解析嵌套表达式以检索每个内部函数

Question

假设我有一个表达式，如下所示：

expression = "LEN(Replace(Lower(UPPER([ProductName]+[ProductName])), 'chaichai', 'chai'))"

所需输出：

['UPPER([ProductName]+[ProductName])','Lower(UPPER([ProductName]+[ProductName]))','Replace(Lower(UPPER([ProductName]+[ProductName])),'chaichai','chai')','LEN(Replace(Lower(UPPER([ProductName]+[ProductName])),'chaichai','chai'))']

我已尝试以下代码，但未获得所需结果：

exp_splits = expression.strip(')').split('(')
for i_enm, i in enumerate(range(len(exp_splits)-2, -1, -1), start=1):
     result.append(f"{'('.join(exp_splits[i:])}{')'*i_enm}")
print(result)

我的代码的输出：

["UPPER([ProductName]+[ProductName])),'chaichai','chai')", "Lower(UPPER([ProductName]+[ProductName])),'chaichai','chai'))", "Replace(Lower(UPPER([ProductName]+[ProductName])),'chaichai','chai')))", "LEN(Replace(Lower(UPPER([ProductName]+[ProductName])),'chaichai','chai'))))"]

Answer 1

import re

e = "LEN(Replace(Lower(UPPER([ProductName]+[ProductName])),'chaichai','chai'))"

print ([e[i:j+1] for i in range(len(e)) for j in range(len(e)) if e[i:j+1].count('(') == e[i:j+1].count(')') != 0 and (e[i-1] == '(' or i == 0) and e[j] == ')'])

输出：

["LEN(Replace(Lower(UPPER([ProductName]+[ProductName])),'chaichai','chai'))", "Replace(Lower(UPPER([ProductName]+[ProductName])),'chaichai','chai')", 'Lower(UPPER([ProductName]+[ProductName]))', 'UPPER([ProductName]+[ProductName])']

展开的版本：

for i in range(len(e)):
    for j in range(len(e)):
        #Check for same number of opening/closing parenthesis
        if e[i:j+1].count('(') == e[i:j+1].count(')') != 0:
            #Check that (first char is preceded by an opening parenthesis OR that first char is the beginning of e) AND last char is a parenthesis
            if (e[i-1] == '(' or i == 0) and e[j] == ')':
                print (e[i:j+1])

输出：

LEN(Replace(Lower(UPPER([ProductName]+[ProductName])),'chaichai','chai'))
Replace(Lower(UPPER([ProductName]+[ProductName])),'chaichai','chai')
Lower(UPPER([ProductName]+[ProductName]))
UPPER([ProductName]+[ProductName])