - j < minimum:
minimum = l[ i ][ 0 ] - j
min_index = l[ i ][ 0 ]
return min_index
def breakInChunks(temp_s): # main
list_of_additions = [ ]
list_of_opened = list(findall(temp_s, '('))
list_of_closed = list(findall(temp_s, ')'))
if sum(list_of_opened) < sum(list_of_closed) and len(list_of_opened) == len(
list_of_closed):
n = 0
# <WHILE>
while len(
list_of_closed) != 0: # read strings-expressions from the most inner ones to the most outer ones
minimum = list_of_closed[ len(list_of_closed) - 1 ]
j = list_of_closed.pop(0)
for i in range(len(list_of_opened)): # find the closest opening bracket to the most inner closing one
diff = j - list_of_opened[ i ]
if diff > 0:
if diff <= minimum:
pop_index = i
minimum = j - list_of_opened[ i ]
else:
break
starting_index = list_of_opened.pop(pop_index)
# start filling KEYPAIRS
if len(keypairs) == 0: # if KEYPAIRS is empty
keypairs[ f'{starting_index}${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')
解决方案
使用堆栈在嵌套括号的每一级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个级别的累积表达式字符串。当您遇到左括号时添加一个级别。当遇到右括号时，弹出当前级别并将其添加到结果中。此时，替换标记被添加到前一个级别的表达式（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'}
推荐阅读
- python - 如何在 Mac OS X 上正确安装 PyGObjct？
- php - PHP中未定义的变量在url中传递输入名称
- reactjs - 无法在提供程序文件中获取 redux 状态
- sql - SQL中的聚合——如何加速查询
- python - SAWarning：对于延迟加载的属性“Similarity.tag1”，使用 uselist=False 返回多行
- r - 格式未应用于数据框
- google-app-engine - 在 GCP 上用于数据管道的最佳服务是什么
- javascript - 我想在单击正方形时显示一个列表，如何在 javascript 中执行此操作？
- typescript - 为什么打字稿将特定字符串视为类属性中的字符串？
- python - 在神经风格转移代码中获取“NaN”