Python代码报错TypeError: 'int' object is not callable求助

我编写了一个Python高级计算器代码，目标是解析并计算像3(5+3)这种省略乘号的表达式，但运行时抛出TypeError: 'int' object is not callable错误。尝试过str()、map()方法，查阅文档和咨询工具都没能解决问题，相关代码如下：

print ("____________________________")
print ("Advanced calculator selected")
print ("Experimental! May break.")
print ("NOTE: Does not work with variables... yet.")
print ("Key:", "\"()\" mean parenthesis", "\"^\" mean exponent", "\"*\" means multiplication", "\"/\" means division", "\"+\" means addition", "\"-\" means subtraction", sep="\n  ")
option = "3(5+3)" #input("Equation to solve:").strip()
option2 = list(option)
count = -1
c = True
while count < len(option2) - 1:
    if c is True:
        count += 1
        c = False
    elif c is False:
        c = True
    if option2[count] == " ":
        option2.pop(count)
        count -= 1
        continue
    elif option2[count] in [0,1,2,3,4,5,6,7,8,9] and option2[count + 1] == "(":
        option2.insert(count + 1, "*")
        count += 1
    elif option2[count] in ["(", ")", "*", "/", "+", "-"]:
        continue
    elif option2[count] == "^":
        option2.pop(count)
        option2.insert(count, "**")
    elif str(option2[count]).isalpha() is True:
        print ("Option not recognized, restarting")
        counting.Calculator.Advanced()
    else:
        continue
print (option2)
answer = "".join(option2)
print (answer.isascii())
answer = eval(str(answer))
print (answer)
input()

核心问题 #

代码未正确识别数字与括号间的省略乘号，导致最终传给eval()的表达式仍是3(5+3)。在Python中，3(...)会被解析为“调用值为3的函数”，但整数3不是可调用对象，因此抛出TypeError。

问题出在数字判断逻辑：

elif option2[count] in [0,1,2,3,4,5,6,7,8,9] and option2[count + 1] == "(":

option2是字符串拆分后的字符列表，每个元素都是字符串类型（比如"3"），但你判断它是否在整数列表[0,1,...9]中，这个条件永远不成立，因此不会插入*号。

修复步骤 #

1. 修正数字判断逻辑：
   将整数列表改为字符串列表，或直接用isdigit()方法更简洁：

   # 写法1：字符串列表判断
   elif option2[count] in ["0","1","2","3","4","5","6","7","8","9"] and option2[count + 1] == "(":
   # 写法2：更简洁的isdigit()判断
   elif option2[count].isdigit() and option2[count + 1] == "(":
   

2. 优化循环逻辑：
   原代码中c变量的切换会导致跳过部分字符，可能漏处理表达式。可以简化为更直观的遍历逻辑：

   count = 0
   while count < len(option2):
       if option2[count] == " ":
           option2.pop(count)
           continue
       # 增加边界判断，避免索引越界
       elif option2[count].isdigit() and count + 1 < len(option2) and option2[count + 1] == "(":
           option2.insert(count + 1, "*")
           count += 1  # 跳过插入的*，避免重复处理
       elif option2[count] == "^":
           option2[count] = "**"  # 直接替换，比pop+insert更高效
       elif option2[count].isalpha():
           print("Option not recognized, restarting")
           counting.Calculator.Advanced()
       count += 1
   

修复后效果 #

处理"3(5+3)"时，会正确插入*号变成"3*(5+3)"，eval()可计算出正确结果24。

内容的提问来源于stack exchange，提问作者Chief Guy